Counter-Intuitive Probability: The Snake Eyes Riddle
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- Опубліковано 14 тра 2016
- You close your eyes and roll a pair of dice. The casino dealer announces that at least one of the rolls is showing a 1.What is the probability that both dice are showing a 1? Watch the video for the solution. (Snake eyes is a term for rolling a pair of 1s. Note: if you do not roll a 1, the dealer asks you to roll again.)
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I will never understand probability despite having taken a 400 level class on it. I can see how technically this works out if you throw it into bayes theorem, but conceptually I see no reason to believe the probability to not be 1/6
When you select a range of boxes in the spreadsheet a little square appears bottom right. You can click it and drag downwards, it will coppy the contents of the cell to the ones you drag down past :-) Really fast alternative to copying cells and pasting them :-)
hi X!
oh my god. here we shall meet. just wanted to tell you i love your vids
+xisumavoid i used to watch your videos! :D sorry for abandoning you i guess? i dont even watch etho anymore so that says something right
+michael collins HOW DARE YOU NOT WATCH ETHO! Jk. It's your choice. I just watch Etho a lot
+xisumavoid
I think that's what he did (via keyboard commands) to get from 12 rows to 10k rows.
I'd sure feel sorry for him if he did it all by hand :)
I feel like the main issue people have with this is that they don't understand the way "at least one roll is a one" works versus "the first roll is a one".
At first glance, 1/6 is an obvious answer because it sounds like you only care about if the second roll is a one because the first already is. However, as you explain and demonstrate, the issue is that the second could be the one and you care about the first. This presents the different results.
+Noah Dale Here's a fun fact though, if the dice roller always shows you a die with a 1 on it when there is at least one 1, the probability is still 1/11 that both are 1 (or that the other one is a 1), despite you now knowing a specific dice is a 1.
+Santarii You'll have to try tell explain that one to me. If I know that one of the dice is going to be a one, then aren't the odds only dependent on the second die, i.e. 1/6?
+Noah Dale It's because you don't know which of the dice is a 1. The dealer looks at both to make that statement so the dice rolls aren't completely separate events any more.
Noah Dale Start with all 36 possibilities.
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
Now we look at all cases where there is at least one 1.
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(3,1),(4,1),(5,1),(6,1)
In 10/11 of these cases, you will be shown a 1, and the other dice will not be a 1. In 1/11 cases you will be shown a 1, and the other die will be a 1.
If you want to include the information of which coin is shown (which doesn't actually affect this but you can look at it this way instead):
(1,1) is different from the others in that there is a random choice to show either the first or second 1.
Let's list all the possibilities.
(1,1) show first
(1,1) show second
(1,2) show 1
(1,3) show 1
(1,4) show 1
(1,5) show 1
(1,6) show 1
(2,1) show 1
(3,1) show 1
(4,1) show 1
(5,1) show 1
(6,1) show 1
(1,1) still only appears 1/11 times obviously (as do all the others), but (1,1) where the first is shown appears 1/11 * 1/2 times, same for the second being shown.
Let's write them out again with their probabilities
(shown 1,1) P=1/11 * 1/2
(1,shown 1) P=1/11 * 1/2
(shown 1,2) P=1/11
(shown 1,3) P=1/11
(shown 1,4) P=1/11
(shown 1,5) P=1/11
(shown 1,6) P=1/11
(2,shown 1) P=1/11
(3,shown 1) P=1/11
(4,shown 1) P=1/11
(5,shown 1) P=1/11
(6,shown 1) P=1/11
Now, if we are shown the left as a 1, this narrows down to:
(shown 1,1) P=1/11 * 1/2
(shown 1,2) P=1/11
(shown 1,3) P=1/11
(shown 1,4) P=1/11
(shown 1,5) P=1/11
(shown 1,6) P=1/11
5/11 of the opposite not being a 1, 1/22 that the opposite IS a 1.
(1/22) / ((1/22)+(5/11)) =
(1/22) / ((1/22)+(10/22)) =
(1/22) / (11/22) =
1/11
So it's still 1/11.
Santarii See, I understand exactly what you're saying there, it's just... something about how it's presented seems... off.
I get it though. It makes sense.
Bayes theorem can be applied here:
Pr(snake eyes | at least one) = Pr(at least one | snake eyes) * Pr(snake eyes) / Pr(at least one)
Pr(at least one | snake eyes) = 1 (definition of 'snake eyes')
Pr(snake eyes) = 1/36
Pr(at least one) = 11/36
So the desired probability is, as you say, 1/11.
+sinecurve9999 Oh man. I'm studying Bayes theorem and it sucks.
sinecurve9999 Conditional probability could have also be applied.
yay I got this one right, I remembered the frog thing from TEDed
Same here, Presh talked about the variations of the frog riddle and it's basically the same thing except that frogs become dice
Isn't this just a rewording of the Boy or Girl paradox? The issue being, of course, that the initial assumption is slightly ambiguous. If we view the Casino game as being
"You roll two dice with your eyes closed. The casino dealer reads the score on the first die as X, and announces "At least one of the rolls is showing X." What is the probability both dice are showing X?" then we instead end up with the result of 1/6. If we instead view the Casino game as being "You roll two dice with your eyes closed. If at least one die is showing 1, the casino dealer announces "At least one of the rolls is showing 1." If neither die shows a 1 the game ends. What is the probability both dice are showing 1 if the game has not ended?" then yes, we end up with 1/11.
Excellent explanation
+Showsni exactly, that's why both explanations of the frog's riddle make no sense. If we know one of the outcomes based on information about one of the dices only, the probability of getting "what you want" in the second dice is just "what you want" divided by the number of possible outcomes considering only the second dice.
+ooo82650 I still don't see the ambiguity here. The dealer announces that one of the rolls is a 1, but he/she never says which one shows a 1. So you gotta consider all the cases when a roll is a 1, just like in the video.
The ambiguity is in whether the dealer uses information from both dice or not. If you roll a die and get a 1 before rolling the second die, the probability of getting two 1s after you know the first die shows a 1 is 1/6. This is also the case if the dealer checks only one of the dice and tells you that you have at least one 1. However, if the dealer checks both dice and announces that one of them shows a 1, the probability of getting two 1s is just like in the video.
Of course, in this case the dealer should be using information from both dice, so the video makes perfect sense. However, both explanations of the frog's riddle (just search for 'frog's riddle' on youtube if you don't know what that is) make no sense, because we have absolutely no information available for the two frogs that did not sing. They are also samples from the same population, of unkown size. So the fact that there is a male singing near another frog does not increase the probability of the unknown sex to be a female.
ooo82650 You obviously roll both dice at the same time. Don't you know the rules ? So the dealer is kinda forced to check both dice since they are rolled at the same time. I don't see no ambiguity here *in this video*. And I won't discuss about thr frog's riddle here since it's in another video.
