Upping the door count to absurd quantities is definitely the best way of resolving the "but it's a 50/50!" reaction, and that switching is identical to having been allowed to choose 99 doors at the start instead of 1.
so cool that people are already posting “corrections” before it’s even possible to have watched the whole video and listened to what you had to say in the first place
@@delightfulkissboy8926 yes I was; I had absolute 100% confidence that our beloved DKB would arrive at the correct conclusion just as I had, hence my comment with my maths reaching the same conclusion
I'm a stats prof, and I love your videos. You do a great job teaching probability concepts. This one is so great for the very subtle difference between Monty Hall and the Buckshot situation. You laid it out wonderfully!
I did not believe the Monty Hall problem was anything but an edge case psychology thing until your "smushed" visual section. Thank you. I've known about it for years and you're the first person to explain it in a way I could understand.
Delightful explanation as always! The Naive Monty Hall Problem interpretation is sometimes called the "Monty Fall Problem", where Monty trips and falls, revealing a random door, then giving you the option to switch.
The tl;dr version of the Monty Hall problem is because you had a 1/3 chance of choosing the winning door to begin with, switching means you have a 2/3 chance of winning, because with these conditions it's essentially the same as betting against yourself.
I think the biggest challenge with the Monty Hall problem is successfully explaining the Monty Hall problem to someone who doesn't understand it. You showed a pretty good explanation here, and I'm going to share this video with whoever I'm trying to teach.
Assuming this is about the magnifying glas... The Monty Hall requires the "prize door" to be guaranteed not to be shown ergo making it a true 67% chance on swap.
I paused this video a couple minutes in to code up a simulation in python of the monty hall problem just to remind myself that there is a god out there messing with us for the fun of it, only to resume the video & learn that a coding experiment was already thought of & included
I recently got into coding not long ago and this gave me a bit of motivation to continue learning more about it through your explanation tysm you are truly delightful!!! :D
Video Spoiler Ahead: To any not understanding how this situation differs from the Monty Hall problem(but do have an understanding of why you should switch in the Monty Hall problem) I may be able to help. In the Monty Hall problem each initial door has a 1/3 chance of being the winner, after opening a dud, your initial choice remains a 1/3 chance but the probability of the remaining door(switching) now has 2/3rds odds of being the winner. However in this buckshot roulette comparison the initial door starts at 1/3 but because the opened door is fixed(and can be revealed to be the winner) the probabilities of each remaining door update when it is opened. The probability only stays locked at 1/3 in the Monty Hall problem because the revealed doors are guaranteed to be duds. Imagine you play Monty Hall again but now with 10 doors(still only 1 winner), you select one and the host reveals eight duds among the remaining doors. Your initial door still has a 1/10 chance of being the winner because the surviving door takes the odds of the other 8 doors(now carrying a 9/10 chance of being the winner). Why do the odds transfer in Monty Hall but not in buckshot roulette? Because the doors opened are guaranteed to be duds thus not changing the probability of the initial pick. The initial selection is excluded from the dud removal(thus keeping it's initial odds) but the probability of a winning door is still split amongst all doors(and thus transfered to any unopened AND unselected doors). As an aside in a Monty Hall problem with 5 doors, if you pick 1 and the host opens 2 duds, your initial door is still 1/5, the two opened doors are of course now 0, and the 2 unselected unopened doors each have a 2/5 chance of being winners, thus you should switch to one of them.
im coining the burner phone variation you described in the end where the burner phone can just tell you the live round as the delightful kissboy problem see yall in 25 years
I’ve never truly understood or believed the Monty Hall problem before watching your video, words can’t express how thankful I am for your explanation. You’re great dude, keep at it!
