The Three Indistinguishable Dice Puzzle
Вставка
- Опубліковано 5 лют 2025
- Can you find a way to squeeze two dice out of three?
SOLUTION VIDEO: • SOLUTION: Three Indist...
Problem suggested to me by Lucas Garron. Thanks to extra work done by Josh Laison, Jan Van Lent and Tarim (which convinced me it would work as a puzzle).
CORRECTIONS:
Actually, more of a clarification. In the first bit I was using a remainder of 0 as 6, according to the convention that n = 0, mod n. But I forgot to mention it explicitly. Sorry! First noticed by DeadLink.
Music by Howard Carter
Design by Simon Wright
MATT PARKER: Stand-up Mathematician
Website: standupmaths.com/
Book: makeanddo4D.com/
Nerdy maths toys: mathsgear.co.uk/
"You can just use a marker"
Oh... Yeah, absolutely!
**hides saw**
Lol
Hammer
This made me laugh so bad
This made me laugh so bad
@@Bean-Time exactly my thought: we have perfectly good dice inside, why even bother with -the box- that transparent container? >:-]
The marker trick would potentially give you 4 dice in one.
"potentially"?
@@ottojagenstedt9740 Yes?
If you need one dice, ignore the three little ones. If you need 4, take them into consideration.
One wouldn't be fair.
@@567secret which one and why?
@@taududeblobber221 the outer one would be unbalanced i guess because it had the other dice in it
Simulate 1 die: Done.
Simulate 2 dice: Boring, let‘s skip the easy stuff and go one step harder:
Simulate 3 dice! And boy, do I have a clever solution for that!
Simulate 3 dice = Simulate 1 die 3 times
I was wrong. It can be done. This is an edit. Old stuff here...It can't even be done with 1 die...
If we can't distinguis the 3 dices, there are visibly 56 different rolls. out of these 56 different rolls, we must equally distribute the 1/6 probablity for each number from 1-6 for a fair dice. So unless I am making a mistake here. it is impossible to create a fair dice because 56/6 = 9.33 thus there isn't a way to equally divide the 1 through 6 over the 56 visibly unique rolls. Problem solved. So your original assumption was incorrect? Please let me know if I made a mistake. I made an excel if you would like to see it and I checked it twice.
@@aledirksen01 for any throw of the die, remove the middle value.
Consider the Combinations of the remaining two die - 6c2 = 21.
Count the parity of the middle value with the Combinations, you arrive at 42.Just divide your data into 6 equal portions of length 6.
For example :
1= { 1even1 ; 1odd1 ; 2even1 ; 2odd1 ; 2even2 ; 2odd2 ; 3even1}
In my notification, I can see a preview of a reply but I dont see it here. Can you reply again so I can see your entire reply from before? Like from 4 hours ago.
@@aledirksen01 Hi Alejandro. i think the clue is on your solution - 9.33. To the human observer, there are 56 apparent combinations, but there are still 3 ways to arrive at each of these, which yields an integer solution.
Three indistinguishable dice
Except the one who only has half the dots on the two side
How about we say "functionally identical"? , and yes you could argue one weighs less than the others, but we are modelling here, we can imagine they are the same.
@@RobRidleyLive It's a joke
3 indistinguishable dice, except for the one that's distinguishable
There are moves that the dice with the point missing cannot be identified. For exemple, when all three dice have 5 facing up, all 2 are facing down.
You can lift them up, and disambiguate which is the correct die, but moving the die will possibly tamper with the real value.
Take the differences between smallest and middle and middle and largest.
These are the relitive dist liklihoods
{(0, 0): 6, (0, 1): 30, (1, 1): 24, (0, 2): 24, (1, 2): 36, (2, 2): 12, (0, 3): 18, (1, 3): 24, (2, 3): 12, (0, 4): 12, (1, 4): 12, (0, 5): 6}
So here are 6 equally likely sets of differences.
(0,0)+(0,1),
(1,2),
(1,3)+(0,4),
(0,2)+(2,3),
(1,1)+(2,2),
(0,5)+(0,3)+(1,4)
This is a great start. Now we just need to assign them to the correct outcome for two dices. So "2" and "12" should both have an outcome of 6 (out of 216, i.e. 1/36) and therefore be matched with (0,0) and (0,5) respectively. "7", the most common should match with 36, ie (1,2). The problem is that there is only one 30 but we need it for both "6" and "8". Hmm... I have to think a bit more. But, as I said, great start!
@@snurreo For 2 dice, assign any a,a,b or a,b,b to a,b ie 1,3,3=1,1,3 => 1,3
All a,a,a goes to 1,1.
Then there are just 20 ordered pairs left. And there are 20 unordered triples.
Or write down the numbers on each dice and then use (mod 3 )+ 1 to generate which one to cross of
@@marekhlavackovi3677 Good idea, but it only works if you don’t sort them in any way when you write them down. Since you can‘t distinguish which die is which in the sequence, the order in which you write down the dice has to be an evenly distributed random.
@@christianosminroden7878 thats Why you write them down before calculating the mod Or even beater calculate the mod from another trow
Note to manufacturers: please make these multi-dice cubes with a different colour for each of the internal dice. Thanks for the great videos, Matt!
Solution: roll twice
Yeah, that'd do it.
Doesn't count. He specifically asked for a mapping of the three dice to the results from two dice
I doesn't solve the "maths problem" that he was getting at, but your solution is effective and thoughtful! The problem is he said no "out of the box thinking".
Hey I love your pfp! Idk if that’s you or not, but it’s an interesting pastel take on a style typically associated with darker colors! It’s awesome!
I'm not sure how this would solve the problem... if you simply divide the total result by six and round to the nearest whole number, as I assume you mean, you would not have a fair die. Look at the probability distribution for any number of dice greater than 1, and you'll see a bell curve; the more dice you add, the smaller the standard deviation.
Product design flaw: design small dice in different colors
Then you can roll 1, 2 or 3 dice at once with clear and easy to follow rules
Where's the puzzle tho? Taking into consideration the product is meant to be a puzzle
you deserve more likes
And dots on the outside for 4 of course
That's a fabulous solution.
@@BroArmyCommander The product was never meant to be a puzzle. It’s only used for one in this video. I also first thought he meant a physical puzzle
"Where as personally, i never say 'die'."
*slow clap*
youreallinsane Understood that reference
Sabahudin Kovacevic the Never Say Die Record label?
They were the last words of the Princess of Wales, who hated her name being shortened.
No, slower than that!
youreallinsane you obviously don't not since you said that u didn't but since you said "die" instead of di.
Looking at the comments, I feel like this video didn't make the problem clear enough because a lot of people are making the same mistake of taking the average. The point is you need to same *probability distribution*, not just the same range of values. When you roll one die, every number has an equal probability of happening. When you add two dice, that's not the case. There's only one way to get 2 (1 and 1), and only one way to get 12 (6 and 6), but many ways to get 6 (1 and 5, 2 and 4, 3 and 3). The numbers in the middle are more likely than the extreme ones. So anything that involves taking the average won't give you the right (uniform) probability distribution.
My guess at a solution for 2 dice from 3: Arrange the rolled values in size order. Sum them and mod 3. Ignore that die and use the others.
For example if you rolled (3, 5, 2), that's (2, 3, 5). The sum is 10. 10%3 is 1, so we ignore the middle die (we're counting from 0), and our result is (2, 5).
If all values of the sum mod 3 are equally likely, this should result in the right probability distribution. I'm not sure they are, but if all values of the sum mod 6 are equally likely, you'd expect that to be true of 3 as well, right?
Edit: Nope, looks like that doesn't work...
+Robert Miles Oh, I see you used the same exact approach I did. At first I convinced myself that this had to work because the sum mod 3 is choosing a die at random. But unfortunately it's biased.
+Robert Fisher Yeah, I had to write a program to check. Still not sure exactly why it doesn't work, but the numbers are way off
+Robert Miles To see why it didn't work I tried on paper to get a result of 11 from 3 dice:
111 - 1 way
112 - 3 ways (112,121,211)
113 - 3 ways
114 - 3 ways
115 - 3 ways
116 - 3 ways
You would need 11 to come out 6 ways for 6/216 = 1/36.
The last die (1-6) kind of "chooses" which die to exclude. But any rule will fail because 111 only has 1 way while the others have 3 ways.
+Robert Miles I wish this was the top comment because of how many people failed to understand the problem, which is a shame because this is such a magnificent mathematical question for laypeople to try.
If you do that, then 1 will be very difficult to get, but 4 will be very easy to get. The only way to get 1 is all of them being 1. To get 4, there are many possibilities. So the distribution is not the same of a single die.
There are three possible interpretations of 'simulating rolling two dice':
i) You need to know only the total of the hypothetical 2 dice.
ii) You need to know the numbers on the dice (for example 1-6 is distinct from 2-5)
iii) You need to distinguish the two hypothetical dice (for example 1-6 is distinct from 6-1).
So there are probably methods which are solutions for i but not ii or iii, and/or for i and ii but not iii.
well, a solution for iii would also be a solition for ii and a solution for ii would also be a solution for i, right? so just try to find the most general solution you can find i'd say
Using his crude example (4,1,1 -> 3) makes it seem like we are considering the solution for i) only. but it would be cool to see if we could even model 'iii' from it. (probably not though)
The third (and by extension all three) are definitely possible:
You have three possibilities, no repeats, a double, and a triple.