There's a mistake in the dealers declaration- if he says 'if one of the dice is a one' then your calculation is correct.
However if he were to say, after looking at the outcome, 'one of the dice is one' then the the probability of the other being a one is 1/6 (since all possibilities where neither die is a one is effectively removed upon the dealer's declaration)
I'd say this should be 1/11 probability -- reminds me of the frog one from a little while back. There are 11 possibilities, equally likely (You must recognize that "At least one of the rolls is a 1." is not the same as "I looked at only one of the dice, and it was a 1." It's simply the statement that at least one roll is a 1, with no other information. Therefore, any of the outcomes that match that criterion are equally likely. I have been confused by these puzzles before.). The possibilities are 1,1 ; 1,2 ; 1,3 ; 1,4 ; 1,5 ; 1,6 ; 2,1 ; 3,1 ; 4,1 ; 5,1 ; 6,1. (1,1 should not be included twice, as it is only one outcome.) Therefore the answer is 1/11, or ~9.1%
Is this to correct a mistake with the frog riddle since it was so broad? If so I think this was a really good example so well done
#PassiveAggressiveCompliments
When are people going to learn? You can't pose a question like this without explaining what information would have been conveyed in every scenario. There's a different answer depending on whether the dealer was going to choose a random die and announce its value, or if he was just going to say whether there was at least one one.
Doesn't the initial problem say that it's just whether there was at least a one one rolled?
I really liked the spreadsheet simulation, I've always found the answer to these kind of problems a bit strange, but the table explained it and the simulation proved it. Maybe you could do these things more often
+MrYousupertube
No, the table proved it. The simulation merely tested it.
Agreed, the mathematical proof plus the simulation really adds something to the video. More of this please Presh, great work
After some time with not argreeing with your answer to this and the pair of aces, I finally understood what was the problem. It's the way you ask the question. Because I understand both of this riddles are based on the same idea as the Monthy Hall problem, so I tried to play these scenarios like Monthy Hall.
The way you ask the questions it's like you only have one step, you roll the dice and the dealer says "one die shows one", now what's the probability the other also shows one. There's no dependency between the first and the second die. You need two steps, first the shooter goes for snake eyes then the dealer gives him information.
If I apply the way you ask to Monthy Hall, it's like the host says "you need to choose one of the three curtains, but I can already tell you number 3 is definitely not the car". Now you have your choise between two curtains which gives you 50%.
OMFG @2:08 you're finally doing this!!!!!!! i've always wanted you to do this on your videos. thank you for that!
It initially took me a bit to really understand it. I originally thought it was 1/6. The issue is that the dealer will say that that "at least one die is a 1" 11 out of 36 times the dice are rolled. Out of those occurrences both will be 1 only one of eleven times therefore. You can't think of it by just that one roll. You have to think of it in a broader sense.
This is an example of conditional probability. Probability(both 1) = 1/36, probability(at least one 1) = 11/36, so (1/36)/(11/36) = 1/11.
+Darin Brown This is the probability of there being only to there being two ones, that is right. But this the case when the outcome of neither (dice, in this case) is known. As soon as the result of one (dice) is known it needs to be removed out of the probability( since it now is a certainty) and you fall back to az 1/6 probability.
What is compared in this video are the odds of having at least one one to the probability of both being a one, when the outcome of none is known. The premise however is different.
+Mattias Maes The answer 1/11 is correct. The answer 1/6 would only be correct if you were told there were a red die and a green die, and you rolled a 1 for the red die. In that case, the probability of both 1 would be 1/6.
+Mattias Maes No, the answer when you know at least one of the dice is a 1 for sure, is 1/11.
Before you know that at least one dice is 1, the probability is 1/36, not 1/11.
The note in the description is extremely important.
The answer is 1/6. If the dealer announces that one of them is a 1, then they have become independent outcomes. The outcome of 1 does not affect the outcome of the other.
The graph showing the outcomes is analyzed wrong, if one of the dice is a 1, then you would eliminate either the column or the row, as each represents a dice roll. But we already have 1 dice roll, so the answer is 1/6.
"the dealer announces that one of them is a 1" is ambiguous.
The video's problem statement uses "at least one", not
"the chosen-independently-of-what-they're-showing one".
No they don't; the dealer is not announcing the result of the first roll. He is announcing after both rolls have been completed that there is a 1. The answer is indeed 1/11. Don't be in denial, bro.
There is a mistake in the introduction.If you bet after the casino dealer announce, the probability are 1/6 because a 1 is already came out, so you cannot develop your statistics on both dice.Your statistics works in the case you bet "before" to roll the dice and the dealer tell you that the roll is valide only if at least une dice is 1, in this case your considerations are corrects.
+IoDavide1 Yes I noticed the same thing. Odds are always 'perceived'. Before any dice are rolled, it would be perceived that there are 36 total possible outcomes. But after you know 1 dice's result, there are only 6 perceived possibilities at that point.
You are mistaken.
If you had rolled one die and been told it was a 1 then yes the probability of rolling another 1 is now 1/6.
This is not happening though. Both dice are rolled at the same time. Either dice A or dice B could be showing a 1. This happens 11/36 times.
From the remaining options, only 1 of the 11 have BOTH dice showing 1.
Therefore 1/11.
This is explained in the video.
You are correct Oliver, my mistake
Yes, that's right.
I know this things are counter intuitive, but I have figured it out, as i'm really into Theory of Probabilities...
Interesting parts is that if you knew that first dice was a 1 then it is 1/6
or if say dice were red and blue and dealer announced only if red is 1, then again 1/6
But otherwise 1/11.
This is only true if the casino dealer only ever picks to tell you that you have a 1, and made you roll again when you don't get a 1. If he could pick any number you rolled then in all the possibilities where you rolled a 1 and another number (2, 3, 4, 5, or 6) there would only be 50% probability that the dealer would tell you that you rolled a 1 and there would be 100% chance that the dealer tells you that you rolled a 1 when you role a 1 and another 1.
So, 50% * 10 = 5
That's the number of times you will be told that you've rolled a 1 when you've only rolled one 1.
(there are 10 ways to roll a 1 and another number)
100% * 1 = 1
That's the number of times you'll be told that you rolled a 1 when you rolled two 1s.