An alternate way of explaining the subtle difference, in case this helps anybody: For simplification, we'll similarly assume you always pick door 3. There are 6 total cases that can happen: Case 1 - Door 1 has the prize and door 1 got revealed Case 2 - Door 1 has the prize and door 2 got revealed Case 3 - Door 2 has the prize and door 1 got revealed Case 4 - Door 2 has the prize and door 2 got revealed Case 5 - Door 3 has the prize and door 1 got revealed Case 6 - Door 3 has the prize and door 2 got revealed In the non-Monty Hall situation where the door that gets revealed is always random and it's possible to reveal the prize, all 6 cases are equally likely, specifically with 1/6 probability each. There's only a 2/3 chance you get to the 1 prize/1 non-prize situation, BUT if you did, it is a 50/50. Because there were 4 equally likely cases that get you to that point, and in 2 of them, you started with the prize. In the Monty Hall situation where a non-prize is always revealed, you instead only have 4 possible cases (case 1 and case 4 are no longer possible). HOWEVER, they are also no longer equally likely. If door 1 has the prize, you will ALWAYS get case 2. If door 2 has the prize, you will ALWAYS get case 3. If door 3 has the prize, you can get EITHER case 5 or case 6. Meaning cases 2 and 3 have probability 1/3, but cases 5 and 6 only have probability 1/6. Because of that, comparing the probabilities now gives you a different result that you are twice as likely to land on the prize by switching. Fun side note: This means Kissboy's code is not actually a true Monty Hall. As written, if door 3 has the prize, you will ALWAYS get case 5. A vigilant observer could notice that if door 2 gets revealed, you will win 100% of the time by switching, and if door 1 gets revealed, it is a true 50/50. It still ultimately results in the same 2/3 probability of winning by always switching, but in this case a strategy of "stay if door 1 is revealed and switch if door 2 is revealed" ALSO wins 2/3 of the time.
Delightful Kissboy back in the game with the (literally) mindblowing thumbnail. Delightful Kissboy is all you need in life. I have no credit and no money in my bank account. My wife left me and took the kids but it doesn't matter. I have my Buckshot Roulette videos and no one can take that away. Bless you Delightful Kissboy, your vids are amazing.
this is the first time I actually understood the Monty Hall problem. Like people were always like "yeah you should switch" and I always asked "why" and the only answer I got was "bc its better" or some explanation that I was just like "nope thats not it".
This reminds me of that game show called Deal or No Deal. There are like 25 cases with money in it that ranges from $1 to one million and the contestant picks one and gets offered money from the “Dealer” based on how likely the case they picked has the million dollars. If they make it to the end, they can choose to swap their case with the last one and I always thought that sticking with the case you picked was dumb because even though there now is a 50/50 chance to win big, it was a 1/25 chance at the start so it still would be more likely that you didn’t pick something good. I always think about that.
The chance of multiplayer survival is very similar to the Liar's Bar game if you have a rank (in that game it shows) you are at a 50/50 chance, you'd probably be fine in Buckshot if you only steal and shoot the one that just had a turn or alternatively steal anyone's player jammer and use it on the one that just shot before you. I do adore Monopoly and you can trade there with 100% of your wit and figuring out the odds is just so exhilerating you wouldn't believe. Having 3v1 is bad, shooting the one that shot you, or someone else, does a great deal in building trust.
new buckshot item: tiny goat trinket. tells you the location of a randomly chosen blank shell. no restrictions on which shell it can be, including the current one.
fun fact: the way buckshot roullete shows you the last shell regardless of if it wins or lose is nearly identical to the way it was proposed that Deal or No Deal should resolve the monty hall problem when it aired for a second time. essentially, the house should always open the leftmost of the remaining doors, regardless of content. deal or no deal instead just made their show more complex, with more doors and mini prizes you could take instead of risking opening doors, and all sorts of silly things like that which complicate things and confuse contestants. however, there are some episodes where they reveal a door only to reveal the grand prize, and the contestant is left with the bittersweet task of trying to determine whether the door they have is worthless or is a lesser prize.
The best way I've ever seen this problem integrated into a video game is a level in the game" Zero Escape: Zero Time Dilemma". It really helps you visualize the problem. Dont listen to people that say it's a bad game, the whole series is great for nerds who like puzzles. (Edit: And yet my dumbass initially thought the problem *was* represented by the burner phone, but I think you convinced me it's not, the phone is not discriminating which shell)
The most intuitive explanation i’ve used for the monty hall problem is just that imagine if there were 50 doors instead of 3 doors. Imagine they opened every single other door but the one you picked and one other door. Would you switch?
I came up with a rephrasing of the Monty Hall problem that makes it more obvious: There are three boxes of different weights. To win you need to pick the heaviest box. You pick box A. The host then places box B and box C on a scale and you see that box C is heavier than box B. Should you switch your choice to box C?