For no repeats, there are 6!/(3!*(6-3)!) combinations you can actually separate, which gives 6*5*4/6 = 20, and each of them occurs 3! (how you sort them) = 6 times.
For doubles, there are 6 (unrepeated digit) * 5 (repeated digit that's different to the first) = 30 different things you can separate, each of which occurs 3 times (position of the unrepeated digit)
And for triples, each occurs once and there are six of them.
For iii, you need 36 buckets to place those things in, which have to have the same size: Put the no-repeats into the first 20 buckets, the single-repeats into the next 15 (=30/2) buckets, and fill the last bucket with the triples. Now each bucket occurs six times. Alternatively, you can have the last two buckets with one single repeat group and three triples, but apart from reordering the buckets there are no other solutions.
As for a nice formula, haven't found one yet.
The way I understood it was that you only need to know the result, but that the frequency of any individual result should be the same as in a two die roll. The result 7 is much more likely than a result of 2, for instance.
You definitely don't need iii. To simulate something you are using different mechanisms to produce results as close to the real mechanisms result as possible.
Knowing whether it's 6-1 or 1-6 doesn't matter if your new mechanism outputs "7" as the result just as often as those two dice would.
"But I don't want a math solution!" - a person who has a marker
I've found a solution which doesn't use the big table you suggested, but uses a simple trick and a smaller table.
1. Add up all the numbers
2. Take that number mod 6.
3. There will be 1 or 2 numbers between 2 and 12 whose mod 6 is the same; one of those numbers is your result.
If mod 6 of your number equals 0 the result is either 6 or 12. If you rolled the numbers 1, 2 and 3 pick 12, else pick 6.
If mod 6 of your number equals 1 the result is 7.
If mod 6 of your number equals 2 the result is either 2 or 8. If you rolled the numbers 1, 3 and 4 pick 2, else pick 8.
If mod 6 of your number equals 3 the result is either 3 or 9. If you rolled three different numbers pick 9, else pick 3.
If mod 6 of your number equals 4 the result is either 4 or 10. If you rolled three different numbers pick 4, else pick 10.
If mod 6 of your number equals 6 the result is either 5 or 11. If 2 is the lowest number you rolled pick 11, else pick 5.
Some of these rules are quite arbitrary, but I checked all of them and they are the simplest rules, that will give you the correct probabilities.
"Whereas personally, I never say 'die'"
Bravo, I'm going to have to remember that!
Hi matt, thanks for ruining my afternoon ;-)))
Here's my solution.
I'll try to show an equiprobable mapping between the three dice outcome and numbers 0 to 35. Probably not the most elegant one, but it's not a lookup table and it works (unless I'm wrong, of course ;-)
By taking integer division and modulo it's then straightforward to map this number in [0, 35] onto the couple of dice values (a, b), with 1
When he started talking about lateral thinking I was definitely assuming he meant smashing the big cube, not writing on it lmao.
Yes. Yes indeed. Sawing the cube in half... I was really baffled by the mere idea of a non-destructive solution :D
Same
My solution is to ask SingingBanana to post a video on how to find the numbahs.
+Mario G. Cardiel Let me know how that turns out!
+standupmaths If you have 216 combinations isn't it impossible to get an even distribution for the numbers 2-12 (What you can roll on 2 dice) Since you have 11 numbers from 2-12 and 216/11 is not an integer? 216/12 would be an integer however that assumes you can roll 1 on 2 dice which you can't. Someone correct me if I am wrong please. Thanks.
+Justin Robinson Nope. Possible. Apart from the ugly table solution, I have another solution that basically...... Spoiler below......
....Spoiler....
Means you use the first method to remove either the largest, the smallest or the middle die, then sum up the remaing.
If 3D6 -> 1D6 = 1 or 2, per the first problem, then you remove the smallest die, and sum up remaining 2D6.
If 3D6 -> 1D6 = 3 or 4, then you remove the middle die, and sum up remaining 2D6.
If 3D6 -> 1D6 = 5 or 6, then you remove the largest die, and sum up remaining 2D6.
It is mathematically equal to rolling a 2D6, the same normal distribution :)
+Justin Robinson two dices is not an even distribution either, as there is less probability of getting a 2 or a 12 than any other number
+Justin Robinson Your mistake is that you assume we need to get an even distribution. Wn need to get a distribution like this: 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36. Since 216/36=6 it should be possible.
When you add two dice, you get a non-uniform distribution (which is great for RPG), but I was surprised to find that when you take the mod, the result is uniform. I did the spreadsheet. I was sure I had him.
+Christie Nel It's not too surprising. The sum of two dice is a triangular distribution, and doing the mod 6 will make you add two linear functions with opposing slope, which will give you a constant.
Alcesmire I'm glad someone has the explanation, though I don't understand which two linear functions you mean have opposing slope or how mod implies addition. Care to clarify?
Here is another solution that gives the 2D roll breakdown and not just the total.
For any triple: Result is a pair of 1's
If there is a double: Remove the duplicate to get result
If there is a 5: Remove the 5 to get result
If there is a 1: Result is a 5 and (TOTAL - 5)
Otherwise: Result is a pair of (15 - TOTAL)'s
If you need an ordered result, then anything resulting from removing the duplicate is low, high. Otherwise it is high, low.
+Fr Stephen I'm a little surprised this works, but congratulations
Verified: repl.it/CHvI/1. Amazing work. How did you find it?
+Fr Stephen I vote this to be the offically best answer. (having read countless)
+Thomas Kwa
Thanks. I started with a 6x6 grid of the possible results and just tried to use the simplest rules to produce each result 6 times (216/36).
The triples were always going to need their own rule, so I saved them as a wild card. I used the pairs rule to produce half the unpaired results. Removing 5s produced the remaining unpaired results that didn't have a 5 (I initially tried removing 6s, but I couldn't easily map the last bit).
To produce the last 11 results (pairs and [5,x]'s) I had to write them all out and use trial and error.
If you want to see how all 36 results were produced here is a spreadsheet.
drive.google.com/open?id=1evdkEN2ChUclRCxAX9W0CofvcJtbuAf6q56M_cMJ8KU
+George Lionon I second that motion! Excellent solution, FrStephen, bravo!
1. If all dice roll identical --> double-6, e.g. 333 --> 66 and 444 --> 66.
2. If there are exactly two distinct values, omit the double value, e.g. 242 --> 24 and 551 --> 15.
3. If there are three distinct values and one of them is a 6, omit the 6, e.g. 625 --> 25 and 362 --> 23.
4. If there are three distinct values and none of them is a 6, place the values on a "5-hour clock" 1 -> 2 -> 3 -> 4 -> 5 -> 1.
4a. If the values are all consecutive on the clock, the "last" one is double, e.g. 234 --> 44 and 521 --> 22.
4b. If the values are not all consecutive, turn the two that are consecutive into a 6, e.g. 124 --> 46 and 513 --> 36.
Tested and approved: (Code is written in D, can be compiled with gcc or just google for "Online D Editor" to run it. Should be fairly easy to translate into another language too)
import std.stdio;
public void main() {
int[5*5+6] result; // Array to count the occurences of 2 dice results
// It's almost treated like a 2 digit base 6 number, just abusing the
// fact that the digits are sorted to easily get rid of 6 integers that
// would always be 0 without doing all the calculations to make this
// array actually dense
// a, b and c are the ordered results of the 3 dices (a 15.
} else if (a == b || b == c) {
weight = 3;
if(a == b) {
d1 = b;
d2 = c;
} else {
d1 = a;
d2 = b;
}
// 3. If there are three distinct values and
// one of them is a 6, omit the 6,
// e.g. 625 --> 25 and 362 --> 23.
} else if (c == 6) {
d1 = a;
d2 = b;
// 4. If there are three distinct values and
// none of them is a 6, place the values on a
// "5-hour clock" 1 -> 2 -> 3 -> 4 -> 5 -> 1.
// 4a. If the values are all consecutive on the
// clock, the "last" one is double,
// e.g. 234 --> 44 and 521 --> 22.
// 4b. If the values are not all consecutive,
// turn the two that are consecutive into a 6,
// e.g. 124 --> 46 and 513 --> 36.
} else if (b == a+1) {
if(c == b+1) {
d1 = c;
d2 = c;
} else if (a == 1 && c == 5) {
d1 = b;
d2 = b;
}
else {
d1 = c;
d2 = 6;
}
} else if (c == b+1) {
if (a == 1 && c == 5) {
d1 = a;
d2 = a;
} else {
d1 = a;
d2 = 6;
}
} else {
d1 = b;
d2 = 6;
}
writefln("%d%d%d -> %d%d", a, b, c, d1, d2);
result[5*(d1-1) + d2-1] += weight;
}
}
}
writeln();
for (int d1=1; d1
@@bultvidxxxix9973 Wow, thanks BV! After I came up with this solution I saw the video from 4 years ago where Matt picked his #1 choice. My solution is similar in many ways, but slightly different. I really like the one he chose for #1 though.