1/(1+5) = 1/6
But, like I said, if the dealer chose to always tell you when you had a 1 and made you roll again whenever you didn't then it would be 1/11. But that would be a dull game because the chances of rolling at least one 1 are 11/36.😒
I disagree that each out come is equally likely in the given probability as stated at 1:32. In general yes but once you hear that one die is a 1, that changes your odds. Consider the fact that there are two dice, so call them Die #1 and Die #2. Like you did in the frog problem, let's also keep track of which die was the announced die. If you look at the table above, say the columns represent Die #1 and the rows represent Die #2. Now in 10/11 scenarios, it is clear what die is being announced as the 1. However, when both Die #1 and Die #2 read 1, you actually have two distinct possibilities. Either Die #1 was the announced die, or Die #2 was the announced die. Thus you actually have 12 equally likely possibilities and 2 of which yield snake eyes. Therefore, you have a 1/6 chance of getting snake eyes. Basically when you roll two 1s, the dealer is twice as likely to announce a 1 as if you roll just one 1.
If the rules of the game work such that whenever a 1 is rolled the Casino dealer announces: "at least one 1 was rolled," and under any other circumstance, the Casino dealer does nothing, only then does the probability described in the video make sense. Because if the dealer can announce the amount of 2s or 3s or 4s etc. showing, then that changes the problem entirely. If on a roll of 1, 2 the dealer can announce: "there is at least one 2," then we should follow the probability calculations as I described in the previous paragraph because a roll of 1,1 is twice as likely to produce the statement "There is at least one 1 showing" then a roll of 1,2.
Love the channel and keep up the good work!
This is incorrect. When you say there are two distinct events for both dice returning 1, you've assumed that those two events are just as likely as the other possibilities - they're not. When two dice are rolled, the probability of rolling two 1s is NOT equal to the probability of rolling one 1 and one 2, for instance. It was already less likely to roll two 1s than a 1 and a 2, for instance. The fact that you are told that one of the dice is a 1 doesn't magically change that. What this boils down to is that there is a difference between:
- A die is rolled and returns 1. A second die is rolled. What is the probability that both dice returned 1? (A: 1/6)
- Two dice are rolled. You are told that at least one die has returned a 1. What is the probability that both dice returned 1? (A: 1/11)
***** That isn't what I was saying. Obviously rolling a 1 and a 2 is twice as likely as rolling two 1s. My point is based on the fact that you have to consider what would happen if the dealer doesn't roll a 1. Say the dealer rolls a 4 and a 5, then what does he do? If he announces either "there is at least one 4" or "there is at least one 5," then the roll influences what the dealer announces. As such a roll of 1,1 is twice as likely to make the dealer say "there is at least one 1" then a roll of 1,2.
So when you look at all of the possibilities in which the dealer announces "there is at least one 1," you can see all eleven of them shown in the graph in the video. However, if the dealer can announce a number other than 1, then 10 of those rolls are half as likely to produce the statement, "there is at least one 1," as a roll of 1,1. Then by doing some simple math you can see that if the dealer announces "there is at least one 1." the odds that there are two 1s is 1/6.
+hailmary7283 That's only the case if the dealer bases whether-or-not to say "there is at least one 1" on _more than_ whether-or-not the statement I just quoted would be true.
Was looking for this comment. Hit the nail on the head.
@@loganwaggoner8654 Totally forgot about doing this puzzle 7 years ago haha. Thanks for the reply, it was fun to revisit this.
I must say Presh, it looks like you’ve ... excelled, yourself.
I thought this one was pretty simple, since you can just write out all the probabilities.
If you roll 2 dice there is a 30.5556% chance that at least one of them will be a 1. If you roll 2 dice there is only a 2.7778% chance that both of them will be a 1. 30.5556% is 11 times bigger than 2.7778%, therefore, only 1 in 11 of the times that you roll at least one 1, you will roll 2 ones.
1/11, because the possible combinations are 11, 12, 13, 14, 15, 16, 21, 31, 41, 51 and 61 = 11 combinations. 11 doesn't show up twice because it doesn't matter if the first or the second dice rolled a 1.
Best and simplest explaination
The correct answer depends on your assumption about the dealer's behaviour.
If the dealer has to announce what one of the dice is on every roll (regardless of whether he sees a one or not), then the probability is 1 in 6 that both dice are one. Think of it this way, the chance of getting a double is 1 in 6, and the dealer has to announce one of the dice. So the probability of both dice being what the dealer announces is 1 in 6.
However, if you make the assumption that the dealer announces what one of the dice is whenever, and only when, at least one of them is a one, then probability of both being one is 1 in 11.
Given the way the question was asked, it is perfectly reasonable for someone to make the first assumption about the dealer's behaviour and give the correct answer that the probability is 1 in 6.
That's exactly what I thought :)
the information given has different ways how to set the rules
A year ago I would have gotten this wrong. I credit channels like this and Numberphile
and TED-ED
If we consider that the opposite side where "1" is located, is always "6", then if the top face of the die is not 6, we always have a "1" in other showing faces, that is if we calculate the probability of the top face not having "6", we are calculating the other showing faces with "1" on them, so Pr(6 not on top) = 1 - 1/6 = 5/6, then we have 2 dice and because their outcomes are independent to each other, so the probability of the 2 die showing "1" is (5/6)*(5/6).
1/6 is also not correct, because you can imagine that rolling snake eyes is half as likely as rolling one 1 and one 2
Why do you consider 2,1 and 1,2 like different cases? if you know one die shows a 1 these two cases are identical.
The probabilities are:
1,1
1,2
1,3
1,4
1,5
1,6
Each with a probability of 1/6.
Oh that's an interesting point. I hope someone has an answer to this.
+Zakaria Rakhrour The outcome of each die is an independent event. There is a 1/36 chance of rolling 1,2. And there is a 1/36 chance of rolling 2,1. So the chance of rolling a 1 and a 2 is twice the chance of rolling a snake eyes.
If you don't believe me, imagine one die is red and the other is blue. You could roll 2 on the red die and 1 on the blue die. Or you could roll 1 on the red die and 2 on the blue die. Both are equally likely and both are separate events that occur 1/36 of the time, and both count as rolling a 1 and a 2. Clearly, the color of the dice doesn't effect the probabilities of the rolls. So even if both dice are white and indistinguishable, the chances of rolling a 1 and a 2 is 2/36.
mellamobob When you put it that way, it makes perfect sense!