One time when playing Multiplayer I was 1v1ing someone, there was 1 live 2 blanks, I had an inverter and no way to kill, and the other guy was basically guaranteed to kill me if I let him have the gun. I figured the only way to win was to shoot myself 3 times in a row so I can get better items. This led to a predicament where I essentially had to guess where the live was and invert it so it turns into a blank. I actually managed to clutch that shit, but it made me wonder if it actually matters which shell you actually invert. The answer is essentially no. I tested this in a simulation but mathematically the reason it doesn't matter is because while the further you go into the chamber the more likely the next shell is to be the live you can only actually get deeper into the chamber if all of the previous shells were blanks. So while saving the inverter until the third shell technically makes it guaranteed you will invert the live, that's only in cases where the first bullet isn't a live (2/3) and the second bullet isn't a live when you actually get to it (1/2). (2/3) * (1/2) * 1 = .33 or a 33% chance of getting to the last bullet and having it be live. Hope this made sense lol.
Makes sense to me, very cool! There are a lot of scenarios like this I've found and have been tricked by on occasion, where it feels like you have control over the likelihood of an outcome, but actually don't. When in doubt, code it out 👌
Didn't even realize I was doing that lol. I think it's bc of the code I've been writing recently. I treat the shell in position 0 of an array as the last shell so I can count down toward 0, just found that way of implementing it easier. So I think it's engrained in my psyche rn lol
*Objection!* I made my own program to test whether it was still relevant even with the phone occassionally selecting the right answer. After testing every single possibility.. It said that it gave the same average whether you changed your answer or not. Excuse me while I sit in a corner of shame.
So if there are 3 doors. Door A, B, and C. Me and my buddy can pick any door, even the same door. The host knows which one we each picked, but my buddy and I won't know what the other picked. I pick door A, my buddy picks B. The host then removes door C, because it did not have a prize. We are then asked if we want to change doors. We both would have a 60% chance of being right if we changed doors?
Interesting question. 1/3 of the time, your bud will pick the winning door. Given that, 1/3 of the time you will pick the winning door too, 2/3 you pick a losing door. So 1/3 of the time, switching will win 2/3 of the time, overall odds 2/9. 2/3 of the time, your bud will pick a losing door. Given that, 1/3 of the time you will pick the winning door, 1/3 of the time you will pick the same losing door as your bud, 1/3 of the time you will pick a diff losing door than your bud. In the last case, the scenario becomes invalid bc there is no losing door for the host to eliminate, so we remove this as a possibility, meaning there's a 50% chance you pick the winning door and a 50% chance you pick the same losing door as your bud given they picked a losing door. So 2/3 of the time, it's a 50/50 winning or losing by switching, overall chance 1/3. That means by switching, from our perspective, we have a 2/9 + 1/3 = 5/9 (55%) of winning, and so does our friend from their perspective. Obviously in your scenario, one player will lose and one will win, but given the hidden info, each from their own perspective is slightly more likely to win by switching.
thankyou for this video! i was discussing this actually with a friend, she thought it was monty hall - i thought it wasnt. however i couldnt figure out how to prove it wasnt.
I haven't watched the full video yet, but I remember learning about this years ago! I just forgot what it ends up being... Am I still a clown even though I knew it wasn't a 50/50..? Love your videos by the way! Very entertaining! And an oddly comforting voice.
I feel like this game in itself is a Monty Hall problem since every turn you should shoot the dealer or in multiplayer the person ahead of you. Speaking of which when you gonna do an analysis guide for multiplayer?
the code isn't exactly what you described, as there is a specific rule how to choose the opening door, so if the second door is opened, you know that it is in the first one, because else the first one would be opened. when the first is opened, it is a 50/50 case. the reason why the first explanation, that you said was wrong doesn't work, is because the 3 had only a half chance of choosing that door, not a full.