I wonder if anyone has tried to find a nice way to go from 4 dice to 2 or 3. (The solution for 1 die works for any number of dice.)
So we need corresponding probabilities to match, and for the mapping to be well-defined (we can't have one roll fit two different cases) and cover all cases.
I had a hard time getting it to add up until I split (2) into two cases, 2i and 2ii.
1.
(two dice) Rolling two 6s: One in 36 chance.
(three dice #1) Rolling three of the same: Also one out of 36 (because for each value you might get on Roll 1, the chance is 1 in 36 that the rest match)
2i and 3.
(two dice) Rolling two distinct but no 6s: 5*4 out of 36.
(three dice #2i) Rolling double and a single, but no 6s: 5 ways to roll the double, followed by 4 ways to roll the single. But this assumes that Die 3 is the distinct die, so again times 3 ways total, out of 6*6*6. 5*4*3 out of 36*6, we need another 5*4*3 out of 36*6.
(three dice #3) A 6 and two other distinct numbers: 3 locations to roll the 6, and then 5*4 ways to roll the other two distinct. 3*4*5 out of 6*6*6, as required.
4a.
(two dice) Doubles, but not the number 6. There are 5 ways out of 36 to do this.
(three dice #4a) There are 5 clock triples, one for each starting number. For each clock triple, there are 3!=6 orders in which you could have rolled the sequence. 6 orders * 5 starting numbers out of 6*6*6, or 5 out of 36 as requested.
4b and 2ii.
(two dice) 6 and something else. 2 locations for the 6, and 5 possible numbers for the other die. 2*5=10 out of 36 outcomes.
(three dice #4b) There are 5 possible ways to roll the "something else" (such as 1), and then there is only one clock pair (in this case 3,4) that does not include or connect to it. There are 3!=6 orders in which you could roll these three numbers. 5*6 out of 6*6*6, or 5 out of 36.
(three dice #2ii) Rolling two 6s and something else, or 6 and two copies of something else: 5 ways to choose "something else", 3 places to place the single, and 2 ways to choose which of the two numbers is doubled. 5*3*2 out of 6*6*6, or another 5 out of 36.
Solution for your cube: Ignore the die where the 2 has only 1 dot.
+George Lionon 3 indistinguishable dice
My immediate thought was "take the top left (in regards to the roller), or exclude the top left"
but noooooooo, it has to be a purely "mathematical" solution.
you damn mathematicians, inventing problems just so you can feel good about solving them
I had the same idea...
The trouble with any sort of positional exclusion is that you can't always depend on there being a clear positional distinction. What if the three dice are in an "L" shape with two of them equally sharing the position of "top left"? Or what if the outer cube lands on a diagonal so it's hard to determine "top left"? The solution has to handle indistinguishable dice.
@@rosuav Here's a positional scheme that works: Each die can be uniquely described by the position of its center of gravity. Since we're working in 3d space this position can be represented as a 3-tuple, (x, y, z). Thus we have a set of three 3-tuples ((a, b, c), (d, e, f), (g, h, j)) and we just need to find an unambiguous sorting scheme.
Here's one: Sort by x, breaking ties with y, then breaking those ties with z. (Lexicographical order.) In other words, first sort by highest northern-ness (highest x). If there's a tie, break ties with western-ness (y). Finally, break those ties with altitude (z).
Now if there's a tie between x, y, AND z coordinates, then the positions are the same. Since we're dealing with dice this is not possible, as it's not possible for dice to somehow clip into each other and have the same center of gravity. Therefore one coordinate must be different and there must be a way to order them. So for any roll of dice there is a unambiguous way to order them.
Thus, to get the result of rolling n die, simply take the first n elements of the dice as sorted by position as described above.
(I'm pretty confident that you can prove what I said with set theory and maxima and minima and well-ordering, but I'm not well-versed enough to actually do that /shrug)
@@phlaxyr That implies that you can perfectly measure each one's position, regardless of its rotation, regardless of the position and rotation of the outer cube, etc, etc. It's VERY hard to guarantee this when you can't measure the CoG of each cube individually.
IOW you're dealing with the easy part (finding an unambiguous sorting scheme) while glossing over the hard part (finding the (x,y,z) of each cube).
@@rosuav Yeah I suppose in practice it might be hard to distinguish between a very slight positional difference and a tie, and that difference could change the ordering significantly. But theoretically though
Here are some rules that work and give the exact probability of a 2D6
- for any n:
(n,n,n) -> (1,1)
- for n>1:
(1,1,n) or (1,n,n) -> (n,n)
- for n>1, m>1 and mn:
(n,n,m) -> (1,n)
- for p
NEEERRRRDDD!
+lhyan 560728 This one is particularly great for programmers: it always
gives the dice in ascending order, uses 1 sort and up to 4 additional
comparisons, and the sum n+m+p can be calculated as part of
the sort.
+unekdoud That doesn't matter, you have to sort them anyway, because dice are indistinguishable and (122,212,221) all become (122,122,122).
+lhyan 560728 I tried to code it up, but didn't immediately get it to work. I got some 7's and 8's and the frequencies weren't right. It would be useful if you could rewrite it with if statements to make sure I interpreted the rules correctly.
+ahzobo If it helps, here is the order of conditionals which I have: let the numbers showing be a,b,c sorted in ascending order.
1. If a==c, return (1,1).
2. If a==b or b==c, check if a==1. If so, return (c,c). Otherwise, (1,b).
3. If a
Here's my solution to the second problem (only gives the result)
If you have a triple, that becomes a 2
If not, look at the sum of the dice:
4 and 10 map to 6
5 maps to 10
7 and 8 map to 7
9 maps to 5
11 and 17 map to 8
12 maps to 9
15 maps to 11
16 maps to 12
If the sum is 6, 13, or 14, see if it has a 2:
A 6 or a 13 that have a 2 map to 10, a 6 or 13 that doesn't have a 2 map to 4
A 14 that has a 2 maps to 11, a 14 that doesn't have a 2 maps to 3
The basic rule is pretty simple, but the distribution is kinda terrible
I think when he said a solution without a lookup table this is what he meant
My attempt at refining this:
Any triple is 12 (2 works mathematically, but I feel that triples should be 12)
If you have doubles, 1s or 5s make 6, 3s or 4s make 7 and 2s or 6s make 8
That leaves all dice different, in which case we look at the total:
7 is 7 and 10 is 10 (so 7 is either double 3, double 4 or total 7 with no double)
I was pretty happy up to this point, but now it gets very arbitrary:
6 is 2
8 is 3
9 is 4
11 or 14 are 5
12 or 15 are 9
13 is 11
Hopefully the probabilities work out, but I rushed it a little so wouldn't be surprised if I messed up.
@@annoyingfoothold they work out, now we just need to figure out some mnemonic for all those pairings... :-s
Those are not outcomes of two dice. You need to distinguish 4,3 from 5,2
"I never say die" Epic.
Making a giant list just to roll two dice? That looks like a classic Parker Square to me.
Any 4 dimensional beings watching this are going to be super confused as to why you can't just remove the extra cubes
This just blew my mind
No, they're amazed that we can roll it without the extra cubes falling out.
Andrew Turner Technically not since we arent rolling it in the forth dimension.
Bl4ck St0rm There are three dimensions, not a first, second, and third dimension. Or can you point at them in order? You can define any direction you want as the first, second or third dimension. The same would be true if there were four dimensions. You could point in any direction and define that as the fourth dimension, but that’s just made up.
Part of the problem here is that when you roll two dice some results are more likely than others, and in a lot of games doubles matter.
0:47 Someone should die for writing that pun. Ba-dum-dum-ching.
No dice.
+standupmaths Darn. I guess I'll just have to roll with it.
+doggonemess maybe you both can add up your problems and come up with a solution. ;)
+doggonemess Agreed
That pun is pun-y. Why would you dice-ide to make this? I'm dice-appointed.
the manufacturer should make one of the dice a different colour. while maths is the solution to this set, engineering is the solution to all future ones.
+kyphilburg This would solve both solutions in a single, simple way. Bravo.
Although this would fix the problem for later sets, it wouldn’t be as elegant as having the larger cube be a die in and of itself- 1 die is the exterior or sum(interior)%6, 2 dice is (sum(interior)%6)+ exterior, 3 is the interior, and 4 is the interior + the exterior
If you only colour one how do you simulate two dice?
@@thatchapthere the two that aren't coloured. So 1 blue and 2 white dice would give you 1 (blue only), 2 (white only), and then both blue and white for 3
@@davidcogdill300 that is elegant indeed
You weren't able to find a solution? That's such a Parker Square move...
+Suave Atore what does this mean?!?!?
+Bruno “Richard” Bronosky check out numberphile latest vid
+Suave Atore got it!
The comments trying to legitimately help solve this riddle while I’m over here reading the title at “The Three Indigestible Dice Problem” thinking to myself, “aren’t all dice indigestible?”
I don't think I caught the "Never Say Die" pun the first time I saw this when it came out
I've spent the my day making a table and calculating all probabilities before continuing the video.
And then with no calculation you just absurdly said:
mod(6)
Which happens to be 100% with all results having the exact same probability
It makes sense. On two dice, if one die is any arbitrary result, then the other die has an equal probability of being any of the 6 results, if you mod six. Adding a third die is just a proof by induction.