He treats the ordered pairs (1,2) and (2,1) as different cases because the unordered pair {1,2} has twice the probability of occurring as {1,1}
normally there are 36 possible combinations. Knowing that at least 1 is 1 limits this to 11 solutions, in which only 1 pair would be 1,1. Hence the probability is 1/11
This was easy. We must keep in mind that there are two dice and the 1 can be rolled on each of them, so the probabilities are 1 and X and X and 1, which gives the correct answer. I "calculated" the odds of 1:10 or 1/11 by simply writing down all possible outcomes, marking each time a 1 is there and the one case with both 1s, counted the lines and got the answer.
There are 10 scenarios where at least one die shows a 1. Only dieA and dieB is 2-6, Only dieB and dieA is 2-6. Add them up and that's 10 scenarios. There is one scenario where both dice show a 1 of course, DieA is 1 and DieB is 1. So that's 11 scenarios where the dealer can make that statement, and in only one of them is the other die a 1, so it's 1/11.
No, the dice are independent of each other. It's 1 in 6 for either independently and since we already know one of them is a 1, this has absolutely no effect on the other die and it has a 1 in 6 probability.
***** I did watch the whole video, he is wrong because you are not solving for probability of two dice both rolling a 1, you already know one die is 1. You are only solving the probability of the other die which is independent of the first, ergo 1/6.
***** After being informed the first die is a 1, the other die, being independent of the first has a 1/6 chance of being a 1. The video does not prove the case correctly.
You sound just like me 3 years ago. Once I passed "Probability and statistics" class, I realised how wrong I was all the time.
The correct answer is indeed 1/11. No doubt about that. @chadachada123 already explained it pretty good.
And no, you are not informed that the first dice is a 1. You are only informed that one of those two is 1. That is a big differene, even though it doesn't look like that at first sight.
He doesn't say that the first die is a one. The one could be either the first or the second die, and those are 2 different outcomes. Both dice being 1 is only one outcome. The first die could be 2-6, and the second die a 1, and there would be at least one 1. That's 5 possibilities. Or, the first die could be 1 and the second die could be 2-6.That's 5 more possibilities. Or, the first die could be a one and the second die could be a one. That's only one (not 2) possibilities. (You can change the order you state the outcome: the second roll is a one and the first die is a one, but that's just saying the exact same thing a different way, which doesn't add an actual outcome.) Eleven possibilities, only one 1,1 outcome, so 1/11 chance. This might be easier to think about with 2 coins. Outcomes can be: HH, HT, TH, TT. If you toss both and know that at least one is a head, then there are 3 possibilities: HH, HT, and TH (similarly, if you roll 2 die and know that one is as 1, then the possibilities are 11,12,21,13,31...total 11 possibilities) If you know that one is a head, then the chance that the other one is a head is 1/3 (similarly, if you know that one die is a 1, then the chance that the other die is a 1 is 1/11). Notice that HH and 1,1 are only listed once, not twice. There is only one way to toss HH, just as there is only one way to roll 1,1. But there are 2 ways to toss a head and a tail, just as there are 2 ways to roll any 2 different numbers.
Pretty straightforward. Got it quickly.
I like your videos a lot, keep them coming. One point, however. If you're going to be correct and say die for singular dice (not really that important) then do it consistently throughout the video. You wouldn't want people to comment... oops.
no there's a 1 in 6 for each dice the result of the first die does not influence the result of the 2nd die
But in the question it wasnt that the 1st dice showed a one, it said any of the both dice showed a one
I always thought that if 2 outcomes were, in the situation, are the same(like 1,6 and 6,1 since one is a 1 and the other is a 6) then we count it as one.
I could be wrong, but if you use identical dices and roll them at the same time, wouldn't it change the outcome of your experiment?
It has been a while since I had stochastics class, but I think I remember something like this.
I think this is wrong. Being 1,2 or 2,1 is the same thing so in fact since 1 dice is already fixed in 1, the other can only oscilate through 6 independent possibilities.
The way the question is asked, is knowing P(1|1) which is independent from whatever dice is 1. The probability that the other dice is also 1 is 1/6 in fact.
+Rodas nePervilo I think I'm with you on this. It's no different than if only a single die was up and you were asked the odds of it being a 1. The dice have zero influence on each other, and the value of die #1 has no relevance to the value of die #2.
+Rodas nePervilo Yeah, his labels are wrong so it messed up his math. should be dealers die and rollers die, not die A and Die B
So you're saying that 1,2 and 2,1 are the same thing? First of all the fact that ' 1 : 2 ' is written differently from ' 2 : 1 ' should give away the fact they're different. But you can also do the experiments yourself with real dice if you wish, and you will obtain that you're twice as much likely to get ( 1:2 or 2:1 ) as ( 1:1 ).
I''m not saying that you're not twice as likely to have 2:1/1:2 than 1:1, i'm saying that no matter what , one of the dice is already fixed in one, so the only possibility would be let's say 1:2 and not 2:1.or vice versa., for each dice the probability of having a 1 is 1/6 and the same for the other dice . The probability that the second die has the same value as the first is 1/6 no matter what.
*****
From what I understand you are counting (1,1) in the (1,y) cases, then when you consider the (x,1) cases you also count (1,1). In other words you are counting it twice.
Problem is : (1,1) is 'shared' between the cases with (x,1) and with (1,y) : you can't count it twice.
I got 1/6 by Monty Hall style logic.
There is 1 way to get (1,1) and 2 ways to get (2,1) up to (6,1) which gives the 1/11 that was given as the answer. However, you do not know the casino had to tell you there was at least one 1, and indeed it makes sense since before the statement, it was likely you weren't seeing a 1 at all. As for example the casino might have said "there is at least one 2" if the roll was (2,1), we only take half the probability for the non-matching dice leaving 6 possibilities all equally likely. Note that if it is (1,1), the dealer has no choice but to say there is at least one 1.
I play a lot of bridge, and there is something called the Principle of Restricted Choice which works on this logic.
Edit: I made a spreadsheet based around whether or not the assumption that dealer can only say "there is at least one 1". It shows around 1/11 with the assumption in place and around 1/6 when you don't make the assumption. docs.google.com/spreadsheets/d/1vp4VnTcMlh1dzoeuTGBE4xbT_ZczhVvHz8gg3VU2kco/edit?usp=sharing
Classic example of the Boy or Girl Paradox. The note in the description is non-trivial and should be included in the video.
Example: if the dealer's rule is to pick one of the visible dice at random and announce that there is at least one X, then there is a 1/6 chance that the second die has the same value.
The case displayed only works if the dealer's speech space is reduced to [at least one 1, no 1's].
Poorly worded question. If you are going by Counter-Intuitive Probability, this is 100% correct, but if you were to tell me one of the dice rolls was a 1, I'd be able to correctly guess the other roll 1/6th of the time.