True, although the sim doesn't use that knowledge, so the numbers are still accurate. But right, should choose a random door in the case that the picked door is the winner
You're right, this was an example of how to look at the problem incorrectly and conclude it's still a 50/50. It can seem like scenario 1 is impossible bc the host revealed there's no prize behind the left door in my example
On FB a while ago, you could put in a custom pronounciation for ur name and a lil robot voice would read it out loud if you clicked it. For some reason I thought it would be funny to set mine to say "Delightful Kissboy" in the silly voice, and it made me laugh, so I kept using it as a username lol
below is a copy paste of my reply to a past comment on 1 of your videos claiming there's Monty Hall: I can see the analogy you're making, but you have an incomplete understanding of the Monty Hall problem as per your analogy, let's talk in terms of 1 live 2 blank (1 car 2 sheep), your "choice" is the chamber, and the phone reveals a blank (the host reveals a sheep). it's almost a perfect analogy, but as you yourself said, the phone can tell you where the odd shell out is. you think it doesn't affect the math, but it actually makes all the difference. this is the key detail that makes the Monty Hall problem the infamous Monty Hall problem. more specifically, the host always shows a sheep, and if the contestant chose the car, the host shows 1 of the 2 sheep at random (let's call this 'normal Monty Hall'). this is mathematically different from the host opening 1 of the 2 others doors at random regardless of Car/sheep (let's call this 'random Monty Hall') to show you the difference, let's first re-examine normal Monty Hall. let's run 6 simulations. because (C)ar/(s)heep placement is random, they go: Css, Css, sCs, sCs, ssC, ssC. without loss of generality, let's say the contestant chooses the leftmost door 1. following the rules of the show, the host opens 1 of the 2 other doors (represented w/ parentheses): C(s)s, Cs(s), sC(s), sC(s), s(s)C, s(s)C. now, the above are just hypothetical simulations. when Monty Hall is simulated in real life, we can observe and use information to speculate on the true nature of each particular simulation. for example, let's say Monty Hall starts airing again in 2025. a contestant chooses the leftmost door 1, and the host proceeds to open the middle door 2 with a sheep behind it. this information proves to us that this current simulation isn't Cs(s) or sC(s), so it can be C(s)s or s(s)C. we knows that simulations go s(s)C twice as often as C(s)s, because half of Css simulations go Cs(s). therefore we speculate there's a 2/3 chance this simulation is s(s)C, so the contestant should switch now, let's examine random Monty Hall. let's run 6 simulations: Css, Css, sCs, sCs, ssC, ssC. without loss of generality, let's say the contestant chooses the leftmost door 1. the host opens 1 of the 2 other doors at random regardless of Car/sheep: C(s)s, Cs(s), s(C)s, sC(s), s(s)C, ss(C). let's say random Monty Hall starts airing. a contestant chooses the leftmost door 1, and the host proceeds to open the middle door 2 revealing a sheep behind it. this information proves to us that this current simulation isn't Cs(s), s(C)s sC(s), or ss(C), so it can be C(s)s or s(s)C. different from the Monty Hall problem, simulations go either way just as often. half of Css simulations still go Cs(s), but so do half of ssC simulations go ss(C). therefore we speculate there's a 1/2 chance for either, and switching isn't better I hope you understand now
The physical position of the door isn't what's important to the statistics. it's the simple fact that the host will always open a door that doesn't have the prize. Since there is only a 1/3rd chance you initially picked the correct door, there is a 2/3rd chance the remaining door is correct. Simple as that.
@ we are in complete agreement: physical position is irrelevant and the system is symmetric; I take advantage of this fact to run simulations without loss of generality
I initially thought your comment was going to be a failed refutation of kissboys video until I read the entire thing and realized I misread the intro. eh whoops.
This is why I don't think the general public can use statistics, if someone convinces you using statistics alone step back and rethink. Statistics are nothing but manipulation tools presented as fact when they lack full context. Simple things like being able to see what is behind 1 door at the start completely differentiates things just like how paying 2$ for a candy and 1,000$ for a TV are both paying 100% of their current price. You're both paying 100% but you're not paying the same amount.
Ok you explained the monty hall problem to me and I dislike it even more now. It sounded illogical before and now i hear the only thing making it work is your initial door not being checked for being a false or true. I could compare it to paying the only hooker not tested for stds.
Here is why it works: if your first guess was incorrect (2/3), then the host will always remove the remaining incorrect door. This leaves only the correct door. In this scenario, switching will give you the win. If your first guess was correct (1/3), then the host will remove a random door, and you will switch to an incorrect one. So what’s essentially happening is that, without switching, you would have a 2/3 chance to be wrong and a 1/3 chance to be right. Switching doors also switches these odds, so now you have a 2/3 chance to be right and a 1/3 chance to be wrong. In other words: if you stay with your choice, you are betting that your choice was correct. If you switch your choice, you are betting that your first choice was INcorrect (which is a 66% chance).