@@DracoSuave this is so straightforward it actually made me angry. Thanks!
@@DracoSuave I was just scanning the comments for a proof, thanks :)
A roll of 10 occurs 27 times out of 216. We want a partition of 36, so we need nine more. 5 and 16 each occur 6 of the 216, 4 and 17 occur 3 times and 3 and 18 occur once. A roll of 11 is 27 as well.
From 3 to 18 the frequency in 216 is: 1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1.
This is really old, but I am pretty sure he means (mod 6) + 1
I think what he means is "mod 6, but if you get 0 call it 6" - but I believe your method would also produce the desired 'fair' result.
I admit I haven’t looked closely at it, but this solution bothers me because it’s impossible to get 1 or 2 on the “low end”, which would seem to skew the results toward 3, 4, 5, or 6. Why am I wrong?
Some results are more likely than others because there are more possible results which add to the number. This happens in such a way that the mod 6 sorts the results to evenly sized buckets with 36 results going to each. For example, 1 is the the result (using mod 6 + 1) of the 10 ways to roll 6 on three dice, the 25 ways to roll 12 and the 1 way to roll 18. 2 is the 15 ways to roll 7 and the 21 ways to roll 13. And so on.
@@MrZcar350 Thank you that makes perfect sense :)
Mod 6 works because:
The first die has results evenly distributed between the 6 buckets.
Each following die merely shifts the results by 1-6 buckets - meaning the results stay evenly distributed between the 6 buckets.
Just always take the one with the broken 2...
Ok but if it rolls with the 5 up, where is your God then?
@@АндрейСтанжицкий I'm atheist. The outer cube is see through so just hold it up and look at the 2
@@sandraaiden8587 legit. The God reference is just a figure of speech tho
Okay, this is a little technical, but bear with me
**ahem**
Roll Twice.
+Deven McKee Thank you. but he said no technical answers
+Deven McKee good one.
LMAO, well played
+Shehab Ellithy
Roll once, (sumOfDice * 2) %6 ?
Actually, I'm curious about this first answer, because if he somehow rolls 18 with the 3 dice, the modulo answer will be 0.
+Erick Bluesun For one die roll result It should have been: (sumOfDice % 6) + 1. This will give you number between 1 and 6, as modulo will result in numbers 0 through 5.
Here's a simplified rendition of the compressed look-up table I posted some time ago for converting an un-ordered roll of three dice to a number representing a roll of two dice (assuming 6 sides each, numbered 1 through 6)...
2 iff {1,1,1}|{2,2,2}|{3,3,3}|{4,4,4}|{5,5,5}|{6,6,6}
3 iff {1,1,2}|{1,1,3}|{1,1,4}|{1,1,5}
4 iff {1,1,6}|{1,2,2}|{1,3,3}|{1,4,4}|{1,5,5}|{1,6,6}
5 iff {2,2,3}|{2,2,4}|{2,2,5}|{2,2,6}|{2,3,3}|{2,4,4}|{2,5,5}|{2,6,6}
6 iff {3,3,4}|{3,3,5}|{3,3,6}|{3,4,4}|{3,5,5}|{3,6,6}|{4,4,5}|{4,4,6}|{4,5,5}|{4,6,6}
7 iff {5,5,6}|{5,6,6}|{1,2,3}|{1,2,4}|{1,2,5}|{1,2,6}|{1,3,4}
8 iff {1,3,5}|{1,3,6}|{1,4,5}|{1,4,6}|{1,5,6}
9 iff {2,3,4}|{2,3,5}|{2,3,6}|{2,4,5}
10 iff {2,4,6}|{2,5,6}|{3,4,5}
11 iff {3,4,6}|{3,5,6}
12 iff {4,5,6}
In this table, the order of the rolled numbers is given as lowest to highest but represents all arrangements of those three numbers. I really don't think a simple mathematical formula solution is possible, because there is no reliable way to distinguish between the same numbers rolled in different orders.
why not just split them into specific results?
1-1 iff {x,x,x}
1-2 iff {1,1,2}|{1,1,3}
1-3 iff {1,1,4}|{1,1,5}
1-4 iff {1,1,6}|{1,2,2}
1-5 iff {1,2,3}
1-6 iff {1,2,4}
2-1 iff {1,2,5}
etc
Ask six mathematicians what they think you got, pick the number that three of them are nearest
I feel like I'm missing something, but I don't think his solution for the first part gives the the same probability for each value. As it is, a 1 and 2 values (each) only represent 2 of 16 potential values from the 3 dice rolls (which is a range of 3-18) , where the other numbers each have a 3/16 chance. Looking at it from another angle, the only way to get the value 1 is 7 mod 6 and 13 mod 6, but to get say 3, you can have 3 mod 6, 9 mod 6 and 15 mod 6.
+Khasar You can't reason in terms of 3/16 or anything/16, because the 16 combinations of dices don't have the same probability of occurring. For example, a sum of 18 can only occur once, when all the dices come out as 6. Same for the sum of 3. But a sum of 12 can come out in a multitude of instances: 156 (also 165, 516, 651, etc), 246 (and friends), 336 (and friends), etc. You need to look at all the 216 combinations.
Calin Jebelean Ah, Gotcha.
+Khasar Last time I checked the comments of this video, many people had posted things like Excel documents used to write out all 216 arrangements and count the frequency of each roll. Somehow, by adding the frequencies of the same post-modular terms, they all equate to 36/216. For example, a roll of 3, 9, and 15 have a probability of 1/216, 25/216, and 10/216 respectively, while a 7 and 13 have 15/216 and 21/216.
More interesting, though, is that I recall someone saying that for any number of dice this method equally distributes probability. I just checked for myself and it appears that Matt's method does work for any numer of dice as well as any number of sides.
+DrunkMasochist Well, there's a physical way to think about it too; when you're just summing the dice together, the identities of the individual dice don't matter (i.e. you could use differently colored dice just as well).
So what if instead of rolling the original cube, you took one die out, then rolled the cube (with two dice left) at the same time as your third? How does that compare to rolling the cube as is (with three dice) and one extra die of your own?
+Khasar - I wasn't sure if the probably of the 1-die solution was correct so I wrote a loop to count all the combos and each number occurs 36 times, so mod 6 is fair. Now, when rolling 2d6 the probably of 7 is much greater than 2 or 12, so does the puzzle want us to maintain the same probabilities?
I've noticed that, if this object was marked on the outside as a dice, and one of the dices was a different color, you could use this one dice to simulate a throw of anywhere from 1 to 4 dices:
1 dice uses the markers
2 dices uses the same color smaller dices
3 dices uses all the small dices
4 uses the marking plus the smaller dices
5 dices 4 + mod trick
the mod dice would not be independent of the other dice so the chances of different combinations would be different from 5 independent dice (equal chance of any combination).
@@artembaguinski9946 Do the mod trick with the 2 same colored small dices. I think it should work out.
@@busteraycan no because the virtual dice won't be independent from the two dices it is based on.
"Whereas personally, I never say die"
Goonie status confirmed.
I’m quite late, but 216/3 is 72 different combination ignoring the order you read numbers in, then you could build a tree, an easy structure to look up, for the different cases, start by looking if there is a 1, if true look for the branch that admins a 1, if true, then oggi for the third one and you have the solution! If there is no 1, go for the 2, there is no 2? Go for the 3 branch and so on!
Solution:
Step 1: Role cube
Step 2: Count total number on dice and divide by 10
Step 3: place cube on hard surface
Step 4: whack cube with hammer
Step 5: role one of three smaller dice
Step 6: use number rolled as final number
"Personally, I never say die" ... that smirk... :P
Nevah shay Nevah Mish Monneh-Penneh
"...personally, I never say 'die'..."
But you just did. Twice.
+MarioFanaticXV ; Pay attention to the adverbs, he said he would "personally" never use the word 'die', while this video would qualify for "professionally" using the word 'die' which is still permissible.
I decided to take a swing at this without looking at anyone's answers first. What follows is a working solution but not as simple as it could be.
I started by recognizing that there are 6 times as many combinations possible rolling 3 six sided dice as there are rolling 2, of them, so one solution would be to make a group of six results with three dice map to each result with two dice.
Counting the orders that a particular set of resulting numbers can be arranged in, there are one each of results where all three numbers match, three each of results where two numbers match, and 6 each of results where none of the numbers match. So I grouped the ones with all three numbers matching into a single group of six results and arbitrarily mapped that to rolling a pair of ones on two dice.
I then went on to arbitrarily match up the remaining groups with other numerical results, such that the proportions match the target odds.