Keeping the A and B labels. Interestingly enough, If we are only told that A is a 1, (if and when) then we can be assured that B has a 1/6 chance of also being a 1. Is that right? For whatever reason this is very hard to grasp. I understand the table you made. I also understand that your result is correct.
But let's say if we only include the sample of when, "A" is a 1 ("one"), regardless of B, but also recording B. Then I'd expect that there is a 1/6 chance of "B" being 1 ("one") therefore a 1/6 for both being a 1. I know that these are inherently different questions, but as I said it's very abstract and it's very hard to see the difference.
+MindYourDecisions
I foolishly did 1/36 cuz the 2 dice being 6^2 and the snake eyes being only one...
i foolishly missed or forgot the part where the sample size is where ATLEAST one of the 2 dice is 1
:'(
*Takes one look at the question.*
"My Monty Hall sense is tingling."
When u place ur bet, it is given that the first die is one. Therefore, pr(die rolling a 1) is 1:6
i got this. we need to be clear whats the given probability. the questions says one of the dice is 1, vs die A is 1.
the resultant probability will be 1/11 and 1/6 respectively.
It depends on how the situation comes up. If the dealer has to announce something, the odds isn't worse than 1/6 (36 pairs, if the dealer announce at least a 1, then you remove the 25 without a 1, and ALSO weight the results with only a 1 by 1/2 each because you would only get to this world half the time [i.e. if the rolls are 1 and 3, dealer would announce a 1 1/2 the time, and 3 1/2 the time])... If the dealer rerolls if there are no 1, the video is right.
+YK L "If the dealer has to announce something, the odds" are still 1/11, since in 25 cases, nothing stops the dealer from announcing "None of the rolls is a 1". If we introduce different probabilities of them announcing "at least one of the rolls is a 1" according to which way that happens, then we can't deduce _anything_ about the conditional probability of a snake eyes. (For example, it could be that if the rolls are both 1 then the dealer announces "at least one of the rolls is a 1", else the dealer announces "water is wet".)
I have to amend my statement a little. If the dealer has to choose one of the roll and announce it, if there is a choice, he does it randomly.
Once the one shows up the chances ARE one in six that the other die will be a one. Throw the first die, and note that it is a one, and three minutes later throw the other. This probably will be one in six. If it is not known what either of the dice will show, the odds will be one in thirty-six. I have no idea where you get one in eleven.
He gets one in eleven by assuming that the dealer will announce "At least one
of the rolls is a 1." whenever that's true, not just when the first die is a one.
That's not what the question says. You roll both dice at the same time.
Hmmm, what he just said doesn't make sense "What is the probability that both rolls show a 1". At least the description says "both die show 1". As the dealer didn't announce which die showed the 1, then the odds are going to be lower that 1 in 6.
I need to learn how to effectively use spread sheets...any helpful tips?
Lol at some of the people in the comments who still think it's 1/6 despite both a mathematical and simulated proof being given
yea, well something still doesn't seem right here... I flat out refuse to accept that in his mathematical proof he had a frame for 1;2 AND 2;1 but he didn't have 1;1 twice.
if we changed the question that 1 dice is a 1, what is the odds of the other one being a 2? what would the result be?
Kaleb Bruwer Ummm what? Why would you have 1:1 twice when you can only it it one way?
Ping Pong Cup Shots I have a point I'm trying to prove, so please answer the question I asked.
Kaleb Bruwer that's a bit hypocritical of you since you didn't answer me question either. Also, if you answered my question you would have also answered your own question.
Anyway, were it to be where one is 1 and other is 2, it would be 2/11 because there's 2 ways to get a 1 and 2 and there are 11 equally likely outcomes
+Kaleb Bruwer i found it confusing too at first, but just try to imagine that the two dice are different from each other. Like one is red and the other green. Then you would have two different rolls for 1;2, 1 could be the green dice, and then 2would be the red. But it could also be the red dice with 2 being green. So 1;2 and 2;1 are indeed two completely different rolls.
This is the first one I've ever gotten :-)
Laws of logic apply to the die that is considered the 1. Once the 1 is determined, it is immediately dismissed and accounted for. It then becomes a single trial of the second die, which has a 1/6 probability of being another 1. Call it what you will, but it's still 1/6. As the trial only applies to one die.
You might be interpreting the problem statement as at least one of
The dealer looks at one of the dice, lets n be the value that die shows,
and announces that at least one of the rolls is a n.
The dealer looks at one of the dice. If _that die_ is a 1, then
the dealer will announce that at least one of the rolls is a 1,
else the dealer won't announce that.
.
In either of those cases, 1/6 would be the right answer. However, the video
is assuming the casino dealer will announce "At least one of the rolls is a 1."
_whenever_ [the sentence quoted within this sentence] is true. Is your argument
that 1/6 will still be right for the video's interpretation of its problem statement?
No, it's not 1/6. Your probability skills are very weak.
For a tournament backgammon player this question is as simple as two plus two. 11 rolls in 36 contain at least one ace, but only one roll consists of two aces. :)
+rolline Right, but we are ignoring any roll that doesn't have an ace. so any time you bet you already know one die is an ace. you are only betting on the outcome of one die.
Great riddle! It's like the opposite of the Monty Hall Problem. We know the probability of throwing snake eyes is 1/36, but the we are given the information that one die is a 1, so we disregard it and work out the probability of a single die being a 1, which is 1/6. And also wrong :-) I'm going to try The same thing with 10 coins.
You still haven't fixed this video? It's completely wrong but it's still here.
No, it's 100% correct. You're just an idiot.
It's still pure rubbish. Even their "simulation" is pure rubbish. You'd understand this if you weren't a moron.
So here's the situation. You roll 2 dice. One isn't rolled, it's a 1, always. So that die has no randomness in it at all. Just read that once more. No element of randomness at all. P1 = P(die1 = 1) = 1, P2 = P(die2 = 1) = 1/6. Alright? Note that it's irrelevant which die is the non-random one. There still is a die that isn't rolled, doesn't matter which one.
You've convinced to believe by that video above that the die without any uncertainty in it affects the other die's probability to hit a 1.
Now let's the change the game a bit. I'm going to replace that guaranteed '1' die with another die that has a '1' on every side. This time we'll save the croupier's time a bit and he can go take a coffee break. Our game hasn't change other than the croupier's sipping coffee, smoking a cig and bitching his life to other workers in their break room. Still think it's a 1/11 chance that the other die will roll a '1'?
Okay, let's change the game again. Let's replace the other guaranteed-one-die with a pineapple (you can pain a big red '1' on it if you wish). The croupier comes back from his break and we'll throw the die and the pineapple (you can blow on them for good luck). After the throw (which we'll do blindfolded, of course) the croupier (his name is Jim) announces that the other thrown item was a pineapple. What's the probability that the die we threw is a 1? Think hard.