Imagine 100 doors with one prize. You pick a door, the trolling quiz master closes 98 doors and offers you switch to the remaining door.
Upping the door count to absurd quantities is definitely the best way of resolving the "but it's a 50/50!" reaction, and that switching is identical to having been allowed to choose 99 doors at the start instead of 1.
I think it would be more troll if 99 doors contained the cash prize. And they offered you a dollar to change your choice.
Ah I get it now. Like I got it before but now it's like completely understood
@@ShirubaGin that's the example I use to explain to people. They still end up saying it's 50/50 when we get to 3 doors...
so cool that people are already posting “corrections” before it’s even possible to have watched the whole video and listened to what you had to say in the first place
I think they were more trying to predict the conclusion 👌
@@delightfulkissboy8926
yes I was; I had absolute 100% confidence that our beloved DKB would arrive at the correct conclusion just as I had, hence my comment with my maths reaching the same conclusion
@@delightfulkissboy8926 Fair enough, just wish they’d actually hear out your points before insinuating that there’s gaps in your analysis lol
@@KasioGamesseems like you're saying what was in their comments before reading them all
@@FFKonoko
CHA-RUE
I'm a stats prof, and I love your videos. You do a great job teaching probability concepts. This one is so great for the very subtle difference between Monty Hall and the Buckshot situation. You laid it out wonderfully!
Thank you! Really appreciate that
God I'd love to watch that lecture
I'm in an intro college mathematicsand you made this way more interesting than that 🎉 @@delightfulkissboy8926
I did not believe the Monty Hall problem was anything but an edge case psychology thing until your "smushed" visual section. Thank you. I've known about it for years and you're the first person to explain it in a way I could understand.
Great to hear!
Delightful explanation as always! The Naive Monty Hall Problem interpretation is sometimes called the "Monty Fall Problem", where Monty trips and falls, revealing a random door, then giving you the option to switch.
Didn't know that, that's great lol
delightful explanation for a delightful kissboy
The tl;dr version of the Monty Hall problem is because you had a 1/3 chance of choosing the winning door to begin with, switching means you have a 2/3 chance of winning, because with these conditions it's essentially the same as betting against yourself.
I think the biggest challenge with the Monty Hall problem is successfully explaining the Monty Hall problem to someone who doesn't understand it. You showed a pretty good explanation here, and I'm going to share this video with whoever I'm trying to teach.
Assuming this is about the magnifying glas... The Monty Hall requires the "prize door" to be guaranteed not to be shown ergo making it a true 67% chance on swap.
Genuinely carrying me through probability this semester thanks dkb
Appreciate the new buckshot thought experiment video Sensational SmoochMan 🙏
One of the only youtubers to make powerpoint presentations engaging
I paused this video a couple minutes in to code up a simulation in python of the monty hall problem just to remind myself that there is a god out there messing with us for the fun of it, only to resume the video & learn that a coding experiment was already thought of & included
Delightful Vid Boy!
smart
I recently got into coding not long ago and this gave me a bit of motivation to continue learning more about it through your explanation tysm you are truly delightful!!! :D
Hell yeah, keep it up
one could even say he is a kissboy
Video Spoiler Ahead:
To any not understanding how this situation differs from the Monty Hall problem(but do have an understanding of why you should switch in the Monty Hall problem) I may be able to help.
In the Monty Hall problem each initial door has a 1/3 chance of being the winner, after opening a dud, your initial choice remains a 1/3 chance but the probability of the remaining door(switching) now has 2/3rds odds of being the winner. However in this buckshot roulette comparison the initial door starts at 1/3 but because the opened door is fixed(and can be revealed to be the winner) the probabilities of each remaining door update when it is opened. The probability only stays locked at 1/3 in the Monty Hall problem because the revealed doors are guaranteed to be duds. Imagine you play Monty Hall again but now with 10 doors(still only 1 winner), you select one and the host reveals eight duds among the remaining doors. Your initial door still has a 1/10 chance of being the winner because the surviving door takes the odds of the other 8 doors(now carrying a 9/10 chance of being the winner). Why do the odds transfer in Monty Hall but not in buckshot roulette? Because the doors opened are guaranteed to be duds thus not changing the probability of the initial pick.