1 of {1,1,1} & 1 of {2,2,2} & 1 of {3,3,3} & 1 of {4,4,4} & 1 of {5,5,5} & 1 of {6,6,6} = 6 of {2}
3 of {1,1,2} & 3 of {1,1,3} & 3 of {1,1,4} & 3 of {1,1,5} = 12 of {3}
3 of {1,1,6} & 3 of {1,2,2} & 3 of {1,3,3} & 3 of {1,4,4} & 3 of {1,5,5} & 3 of {1,6,6} = 18 of {4}
3 of {2,2,3} & 3 of {2,2,4} & 3 of {2,2,5} & 3 of {2,2,6} & 3 of {2,3,3} & 3 of {2,4,4} & 3 of {2,5,5} & 3 of {2,6,6} = 24 of {5}
3 of {3,3,4} & 3 of {3,3,5} & 3 of {3,3,6} & 3 of {3,4,4} & 3 of {3,5,5} * 3 of {3,6,6} & 3 of {4,4,5} & 3 of {4,4,6} & 3 of {4,5,5} & 3 of {4,6,6} = 30 of {6}
3 of {5,5,6} & 3 of {5,6,6} & 6 of {1,2,3} & 6 of {1,2,4} & 6 of {1,2,5} & 6 of {1,2,6} & 6 of {1,3,4} = 36 of {7}
6 of {1,3,5} & 6 of {1,3,6} & 6 of {1,4,5} & 6 of {1,4,6} & 6 of {1,5,6} = 30 of {8}
6 of {2,3,4} & 6 of {2,3,5} & 6 of {2,3,6} & 6 of {2,4,5} = 24 of {9}
6 of {2,4,6} & 6 of {2,5,6} & 6 of {3,4,5} = 18 of {10}
6 of {3,4,6} & 6 of {3,5,6} = 12 of {11}
6 of {4,5,6} = 6 of {12}
It's something to start with if anyone wants to make use of the general idea. Also, perhaps an analysis of this work will inspire some idea of something else to try out.
What?! You haven't solved it yourself? Now that makes it a bit of a Parker Square.
rekt
+proefslak :: He's become the mascot for giving things a go. That's something, right?
Eric Fontenot
Yes, that's great!
So compared to that achievement, this whole video is a classic Parker Square...
More like Parker cube.
Here's my solution:
If all 3 dice are the same, then the result is [1, 1]
If 2 dice are equal, let x be the repeated number and y be the other number. First, if y < x, add 1 to y (as a confidence boost, since it's alone and small). Then, if x ≤ 2, the result is [y, y], and otherwise (x ≥ 3), the result is [1,y]
Finally, if all 3 dice are different, let x be the smallest, y be the middle, and z the largest. If x ≤ 2, the result is [y,z]. Otherwise (x ≥ 3), the result is [2, x+y+z-9].
+MrBoxinaboxinabox This works perfectly. Still somewhat like a look-up table, but can be memorized.
+MrBoxinaboxinabox Hmmm... With your method I get this:
2 / 66 times (30.56%)
3 / 4 times (1.85%)
4 / 6 times (2.78%)
5 / 16 times (7.41%)
6 / 18 times (8.33%)
7 / 28 times (12.96%)
8 / 26 times (12.04%)
9 / 24 times (11.11%)
10 / 14 times (6.48%)
11 / 12 times (5.56%)
12 / 2 times (0.93%)
+MrBoxinaboxinabox
This is the only correct answer I have seen in the comments so far. I will post my own solution underneath (although it is a bit longer)
rules in order, if a rule doesn't apply, try the next:
sort dice
1. a (2,3,4) results in a (6,6) [there was an ambiguity with rule 5]
2. if all three dice are identical (x,x,x), it results in a (3,3)
3. if two dice are identical (x,x,y), the result is (x,y)
4. if one of the dice shows a 6, the other two dice show the result: (x,y,6) -> (x,y)
5. if the sum of two dice is 6, then the result is the third dice and 6: (x,y,z), where y+z=6 -> (x,6)
6. if none of the above applies, the result is two equal dice with the value of the sum of the largest and smallest of the three dice modulo 6. So if the throw results in (x,y,z), where x is the smallest and z the largest, then the result is ((x+z)mod6,(x+z)mod6)
The one die solution would be (sum mod 6) +1 otherwise your single die values would be 0 to 5 instead of 1 to 6.
X mod Y = Y is the same as X mod Y = 0
Oh. So you're saying that in this case 0=6.
I guess.. sorta. But you have to define it that way and that won't be true if you do any other math with the result. For the sake of the game, yeah, that's the easier way to have a result.
+ThisIsMeAndThisIsWhatILookLike No, X mod 6 only gives the value 0 thru 5. If you just take sum mod 6 then if you roll a 12 then that's rolling a zero? Zero isn't a valid 1d6 value. It's also impossible to roll a 6. 17 mod 6=5 and 18 mod 6 = 0. Adding 1 to whatever the result of the mod is solves this problem.
He did not say it, but it is obvious that 0=6 in this case..
+jgallantyt yes.
My dad used to have one of those. Albeit his had a white one a red one and a black one. The constant rattling brings back all sorts of memories.
I think there might be a problem with the ( X mod 6)+1 solution, as I'm not sure that would yield an equal chance for each outcome. For one thing, the range of input values are 3 to 18.
I thought the same
X Mod(6)+1 does work, it's fair. Each of the 6 values appears 36 times out of 216.
Hmmm ... it does work. Looking at all the 216 combinations, the number of combinations for each of the 6 states of X Mod(6) is 36. So each one has exactly 1/6 chance of occurring.
this is very late but I tested it myself and it does work, these are the probabilities of each number occurring:
3: 1/216
4: 3/216
5: 6/216
6: 10/216
7 (1): 15/216
8 (2): 21/216
9 (3): 25/216
10 (4): 27/216
11 (5): 27/216
12 (6): 25/216
13 (1): 21/216
14 (2): 15/216
15 (3): 10/216
16 (4): 6/216
17 (5): 3/216
18 (6): 1/216
your probability of throwing each number with the mod 6 rule are:
1: 15+21=36/216
2: 21+15=36/216
3: 1+25+10=36/216
4: 3+27+6=36/216
5: 6+27+3=36/216
6 (technically 0 but whatever): 10+25+1=36/216
Here is a very easy to do solution that works perfectly!
Triples -> Pair of Ones
2-3-4 -> Pair of Twos (Not a pretty exception, but worth it)
REMOVE
Duplicate if there is a pair (lowest die goes first)
Six, if there is a six (highest die goes first)
MAKE
A six from two dice (six goes first) (eg. 145 -> 64)
A pair from three dice (eg. 345 -> 66)
This is very easy to remember: Remove Pair then Six, Make Six then Pair.
The extra benefit is that if you only want the two dice total, you don't have to bother with the MAKE steps. You can just total the three dice!
I have posted this solution before, but it was in response to a previous comment of mine and I think got a bit lost, plus I think this is a better way of presenting it.
I've also made a video response (more of a slideshow) with this solution. Probably doesn't really need visual explanation, but there is a table showing all the mapping at the end. v=cV4aPgMkW7o
+FrStephen I definitely missed this the first time, so it's great that you posted it in its own comment with a simple description. What I like about it is its intuitiveness, even in the choices of which die goes first, and it's simple enough to learn the combinations that produce a double roll, for example.
0:46 Whereas, personally, I never say, "die".
Took some work to get to it, but he got to it!
Man, two dice roll from 3 dice? Easy.
1. Roll the dice.
2. Add all the numbers.
3. Divide by three.
4. Multiply by two.
Thanks for coming to my TEDtalk.
As for one dice solution, why modulo 6? Not everyone knows even what a modulo is. Just add all three numbers and divide them by 3.
all dice are the same -> 2
all dice are different:
6 or 7 -> 3
8 or 9 -> 6
10 or 11 -> 7
12 or 13 -> 8
14 or 15 -> 11
otherwise:
5, 6 or 7 -> 4
8, 9 or 10 -> 5
11, 12 or 13 -> 9
14, 15 or 16 -> 10
4 or 17 -> 12
Still a lookup table, but that's all I've got so far.
One other try:
With the 3 dice you apply the MOD rule and get a number between 1 and 3, that is equally distributed.
The 3 dices are then sorted from lowest to highest, numbered them 1 to 3. The dice that has been indicated by the previous number is removed.
For example:
- draw 6 3 5.
- mod rule gives 2.
- sort dice: 3 5 6.
- 5 is removed, as dice number 2, the result is 3 6
This should be correct because:
As soon as mod 6 of 3 dices is fair, also mod 3 is because it is just like mod 6, twice faster.
Being that number fair, or equally distributed, the removed dice won't be pending towards higher or lower numbers.
The resulting two dice will be fair because we didn't sum or do any other trick, but just found a way to choose which 2 dice make valid and which remove.
Obviously the removed dice is always certain, and everyone will agree on which to remove, unless they have the same number (and that's not a big problem). At the end, everyone will be happy and laughing with incredible joy.
I tried and the distribution seems to end correct. Hopefully it's 100% fair.
+Beppi Menozzi I found the same solution. I do not have an elegant proof yet, only a computer counting proof.
I just made an excel and it looks fair... after all it was just a matter of forgetting sums and work just to identify one dice to remove :)
Apparently the mod 3 variable is sufficiently independent from the eliminated dice.