The odds are 1/11 because of the information we're given. We aren't told the results of a specific die - the dealer says that of two rolled dice, one has rolled a 1.
Now, before we move on, can you tell me the odds of either die rolling a 1 from two dice rolls?
Hint: he showed in the video how many of those outcomes exist.
Sami Tynninen So you completely misunderstood the problem. The dice rolls are simultaneous and the observer is looking at BOTH results at the same time. When this is the case, 1/11 is correct, and it can easily be proven by looking at all possible permutations of the dice that include a 1: there are 11 of them, equally likely to occur, of which ONE is 1-1. So let's see if you can accept 1/11, or if you're going to still be a stupid fool.
If you rolled one die, and it came out 1, and THEN rolled the second die, the odds of getting a 1 with the second die would be 1/6. But, that is a different scenario then what was described.
It's called conditional probability. As both dice are independent, we do not count one die being 1 and the other being 6 as any different than one being 6 and the other being 1. When we don't know either dice, there is a 1/36 possibility of snake eyes, but as soon as we learn the outcome of one dice, there are only 6 states the second one could assume. The right answer should be 1/6.
If "we do not count one die being 1 and the other being 6 as any different than
one being 6 and the other being 1", then that event's probability will be 2/36. Otherwise, your probabilities would add up to less than 1.
Choose x uniformly from {"HT","HH","TH"}. The probability of x being "HH" is 1/3, and since _all_ elements of {"HT","HH","TH"} satisfy
at least one of the letters is H
, conditioning on the event [at least one of x's letters is H] does not change that probability. This is despite the fact that as soon as we learn one of x's letters, there are only 2 possibilities for what x's other letter could be.
Wrong, it's 1/11. The dealer is announcing it after BOTH rolls, and therefore they are not independent.
It all depends on,wheter these dice are distinguishable or not.If you can be sure that one dice is a 1 then the prob of the other also being a 1 is 1/6,that's the case in which they are indistinguishable,s. o they can only be 1.1. 1.2. 1.3 and so on,a result of 6:1 is the same as 1:6 thos the six results .But if you can differ one from another,a result of 1 on dice A, and 2 on dice B is not the same as a 2 on A and 1 on B and thos the 11 results
The problem with this answer is that you treat both dice as seperate entities. Or in probability with grabbing and laying back.cbut the order in which the dice are is still not important. You will always in this have 1 dice showing a one and it does not matter if it is dice one or two. So you always 1/1 have 1 dice showing that 1. Then there are 6 possibilities for the other dice. Without the order of the dice being important and counting them as seperate gives you 1/6. When you do use the order as important you get to 1/11
Similarly, without "the order of the" coins "being important and counting them
as seperate gives you" a 1/3 chance of two coins landing on different sides.
You were just showing off with the spreadsheet!
I think the probability is only 1/11 if you are going for two 1s and you do not loss if none of the dice shows a 1, (you loss if one is a 1 and the other is not) If the dealer announces that one dice is already showing a 1, the probability for the second dice to be 1 should be 1/6. So the odds is 1/6 if you place the bet AFTER the dealer announces at least one of the roll is a 1, but 1/11 BEFORE the dealer announces one of the roll is a 1.
P(n_k)-P(n_1)=P(n-1)
P(n_1)=1/6
P(n_k)=1/6^6
P(n_1)=1/6-1/6^6=0.166
Or maybe 1/6^2=0.0278
Damn it I'm wrong.
Now, I don't have a very great track record in math, and I tend to study paradoxes and philosophy more than mathematical puzzles, but something about this puzzle seems off in my mind. I am not saying that the narrator is wrong and that I am right, so correct me in my comparisons and logical steps if I am mistaken.
So, when the two dice are rolled with my eyes closed, I am assured that one die of the pair is a one. Now, I need to calculate the odds that both dice are ones. Rather than calculate the odds of both as seen in the video, couldn't I just logically say "because I know that at least one of these two dice are displaying a one, I should just calculate the odds of rolling a one on a standard six sided die?
To give an example of what I am thinking of, imagine that you are rolling a single die. You roll a 3. Then, on your second roll, you roll another 3. Then, on your third roll, you roll another 3. If you calculate your odds of rolling that die and getting a 3, it is still one in 6. There was a 1/216 chance of rolling 3 3's in a row, but rolling a 3 on a six sided die is still 1 in 6.
Again, I am no mathematician. This isn't some grand way of saying that I am right or the video is wrong. I am simply looking to confirm or deny this line of thought with a bit of help.
It would seem that way at first glance, yes, but that isn't how conditional probability works. There's a reason that Probability & Statistics is an advanced college-level math course, and not just first-year stuff.
It's subtle, but it's the difference between "given that the FIRST die is a one, what's the probability that the SECOND die is also one" (1:6) and "given that AT LEAST ONE die is a one, what's the probability that BOTH are one" (1:11). In the first case, you can basically ignore the first die and look only at the six possible outcomes of the second roll. But in the second case, you don't know WHICH one rolled a one, so instead of looking at just six possibilities for one die, you have to look at all eleven ways in which at least one of the pair can show a one.
+IronMagus Ah, I appreciate the clarification
but we are sure that at least one of the dice shows one.
And...?
If the probability is 1 in 11, the betting odds would be 10 to 1.
A number of these puzzles are hard. This wasn't one of them. It was straightforward to solve as soon as I read it in the description.
how about you go read a book on manners. while you are at it figure out that "hard" is a measure of "this... one" to "a number of these puzzles"
so.. your statement is completely non-sequitur to his experience.
"stupidly brag about how straight forward"... well... a bit of context; The frog problem given by OP was more complex.
and... I dont know why you chose the word "apathetic". Seems like it might stand, considering he didnt sneer back at you. Maybe he took a higher path... zen of him?
so you are wrong about everything you said... and you were a condescending jerk about it. NJ!!!
"how much thought went into developing probability theory"...
Some ppl find somethings easy even tho there is a ton of theory. Rubik's cubes, chess openings, tying knots... ur a tool.
that doesn't make sense. if there is already a 1 how is possible to have another 1 on a same dice. so what happens if I roll the 2nd dice a second after the first 1st dice is said it's was rolled 1. would the probability be 1/6 then.
The point is that the rolls happen at the same time. If the dealer only looked at 1 die and said it's a 1, it would be different. But he looks at both before announcing.
...Oh!!