The initial selection is excluded from the dud removal(thus keeping it's initial odds) but the probability of a winning door is still split amongst all doors(and thus transfered to any unopened AND unselected doors).
As an aside in a Monty Hall problem with 5 doors, if you pick 1 and the host opens 2 duds, your initial door is still 1/5, the two opened doors are of course now 0, and the 2 unselected unopened doors each have a 2/5 chance of being winners, thus you should switch to one of them.
"Indexing starts at 0"
Lua: hold my 1
was gonna comment the same thing lol. lua is so weird and quirky but we love it
im coining the burner phone variation you described in the end where the burner phone can just tell you the live round as the delightful kissboy problem see yall in 25 years
I’ve never truly understood or believed the Monty Hall problem before watching your video, words can’t express how thankful I am for your explanation. You’re great dude, keep at it!
Really great video, made me understand Monty Hall problem much better. I do think burner phone that shows a blank would be a really cool item.
An alternate way of explaining the subtle difference, in case this helps anybody:
For simplification, we'll similarly assume you always pick door 3. There are 6 total cases that can happen:
Case 1 - Door 1 has the prize and door 1 got revealed
Case 2 - Door 1 has the prize and door 2 got revealed
Case 3 - Door 2 has the prize and door 1 got revealed
Case 4 - Door 2 has the prize and door 2 got revealed
Case 5 - Door 3 has the prize and door 1 got revealed
Case 6 - Door 3 has the prize and door 2 got revealed
In the non-Monty Hall situation where the door that gets revealed is always random and it's possible to reveal the prize, all 6 cases are equally likely, specifically with 1/6 probability each. There's only a 2/3 chance you get to the 1 prize/1 non-prize situation, BUT if you did, it is a 50/50. Because there were 4 equally likely cases that get you to that point, and in 2 of them, you started with the prize.
In the Monty Hall situation where a non-prize is always revealed, you instead only have 4 possible cases (case 1 and case 4 are no longer possible). HOWEVER, they are also no longer equally likely. If door 1 has the prize, you will ALWAYS get case 2. If door 2 has the prize, you will ALWAYS get case 3. If door 3 has the prize, you can get EITHER case 5 or case 6. Meaning cases 2 and 3 have probability 1/3, but cases 5 and 6 only have probability 1/6. Because of that, comparing the probabilities now gives you a different result that you are twice as likely to land on the prize by switching.
Fun side note: This means Kissboy's code is not actually a true Monty Hall. As written, if door 3 has the prize, you will ALWAYS get case 5. A vigilant observer could notice that if door 2 gets revealed, you will win 100% of the time by switching, and if door 1 gets revealed, it is a true 50/50. It still ultimately results in the same 2/3 probability of winning by always switching, but in this case a strategy of "stay if door 1 is revealed and switch if door 2 is revealed" ALSO wins 2/3 of the time.
Delightful Kissboy back in the game with the (literally) mindblowing thumbnail. Delightful Kissboy is all you need in life. I have no credit and no money in my bank account. My wife left me and took the kids but it doesn't matter. I have my Buckshot Roulette videos and no one can take that away. Bless you Delightful Kissboy, your vids are amazing.
this is the first time I actually understood the Monty Hall problem. Like people were always like "yeah you should switch" and I always asked "why" and the only answer I got was "bc its better" or some explanation that I was just like "nope thats not it".
i was just watching the powerball video and this pops up after i finish it 💜
This actually made the problem make way more sense to me. I appreciate it! You're just a great teacher
Another delightful video from our kissboy ❤
This reminds me of that game show called Deal or No Deal. There are like 25 cases with money in it that ranges from $1 to one million and the contestant picks one and gets offered money from the “Dealer” based on how likely the case they picked has the million dollars.
If they make it to the end, they can choose to swap their case with the last one and I always thought that sticking with the case you picked was dumb because even though there now is a 50/50 chance to win big, it was a 1/25 chance at the start so it still would be more likely that you didn’t pick something good.
I always think about that.