+Beppi Menozzi
I had the same idea, but if you do an exhaustive search, it definitely isn't equally distributed. The problem is that the process of choosing which die to ignore is not independent of the chosen dice. For example, getting a (1,2) is 5 times as likely as getting a (5,6).
roll the dice twice
1) Pick a point on your table, roll the dice, take the closest dice to the selected point and take his value
2) Pick a point on your table, roll the dice, take the 2 farthest dices to the selected point and take their value
The Dice: *line up perpendicular to the line that passes through the chosen point and the center die*
@@SJNaka101 then, pick three points in the shape of a triangle, if the dies are perpendicular to the first two, they cannot be perpendicular to the third
I paused the video when you mentioned modulo and was about to tear you out for not considering the bell curve that's a result of adding the values of multiple dice, but I sat and worked it out and the chances of each value using the method described does actually add up to 36/216, 1/6 chance. That was annoying :P
I sat and worked it out, with the aid of netbeans, and this seems to work no matter how many dice you use. Why is this? There doesn't seem to be any obvious reason as to why it should be this way.
Anyway, back to the video.
+HaniiPuppy choose n-1 dice out of the n dice, now imagine if the nth was rolled after all the others, no matter what the values of the first n-1 are, once they are determined, each possible value of the nth will map to a
different final outcome. Therefore the probabilities must be equal. The fact that the dice are rolled simultaneously and are indistinguishable does not change the probabilities. Every die can independently cycle the outcome through all possibilities with equal probability.
+HaniiPuppy I tried the second problem by using Poker/Yahtzee hands, but couldn't find a neat solution.
i.imgur.com/Ef9NYOm.png
Probabilities and hands if anyone else wants to have a crack at making something neater with the idea:
pastebin.com/sgmRufjF
+HaniiPuppy I agree this is strange. August Hale your explanation is very helpful, but this should be explained in the video. Why should 1 dice be the same as sum(n dice) mod(dice count) + 1
+HaniiPuppy the idea I thought up was like this:
do a mod 2 on each die, and xor that (giving you a coin flip)
do a mod 3 on each die, and add that up (mod 3) that would also give you a uniform 3 result (independent of the coin flip).
put together, you get 6 options.
now what I did there, is actually, the same as just mod 6 on the sum (but I agree, when I first heard the mod 6, I too thought it was ignoring the bell, but after thinking more about my solution, I realized they are the same).
1. Do whatever you did for 1 dice
2. Roll it twice
I'm not a big fan of finding complicated solutions to a problem that can be simply resolved.
I would ignore the dice furthest to your right and if it's a tie I'd then ignore the one closest to me.
Simple solution, but technically physical.
Simplest I could find, for simulating 2 dice where the probabilities of the unordered pairs match the actual probabilities of 2 dice:
A,B,B+1 -> A,B where A min(1,max(5,A+B+C-7)),6 where A A,A if A 6,6
I think this could actually be used if you had no other option to simulate 2 dice with 3 identical dice. You can use this also to simulate 2 ordered dice by dividing up the cases for example 1,2,3->1,2 while 1,2,2 and 2,1,1 -> 2,1
If you want to check the probabilities
P(AA) should be 1/36
P(AB) should be 2/36
P(ABC) is 1/36
P(ABB) is 1/72
P(AAA) is 1/216
Remove the middle dice number (whichever dice isn't the lowest or the highest)
This ensures there's no bias for higher or lower numbers (this bias would exist if we remove the highest or lowest)
hey, this is actually a really interesting idea, but i don't think it actually works :(
you'd expect 2 to come up exactly 6 times, given that you have 216 results with three dice, 36 results with two dice, and rolling 2 in two dice is a 1/36 chance. but using your method the only way we can get a 2 is by rolling 1, 1, 1. whenever we roll 1, 1, x, with x greater than 1, the outcome will be greater than 2. for example! i ran a python script with your idea and this is how the distribution would actually turned out:
2: 1 (should be 6)
3: 6 (should be 12)
4: 13 (should be 18)
5: 24 (correct)
6: 37 (should be 30)
7: 54 (should be 36)
8: 37 (should be 30)
9: 24 (correct)
10: 13 (should be 18)
11: 6 (should be 12)
12: 1 (should be 6)
thanks for sharing your idea though! it was fun testing it out ^^ hope you don't mind my response, it wasn't meant as a critique or anything but just wanted to kindly point it out!
But then you’re biased against middle numbers
That radio show sounds like the kind of thing I'd love to listen to! Can you please make it available for people outside the UK like me?
No, UK is wuv, UK is Lyfe
bbc radio shows are available outside uk. just go to the bbc website. you can only listen live or stream for a few weeks after it’s aired however.
@@paintbokx dang, I didn't know that. Oh well.
I cracked this a while ago after the video came out. My solution was to use a symmetric 3-dimensional lookup table.
I came back to this video to comment on something I thought about a while after watching it, but this has nothing to do with the question that was asked. I was thinking about the mention of writing a number on each side of the outer cude, and it occurred to me that a much more useful device could be made as a slight modification of the one in this video. Imagine a hollow transparent dice with markings for 1 through 6 on it, and with three dice inside of it each in a different color, such as a red one, a green one, and a blue one. Rolling that would be rolling 4 dice at once. Pick one to ignore and you've got 3 at once. Pick 2 to ignore and you have 2 at once. Pick three to ignore and it's like rolling just one. You could also agree to apply a bit of specific math to each roll such as subtracting one from each dice and multiplying the result of that subtraction by a unique power of six. So this got me wondering... how hard would it be to make a mechanical device that could be rolled like a dice to internally produce a long binary number, perhaps strung together from a set of 8 sided dice, or even to produce a random base ten number of a given length from which a given number of digits could be agreed upon in advance to be ignored? Something to think about. :)
Subscribed for "Never say die"
I LOVE combinatorics. I feel like I need to write a Python script for this and puzzle it out now...
First half, hmmm, okay, now looking into it, modulo 6 does yield a uniform distribution.
Probability of a 1: 0/216 + 15/216 + 21/216 = 1/6
Probability of a 2: 0/216 + 21/216 + 15/216 = 1/6
Probability of a 3: 1/216 + 25/216 + 10/216 = 1/6
Probability of a 4: 3/216 + 27/216 + 6/216 = 1/6
Probability of a 5: 6/216 + 27/216 + 3/216 = 1/6
Probability of a 6: 10/216 + 25/216 + 1/216 = 1/6
So okay, your first approach of doing modulo 6 is accepted.
Now in the other case it is a normal distribution that should be mapped onto another normal distribution. Hmmm. Give me a few minutes :)
Edit: I was completely wrong, made a slight error in my proof :D
+Alexander Nielsen It's actually easier to see that going mod 6 gives a uniform distribution if you just condition on the values of two of the dice. This also implies that the number of dice doesn't matter. If you throw n dice and condition on the sum of the first n-1 dice, the sum of all n dice mod 6 will always be uniform 1-6.
+Alexander Nielsen You don't even need to calculate the probabilities. The distribution of the first die is uniform, and the other two dice can be considered to "shift" the result by that amount. The second die is uniform, and so is the third. QED.
+edderiofer True :)
+Brian Moehring ah that's more elegant than checking one by one, as I did. Facepalm @ me.
+Brian Moehring ah that's more elegant than checking one by one, as I did. Facepalm @ me.
Taking the modulus already gives us a way to simulate one die. We just need a way to independently simulate the other. Here's my idea:
Half of the numbers on a D6 with more than three dots: 4, 5, 6.
Half of the numbers on a D6 have dots in the middle: 1, 3, 5 (the odds).
Half of the numbers on a D6 are radially symmetric: 1, 4, 5.
Thus, for each roll of a single D6, the properties of (a) having more than three dots, (b) having a dot in the middle, and (c) having radial symmetry are equivalent to a coin flip. And obviously if you flip three coins, half the time at least two of them will be heads, and half the time there will be at least two tails.
So then, just take the property that holds of at least two of the dice. Then your simulated result is the number that has all three of those properties.
Let's say you roll a 2, 3, and 6.
2 and 3 are less than 6, so the "less than 6" property holds for the majority.
2 and 6 have a dot-free center, so the "even" property holds for the majority.
2, 3, and 6 are all radially asymmetrical, so the "radial asymmetry" property holds for the majority.
So then, the number with the properties of being less than 6, even, and radially asymmetrical is ... wait ...
All right, so I guess if there is no number, we'll just take the minority properties. So greater than 3, odd, and radially symmetric. So we get 5.
I'm no longer quite so confident in my answer, but I think that still gives each number a 1 in 6 chance of being generated by this procedure. Maybe.
The green table at the bottom makes it look like my phone is constantly losing signal & getting back online
Yeah that’s what I thought the whole time
What sort of weird phone do you have lol
@@timotejbernat462 it's a notification in the youtube app that shows a green bar when it reconnects
If the number is odd, remove the highest die from the group and then proceed as with 2 die.
If the number is even, remove the lowest die from the group and then proceed as with 2 die.
If if double or triple numbers are rolled (e.g. 3,3,2 or 5,5,5) remove any one of the highest/lowest die.
There are 8 possible even values and 8 possible odd values.
3,5,7,9,11,13,15,17
4,6,8,10,12,14,16,18
Outcomes:
Odd rolls will tend to have lower values
Even rolls will tend to have higher values
Distribution of odd/even rolls should be random
+Dave Thompson Does not work. P(1,1) needs to be 1/36=6/216.
Triples have probability 1/216, doubles 3/216 and triples 6/216.
If you use your method, you'll get P(1,1)>=P(1,1,1)+P(1,1,3)+P(1,1,5)=1/216+3/216+3/216=7/216.