If one of the two dice is a one, then there are eleven possible states (d1 = 1 and d2 = 2-6, d2 = 1 and d1 = 2-6, d1 and d2 are 1), so both dice _do_ matter.
Got it correct immediately! No need to clap
reminds me of "equally likely events" from the frog riddle xD
Why don't you count a 1-1 roll twice, once for each "at least one" 1? ("?"-1 and 1-"?" are different permutations aren't they?) Or does this question only deal with combinations?
+N0tY0ur4v3r4g3N0th1ng
Because a 1 and a 1 is half as likely to occur as a 1 and a 2.. Just as two heads is half as a likely to occur when flipping coins as a head and a tail.
The trouble with this is the premise of the probability calculation is wrong. The theory of which die shows the 1 is irrelevent because the premise presupposes that one of them already shows a 1. One of the dice can be completely discarded in terms of probability because it is guaranteed to be a 1, the only dice roll that matters is the remaining die.
Surely if one of the dice is guaranteed to be a 1, then there is only one variable, which is the remaining die, regardless of which that is. The true premise of the game is if the remaining die can be a 1. The premise that they used requires order as well. Once you know without a doube that there is a 1 among the two rolls, you know the options are 1,1, 1,2 ,1,3, 1,4, 1,5, and 1,6. You can disregard the couterpairs (2,1, 3,1 etc.) as they are irelevant to this. You know that one of them is a 1 already. the only variable is a single d6 die.
The probability is surely actually 2 seperate trees of 1/6, one for each die that ended up as a 1.
Of course, that premise is with the presupposition that the game is only played if a 1 is rolled from one of the dice rolls. If the game is played with any roll of the dice, with perhaps, the lower of the two results being selected as the 'target' in that case, yes the prbability is 1/11 as the results show. Basically the probability comes down to the information the roller has. If you know you are trying to get snake eyes and that that there will be no expectation of payout in eithe rdirection then the probability is 1/6. if you know the premise of the game is to get a doubles then it is 1/11.
+mistriousfrog
Your analysis makes no sense. The problem states that you roll two dice and don't know anything about the result until you're told that "at least one is a 1". There are 11 ways this can happen and only one of those ways results in a pair of 1's. The answer is 1/11. It's that simple.
I don't think the guy from the casino would be smart enough to know that.
You could say 1/11 and he/she would say "wrong, it's 1/6".
+Ramiro de Souza Show him the proof.
When i tried to solve this I thought can't you just see the two die as independent events?Therefore the chance of the each of the die landing on a 1 is 1/6. Surely then you can just multiply those together to get 1/36? why did I go wrong here?
The results 1:2 and 2:1 are repeated, aren 't they? and so are the rest of the cases (?)
If I know that one die rolls 1, then it doesn't matter for me wether it is the first or the second one.
Although this is not the given answer, 1/6 still feels right for me. What am I doing wrong?
+Regino Domínguez They are twice as likely
No, as the dice are not separate occurrences , so you need to account for all permutations, not just combinations
For die A, given it is a fair die, there is a 1/6 chance of rolling a 1.
For die B, given it is a fair die, there is a 1/6 chance of rolling a 1.
I think we can all assume that much is true.
The probability of rolling die A as a 1 and rolling die B as a 1 is 1/36.
The 1/36 comes from the sample space shown, but can also be obtained by multiplying (1/6 * 1/6).
(A|B) is the probability of A given B.
If you are given that the chance of rolling a 1 for B is 100%, then you have (1/6 * 1), which equals 1/6.
This problem can also be inverted to be (B|A).
If you are given that A is rolled as a 1, then the answer is still 1/6 by multiplying (1/6*1).
+Travis Land Choose x uniformly from {MF,MM,FM}, and consider the probability of x being MM.
"If you are given that the chance of" x's left letter being M is 1, then you have (1/2 * 1), which equal 1/2.
"This problem can also be inverted to be" "given that the chance of" x's right letter being M. "If you are given that" x's right letter is M, "then the answer is still" 1/2 by multiplying (1/2*1).
Does it follow that
the probability of x being MM, given that
x's left letter is M or x's right letter is M
is 1/2?
(If you say yes, then what if we replace "x's left letter is M or x's right letter is M" with "x is in {MF,MM,FM}"?)
Yes, the probability would be 1/2 in either case.
If the left letter is M, then we can eliminate FM from the sample space, making it a 50% chance of MF and a 50% chance of MM.
Likewise, we can invert the problem by saying the right letter is M. In that case, we can eliminate MF from the sample space and have MM and FM to choose from, each with a 50% chance.
+Travis Land But his final point is trying to show the flaw in your argument by saying "given that either the left OR the right letter is an M, is the probability of MM out of the whole set still 1/2?" Obviously not - if you pick randomly from that whole sample space (for which the statement applies to all three) the probability is 1/3.
Your first post makes the mistake of thinking that in the dice roll, (3, 5) is identical to (5, 3) but they're not, they're distinct outcomes. It's a very common mistake. Here's the best way I can think of to explain why. Let's use coin flips instead of dice to simplify. You have two outcomes, H and T. Under your reasoning, there are three outcomes for flipping both - HH, TT, or HT, which is the same as TH. So if you flip both many times, you would expect to see HH a third of the time, TT one third, and combination one third. But here's why that makes no sense.
Since the combination outcome can be either HT or TH, that means that 1/6 of our total trials were HT, and 1/6 were TH. Our total experiment, sorted, might look like this: HH, HH, HT, TH, TT, TT. That's six tries, one third with each result, in your scenario. But notice for the first coin, two out of three of its heads were paired with another head, and two out of three tails were paired with another tail. Why would this happen? You should expect that no matter what the outcome of the "left" coin, the other has a 50/50 shot. Your logic is making it look like independent coins have a tendency to "pair up" and get the same result when flipped together.
If I throw together a simulation in javascript right now in the browser for the coin flips, here's what I get: With 1000 trials, I got both heads 279 times, both tails 225 times, and a mix 496 times. Tested 10000 times, I got both heads 2482 times, both tails 2495 times, and a mix 5023 times. It's certainly looking like "one heads one tails" is twice as common as the other outcomes, happening half the time overall. That's because there's actually 4 distinct outcomes: HH, HT, TH, TT.
This is fundamentally wrong because there are only 21 different arrangements with two identical dice. The only time 1/11 is correct is if the two dice are different. IE if you throw a red and a blue dice and the dealer says one of the dice is 1 what are the chances the red dice is 1?