+Dave Thompson This is really simple and clever, but the probability doesn't seem to work out. I wrote a little script to test the first part of the video (add the three, mod 6), which worked, but modifying it to do the even/odd check gives results that don't match the expected two-dice probabilities:
Even = 50.27%
2 = 2.92%
3 = 8.58%
4 = 7.4%
5 = 13.46%
6 = 8.94%
7 = 17.25%
8 = 8.77%
9 = 13.66%
10 = 7.03%
11 = 8.93%
12 = 3.06%
It's possible I screwed up the script, I guess? I wonder if this has to do with the highest die "floating" to the top?
+Rabid Llama I think you're absolutely right... my solution is not a solution. Giving this some thought, I think it is because my idea dis-proportionally eliminates high and low values, while preserving moderate values (e.g. 1, 2, 5, and 6 will be eliminated more frequently than 3 and 4 will be eliminated). This is because I lack a 3rd variable (Odd, Even, ...?) that could select for "most middle value".
This seems borne out by your data with 5 (2+3), 7 (3+4), and 9 (4+5) having the highest rate of occurrence.
I got to say that the giant lookup table was my primary idea to solve all problems here.
Solution: Roll twice, following the solution to the "reduce to one die" version.
Not like he said physically solutions didn’t count or anything like that
First problem: simulate 1d6. Easy.
Second: simulate 2d6 complicated.
Extrapolating the trend, simulating 3d6 will be nearly impossible.
Um-- wait a not --
divide 3 multiply 2, which could bring a decimal amount, then round 1-4 down and 5-9 up. you'd then have a 2 die amount quickly and easy to calculate.
+shelly alison Unfortunately that's not weighted the same way as rolling two actual dice. Here are the odds of each outcome using each method.
When calculating the total based on three dice
2->1/56
3->3/56
4->3/56
5->9/56
6->6/56
7->12/56
8->6/56
9->9/56
10->3/56
11->3/56
12->1/56
When using two actual dice
2->1/21
3->1/21
4->2/21
5->2/21
6->3/21
7->3/21
8->3/21
9->2/21
10->2/21
11->1/21
12->1/21
As a result, your method will give worse odds of rolling a 2, better odds of rolling a 3, worse odds of rolling a 4, better odds of rolling a 5, worse odds of rolling a 6, better odds of rolling a 7, worse odds of rolling an 8, better odds of rolling a 9, worse odds of rolling a 10, better odds of rolling an 11, and worse odds of rolling a 12.
+shelly alison Won't be fair because when you sum the dice, 3 and 18 values will be 1 time in 216 but 10 and 11 will be 27 times in 216. You become mathematically unfair when you sum the values.
Reason: only 1-1-1 equals to 3 but 1-2-6 1-3-5 1-4-4 2-2-5 etc. all are equal to 9 which is unfair.
I wasn't proposing a specific calculation, just a fast simple solution. It's how I would do it playing with my son.
+shelly alison Well then, I was thinking summing the values and taking mod 11 and then adding 2 to the value. It's still unfair but It's more fair generally.
It may not be 100% fair or accurate, it was only a quick suggestion, which would give a 2 die amount. That's all I was saying. I wasn't trying to pick apart anyone's math, just quick and simple enough for a child to figure out on their own. Sorry. I suppose my answer should have been to smash open the plastic case, take 2 dice, and throw the other away. Guaranteed fair now.
3d6 Keep rolling till you get double or triple, each number has a equal chance of happening, but you might have to roll a couple times.
That's certainly not an ideal solution
***** not ideal, but it is fair. And even idiots can use/see that method.
SirJMO Credit where it's due: you did successfully maintain the probability distribution (which a lot of people seemed to struggle with). But your solution does kind of violate the notion Matt brought up where he wants an actual mathematical solution. In reality, rolling them over and over until two of them are the same is basically just the same as removing one of the dice and only rolling two dice, which is a non-answer in my opinion.
Consider the lowest dice: that's the value of the first dice.
The other two, combined with the mod method, give the second dice.
+Beppi Menozzi
I like the idea, but with picking the lowest value, there is no random distribution
2 15
3 25
4 30
5 33
6 35
7 36
8 21
9 11
10 6
11 3
12 1
+Beppi Menozzi
I checked your solution and it overrepresents the smaller numbers. I'll attach my findnings to my document:
nbviewer.jupyter.org/gist/begggah/f530ae7b789a377d24e3f97c32815bfc
just strg +f your name you'll find it towards the end
Very nice analysis, thank you. I wonder if the same happens if you decide to take the first dice as the highest and the second one as the mod?
Simple, you roll it, punch all three dice digits into a pseudorandom number generator to use as the seed, then every roll afterwards for the game will be punched into the same pseudorandom number generator that will pick one of the three options each time! Or set it to pick two options for a two dice game.
A lot of people seem to not understand how dice work.
Divide what you get by 3.
Round to the nearest number.
Just to clarify. The solution should produce tuples of numbers from 1,1 through 6,6 and not merely the sum of two dice ranging from 2 through 12.
Perhaps the way to solve it is to use modulo division to determine the sum of the two virtual dice and then use the sum of the differences of the three real dice to determine the difference between the two virtual dice.
I think the real tricky bit will be if the two virtual dice are required to duplicate the probabilities of a pair of real dice.
+John Kesich what are virtual dice? and modulo division? this doesn't seem so simple
This problem distinguishes people who score well on an SAT, and someone who can actually solve problems with abstract solutions.
so the trick of converting 3-1... and roll it twice
Yeah that would be the easiest solution 😀
I think this is the closest you can get. There are only 2 cases where you have to re-roll.
All 3 numbers match: use that number
Two match: use that number
2 even 1 odd: use the odd
2 odd 1 even: use the even
All 3 odd or even: re-roll
The reason I think you can't do any better than having 2 re-rolls is that all of this so far has an equal chance of coming up, but there are 2 left over.
If you mean that as a solution for the first part (to simulate a single die), then that's already weaker than the answer given in the video...
You are surely right, you need some rerolls
@@lps9767 sorry, no. There are no rerolls _necessary_ for a solution to either of the puzzles.
Well, if you would count having every single outcome to give you the same exact result a solution as well, you would be right. But usually you'd want all of the outcomes to be possible with exactly the same probability
As you have 116 possibilities (or 20+6+30 distinguishable ones). Those numbers do not nicely divide by 6, unleds you substract 2 of them
1: roll a sum of 4 (3/216), 5 (6/216) or 10 (27/216)
2: roll a sum of 3 (1/216), 6 (10/216) or 9 (25/216)
3: roll a sum of 7 (15/216) or 8 (21/216)
4: roll a sum of 13 (21/216) or 14 (15/216)
5: roll a sum of 12 (25/216), 15 (10/216) or 18 (1/216)
6: roll a sum of 11 (27/216), 16 (6/216) or 17 (3/216)
All of these combine to 36/216 or 1/6
+Phillip Marathakis This is kind of a half solution. You whittled down the list in your lookup table from 216 to 16. (sums 3-18). In order for this to be a real solution I think you need to be able to relate the results to the sum categories. Also note you are attempting to convert 3 dice to 1.The puzzle is 3 dice to 2.
+Phillip Marathakis But a 36/216 is not what you want. The distribution for two dice is different. The odds of rolling a sum of 2 with two dice is far different from rolling a 7 with two dice.
+Tyler Reed indeed you are correct those odds are different. That isn't relevant to the fact you are solving the wrong problem. He wants 3 dice converted to 2 dice. Watch the video again.
+squid bait
To be fair, a lot of people probably stopped the video and commented a solution to the first puzzle without having finished the video first. The commenter may not have known at the time of posting about the second problem.
+squid bait Yeah I did what I always hate when I only watched half the video :P
I feel like there is a problem with my solution because nobody has said it yet. Nevertheless, I will say it. It is quite simple.
For simulating 1 die, take the average of the 3 values. Round it if it is a fraction.
For 5, 6, and 3, the 1 die result would be 4.66, which you would round up to get 5.
For simulating 2 die, take the total of the 3 values and multiply by 2/3; then round it. This is essentially having double of the average, or 2 different die results with both having the value of the average.
For 5, 6, and 3, the 2 die result would be 9.33 which would round to 9.
Technically, it is not completely fair since rounding is happening and value is lost, but if a player wanted they could easily just not round. I chose to include rounding because the result should be of a number found on a 6 sided die.
I don't know that the "lookup" solution is that ugly. I don't think you're going to do much better (although there's an attempt below).
Consider what you've got: the triples, of which there are six, which have one combination each; the pairs, of which there are thirty, which have three combinations each; and the rolls of three distinct numbers, of which there are twenty, with six combinations each. You have to map this onto the pairs, of which there are six with one combination each, and the non-pairs of which there are fifteen with two combinations each. For each combination of the latter there are six of the former. The triples pretty much have to be considered together, so we'll say they're all boxcars. If we consider the pairs with their complement, we get six combinations each, so we may as well, and that gives fifteen of six.
So just given all those indistinguishable sets of six, I'll say this is about as good as it's going to get:
-If two of the dice are the same, they map to the two faces showing. (e.g., 1,2,1 -> 1,2)
-If all three are different, and a six is showing, they map to the other two. (e.g., 2,3,6 -> 2,3)
-If they make a triple, they map to midnight.
-If they don't contain 6 or a pair, and they line up, including round the bend bar 6, they're the pair of the middle (e.g., 1,2,5 -> 1,1).
-If they don't contain 6 or a pair, and they don't line up (including round the bend bar 6), one will be surrounded by a gap on either side - this maps to that and six. (e.g., 2,3,5 -> 5,6)
It's a little clumsy, but you wouldn't need a table, at least not after some practice.
+IoEstasCedonta I posted a formula yesterday. Didn't seem to get much attention, though luckily for me someone verified it.
Sorry for the double post but I want to write up my solution in a clean reply.
Below, "s" (small) means 1,2,3,4 and "B" (big) means 5,6. I have 5 different classes of throws:
1. xxx (all three numbers are equal): 66
2. BBB (three big ones but not equal): 55
3. ssB (two small, one big): ss
4. sBB (one small, two big): s5
5. sss (three small but not equal): m6
where m=(sum of throw)%5, if 0 then take 56.
This totally works and it yields both numbers, not just their sum, with the correct probabilities. The big idea I had was rule 3, i.e., sacrifice 5 and 6 to have a fair toss of two "four-sided dice"
Aren't dice 6 sided?
+tgwnn This looks similar to my solution, which is already written somewhere (not equal, though):
Sort the numbers, starting from the highest (since the cubes are indistinguishable, this will give as a unique representation). Different letters in one triple always mean different numbers.
(a,a,a)->(5,5)
(a,a,b) or (a,b,b) ->(b,a)
(6,a,b)-> (a,b)
(5,a,b)->(6,a*b%7)
(a,b,c) (a
+Badass Aviator yea they are. That's why I put it in quotes. The point I was making was that if you toss out the 5 or 6 if it's alone, then we can get the correct probability (1/18 for a!=b or 1/36 for a==b) for any toss ab where they are both at most 4. From then on, I just fiddled around until I was happy with the final result.
+Elchi King (Maddemaddigger) wow, does it really work? it looks really cool. I would say our solutions are about equally ugly :p I can't believe how much time some people on the internet have.
+tgwnn man this is amazing, i just made some excel sheets to try it and it works perfectly
I know one solution, but it's quite hard to find result without calculator. Just make a number with base 6 and translate it to base 10: BASE10 = ([1]-1)*6^0 + ([2]-1)*6^1 + ([3]-1)*6^2. We've got a continuous sequence. Then do mod 12: BASE10 mod 12.
For example, (3 4 1) -> (3 + 4*6 + 1*36) mod 12 -> 63 mod 12 -> 3
+standupmaths The proof ideone.com/t634O3
+SeGaYarkin (3 4 1) should have the same outcome as (4 3 1), because you can't determine an order on the 3 dice. But (4 3 1) -> (4 + 3*6 + 1*36) mod 12 -> 58 mod 12 -> 10.
Paused at 3:03 to calculate a solution programmatically. If you sort the dice results you get a unique key -- so, if you role a 3, a 5, and a 2, call that (2, 3, 5).
You can split the rolls up into twelve groups like this that have the same probability distribution of being rolled as 2-12 do on two normal dice. Here are those groups -- for each group just assume that if you roll something in one of those groups you'd call it a 2, 3, 4, ..., 10, 11, 12 respectively depending on which of the following groups it lands in.
Odds of something from this group being rolled is 1/36
[(1, 1, 6), (4, 4, 5)]
Odds of something from this group being rolled is 2/36
[(1, 2, 6), (1, 3, 6)]
Odds of something from this group being rolled is 3/36
[(1, 1, 3), (2, 2, 4), (2, 2, 6), (2, 3, 3), (2, 5, 5), (2, 6, 6)]
Odds of something from this group being rolled is 4/36
[(2, 2, 5), (2, 3, 5), (2, 5, 6), (3, 3, 5), (3, 4, 4), (4, 5, 5)]
Odds of something from this group being rolled is 5/36
[(1, 2, 4), (1, 3, 4), (1, 5, 6), (2, 3, 4), (3, 5, 6)]
Odds of something from this group being rolled is 6/36
[(1, 1, 5), (1, 4, 5), (1, 5, 5), (2, 2, 2), (2, 2, 3), (2, 3, 6), (2, 4, 6), (4, 4, 4), (4, 4, 6), (5, 5, 6), (6, 6, 6)]
Odds of something from this group being rolled is 5/36
[(1, 2, 5), (1, 6, 6), (2, 4, 5), (3, 3, 6), (3, 4, 6), (4, 5, 6)]
Odds of something from this group being rolled is 4/36
[(1, 1, 1), (1, 1, 2), (1, 3, 5), (1, 4, 6), (3, 3, 3), (3, 4, 5), (5, 5, 5)]
Odds of something from this group being rolled is 3/36
[(1, 2, 3), (3, 3, 4), (3, 5, 5), (4, 6, 6), (5, 6, 6)]
Odds of something from this group being rolled is 2/36
[(1, 2, 2), (1, 3, 3), (1, 4, 4), (3, 6, 6)]
Odds of something from this group being rolled is 1/36
[(1, 1, 4), (2, 4, 4)]
This is still only a lookup table, but at least it has the correct distribution and is quite simple, using only the sum of the dice and whether the roll contains a pair or 3-of-a-kind:
3-of-a-kind = 12
A pair:
If the sum is 4 or 17 = 2
If the sum is 5, 6 or 7 = 4
If the sum is 8, 9 or 10 = 5
If the sum is 11, 12 or 13 = 9
If the sum is 14, 15 or 16 = 10
No equals:
If the sum is 6 or 7 = 3
If the sum is 8 or 9 = 6
If the sum is 10 or 11 = 7
If the sum is 12 or 13 = 8
If the sum is 14 or 15 = 11
+Stein Nornes Interesting solution. This might give the correct distribution for the sum, but it doesn't provide the values for two dice. In some games, like backgammon, the individual values on the dice are important, not just the sum.
+Ocie Mitchell Yeah, of course. Unfortunately this didn't occur to me, since Matt used 4,1,1 -> 3 as his example, not 4,1,1 -> 1,2.
It might be possible to add a rule along the lines of "If the 2-dice sum is ambiguous (in the range 4-10), look at the smallest/middle/largest dice, if it is such-and-such, the difference between the 2-dice is such-and-such."
Finding a rule with the correct distribution would however require time I reeeaaally should be spending on working, not maths puzzles...
Maybe sometime this weekend...
Easy. Roll the cube with all 3 dice inside and pick the number that fits your purposes best.
+Marcus Payne xkcd.com/221/
problem is, that two dice don't have a constant probability spread over them. A sum of 6 is more likely than 2 or 12, contrary to a single dice where 1,3 and 6 are all just as likely. That's what makes it difficult.
My solution would be: find the Z value of the sum of 3 dices, multiply by the sigma of the sum of 2 dices, add it to the average of the sum of 2 dices and take the closest value. In other words: you convert it to the continuous domain and back into discrete, assuming a normal distribution. It's an approximation and I'm too lazy to check if the percentages actually more or less match, but it shouldn't be all that wrong.
and how, pray tell, would you roll a 1
+tim turner Explain to me how you normally roll a 1 with 2 dice?
+tim turner but 2 dices can never equal 1...
+Sean Hazlett he's talking about a small error I made in my second sentence in my reply. The non-relevant part ;).
+Xaab Xaa Applied as exactly as you describe it, this assigns way too low a probability to a 2 result (1/216--should be 1/36) or a 3 result (1/24--should be 1/18). It's pretty clear that fudging the means and variances will never give you the correct distribution, because summing the 3 dice just throws away too much information.
To fairly choose a 2-dice result, first calculate the single die value (3dmod6 +1)
If the result is 1,2 add the 2 lowest dice for the 2 dice result.
3,4 add the highest and lowest.
5,6 add the 2 highest.
For the first question:
Surely probability would have a negative effect on the dice? Just like rolling a 7 is 3 times more likely on a pair of dice than a 12 or a 2.
Think of it this way.
Whatever I've roled on the first die, the second die us equally likely to change the result to any of the six outcomes. Same goes for the third die.
For instance. I roll a 3 on the first die, now the six equally likely outcomes of the second die are as follows:
1->4
2->5
3->6
4->1
5->2
6->3
@@invenblocker Elegant explanation!
Elemental Sheep you mean *_SIX_* times more likely. :-B (Unless you meant three times as likely than rolling either of those other values as a set?)
I paused it at the 1:50 mark to give my guess and as a computer guy I would say you use a mod six plus 1 and that would work. for the second problem you can read the dice left to right or in some way that gives an order then count the evens as a 1 and the odd as a 0 this would give you a binary number like 101 which would be a 6 if a 7 or in binary a 111 or a 0 in binary 000 is rolled you would re roll the dice. granted this is not very clean you end up with two answers that require a re roll. Its not quite close enough so you could call it a Parker Square of an answer.
+FisforFenton well then read the dice to the left and this is your dice. lol , same for 2 dice
For the second solution: use the first solution and just roll twice!
Maybe later. For now I'm just going to put solving this puzzle on my bucket list.