No, because you are finding the likelihood of only one permutation, 2 identical die would work just fine but due to us not knowing which die has that number, doesn't matter if they're unique or not the odds are the same
+BulltonCubing yes 1/6
+Nathan Koenig (Nate) He's not wrong, let's put it this way: If you roll 2 dice there is a 30.5556% chance that at least one of them will be a 1. If you roll 2 dice there is only a 2.7778% chance that both of them will be a 1. 30.5556% is 11 times bigger than 2.7778%, therefore, only 1 in 11 of the times that you roll at least one 1, you will roll 2 ones.
+Nathan Koenig (Nate) Think of it another way: What is the probability of rolling ANY number on one die and a one on the other?
The singular of "dice" is "die"
I just want to say, to those out there insisting that the answer is 1/6, you are incorrect, but that is okay, it is a very confusing problem! Try to hear out the explanations and understand it, and if you get it, don't feel bad about not getting it before, and if you don't get it, hey, that's okay too. And there's nothing wrong with explaining your own reasoning as to why you think your answer is correct, that's likely the best way to find an explanation that makes sense to you. Whatever the case, what I'm trying to say is, don't feel bad if people mock you for insisting that the wrong answer is correct. This can be a very confusing problem, that's what makes it interesting.
Yay! I got this one right! Probability spaces ftw.
Great minds think alike! I laid out table the same way as shown.
+Rich Dobbs Still believe the great thinker made a mistake in caculating the third column...
I don't think we are talking about the same table. I was talking about the table he showed at about 1:09 in the video. I think you are referring to the spreadsheet that he put together.
roll 2 dices: both are showing a 1?
p=probability:
p(showing 1)+p(showing 1)
=
1+1
6+6
=__2__
12
also written 2/12
another example:
p(not showing 1) + p(not showing 1)
5+5
You lost me! Casino has already announced that one of the dice show 1.
I rolled snake eyes closing my eyes and rolling on my first try
feels like a modified Monty Hall problem
A tru mindfuck starts when you try to count the propability of the second dice to be 2... or 4(or 5, or 6)
If at least one of two rolled dice is 1 then there are 11 possible outcomes, (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1) or (6,1). So, since there is only one way to get desired (1,1) outcome, its probability is 1/11.
+Rick Reed
I disagree. 1,2 and 2,1 is exactly the same thing and the same for others. Let's put it this way, you know one of the dice is a 1. The other can only be 1, 2, 3, 4, 5 or 6 with the same probability of 1/6 each.
Rodas nePervilo When figuring the probability of a desired outcome, you always count the number of unique possible outcomes, and divide that into the number of the possible outcomes that meet your criteria. You have to consider each die separately. There are two ways to role each combination containing at least one 1, except 1,1. So, for example, if instead we asked the probability of the other die being a 2 instead of a one. Then the answer would be 2/11 because there are two ways to get that outcome, 2,1 and 1,2. You would most definitely get this result more often than the 1,1 outcome.
*****
I get what you're saying but i don't really agree with the approach. Let's give the example of coins. You throw two coins, and know that one of them is tail. therefore the other can only be head or tail. What you're saying is that the 2nd coin has 2/3 probability of being a head and and 1/3 of being tail, while i'm arguing the second coin is independent and still has 1/2 chance of each result
On another approach, let's consider the question: What is the chance that the two dice have the same outcome? It's easy to check that it's 6/36= 1/6
Why would "knowing" the value of a dice reduce the chance of 1/6 that the other die has the same number? It'd make no sense that the chance of being the same number reduces simply because we know which number is it.
+Rick Reed Was going to disagree, but I may be getting it now. The answer would 1/6 IF the two die were labeled say A & B AND the dealer told you that A was 1 (equally if B was one) But because there are more ways to fail than succeed if either A or B can be 1 the odds must go down.
So then if at the beginning the dealer says that at least one of the dices will be 1 which is the probability to have both dices showing 1? Looks like he is not very useful with sharing the dice information in the video scenario isn't he? Or maybe the solving method is wrong...
Wait... since this is all a probability, would the chance of getting snake eyes 1/36 since theres 36 different ways the dice could end up?
(1/6 * 1/6)
wouldn't* sorry
If you look at it the question, doesn't just ask what is the probability you will get snake eyes?
You can also use Bayes theorem to do this.
Can you please, prove in this way that the riddle that includes two goats and a car, changing the selected door after one door is opened actually doubles our probabilities to win the car. I still do not understand why that.
+Juan Serrano (JKL) The monty hall problem works like this. You pick a door. There is a 1/3 chance it is a car, and a 2/3 chance that it is a goat. Now once you reveal a door with a goat, you have the choice to switch. You'd assume it is 50/50, but when you initially picked doors, the probability that you picked a car was 1/3. meaning that if you switch doors, the probability of you picking a car is the same as the probability that you didn't pick a car in the beginning (since you would then know all of the locations of goats). so if you switch you win 2/3 of the time, since you would have lost. I think if you stay with the same door (based on this video) the odds are 50% for the car but that is because switching doors changes the problem.
Matthew Ciaramitaro what I really want to see is if that work in real life.
it works in simulations of the gameshow. I'm sure if you collected the data of all the people that switched doors from that game show about 2/3 would have won the car
Interesting, works against my reasoning... I need to study this more
I said 1/36 . They did not say that it had to be those specific die
You ignored the other 1:1 in the probability. Is there a rule for when it is applicable to only count one of the duplicate pairs?
+Kate Blank " If the problem is changed from
"The dealer announces that one die is a 3. What is
the probability that the other die is a 4?" " to what?
Let's mark dice with Left and Right. L4R3 isn't the same event as L3R4.However, both dice showing the same number is clearly a unique event.
+Eric Demer
so we have the initial possibilities
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 45
51 52 53 54 55 56
61 62 63 64 65 66
now we can eliminate every case, where there is no 3 in it, and we are left with this matrix
13
23
31 32 33 34 35 36
43
53
63
so now we are counting, we have again 11 cases.
now we need to count every case where the a dice is also a 4
this are 34 and 43
so your probability is 2/11
why can't i use bayes' theorem? i got 1/12 using it
You are wrong. The answer actually is 1/6. The table used repeats possible rolls twice because it assumes both dices can roll any number but we already established that one dice is a one for sure. So the only possible outcomes are 1-1, 1-2, 1-3,1-4,1-5,1-6. No other combination is possible. So you are wrong
I got this only because I saw the Frog riddle.
You're wrong, it's 1/6, he already said one dice was one so it's 1/6 that the other would be one taoo
He only said _at least one_ die was one.
first i was thinking its 1:6 but this is too easy
I still think about it
I was way off. I had the idea of the equation being 1/x*y with X being the number of sides of the die, and Y being the number of die rolled.