I blame Zero Escape games for making me obsessed with topics like these, good video
In the process of learning Godot and GDscript just to comprehend how stupid the Dealer is🔥
love this guy
Accurate username as always sir
The chance of multiplayer survival is very similar to the Liar's Bar game if you have a rank (in that game it shows) you are at a 50/50 chance, you'd probably be fine in Buckshot if you only steal and shoot the one that just had a turn or alternatively steal anyone's player jammer and use it on the one that just shot before you. I do adore Monopoly and you can trade there with 100% of your wit and figuring out the odds is just so exhilerating you wouldn't believe. Having 3v1 is bad, shooting the one that shot you, or someone else, does a great deal in building trust.
your videos are so good i always forget you're still a "small" UA-camr, keep that shit up DKB
new buckshot item: tiny goat trinket. tells you the location of a randomly chosen blank shell. no restrictions on which shell it can be, including the current one.
fun fact: the way buckshot roullete shows you the last shell regardless of if it wins or lose is nearly identical to the way it was proposed that Deal or No Deal should resolve the monty hall problem when it aired for a second time. essentially, the house should always open the leftmost of the remaining doors, regardless of content.
deal or no deal instead just made their show more complex, with more doors and mini prizes you could take instead of risking opening doors, and all sorts of silly things like that which complicate things and confuse contestants. however, there are some episodes where they reveal a door only to reveal the grand prize, and the contestant is left with the bittersweet task of trying to determine whether the door they have is worthless or is a lesser prize.
this video was an interesting date night
The best way I've ever seen this problem integrated into a video game is a level in the game" Zero Escape: Zero Time Dilemma". It really helps you visualize the problem. Dont listen to people that say it's a bad game, the whole series is great for nerds who like puzzles. (Edit: And yet my dumbass initially thought the problem *was* represented by the burner phone, but I think you convinced me it's not, the phone is not discriminating which shell)
The most intuitive explanation i’ve used for the monty hall problem is just that imagine if there were 50 doors instead of 3 doors. Imagine they opened every single other door but the one you picked and one other door. Would you switch?
I came up with a rephrasing of the Monty Hall problem that makes it more obvious:
There are three boxes of different weights. To win you need to pick the heaviest box.
You pick box A.
The host then places box B and box C on a scale and you see that box C is heavier than box B.
Should you switch your choice to box C?
4:40 gave me the same, slightly unhinged feeling as the Click
Professor Kissboy! Thank you for today's lecture! I'm studying up I promise
ah so it is 50/50, I was like "oh no wait is this just monty hall" but in that it specifically just reveals duds not any door
I missed hearing delightful kissboys voice
now we need a monty hall item
This is my new favorite channel
That guy looks like a creepypasta Version of one of the guest faces from roller coaster tycoon
😂
One time when playing Multiplayer I was 1v1ing someone, there was 1 live 2 blanks, I had an inverter and no way to kill, and the other guy was basically guaranteed to kill me if I let him have the gun. I figured the only way to win was to shoot myself 3 times in a row so I can get better items.
This led to a predicament where I essentially had to guess where the live was and invert it so it turns into a blank. I actually managed to clutch that shit, but it made me wonder if it actually matters which shell you actually invert.
The answer is essentially no. I tested this in a simulation but mathematically the reason it doesn't matter is because while the further you go into the chamber the more likely the next shell is to be the live you can only actually get deeper into the chamber if all of the previous shells were blanks.
So while saving the inverter until the third shell technically makes it guaranteed you will invert the live, that's only in cases where the first bullet isn't a live (2/3) and the second bullet isn't a live when you actually get to it (1/2).
(2/3) * (1/2) * 1 = .33 or a 33% chance of getting to the last bullet and having it be live.
Hope this made sense lol.
Makes sense to me, very cool! There are a lot of scenarios like this I've found and have been tricked by on occasion, where it feels like you have control over the likelihood of an outcome, but actually don't. When in doubt, code it out 👌
Thank you delightful kissboy:)
I find it curious how you count from right to left. Your last door or last shell is the leftmost one (10:43) (4:20)
Didn't even realize I was doing that lol. I think it's bc of the code I've been writing recently. I treat the shell in position 0 of an array as the last shell so I can count down toward 0, just found that way of implementing it easier. So I think it's engrained in my psyche rn lol
*Objection!*
I made my own program to test whether it was still relevant even with the phone occassionally selecting the right answer.
After testing every single possibility..
It said that it gave the same average whether you changed your answer or not. Excuse me while I sit in a corner of shame.
Heeey DKB it's been a while, love your content cheers
3:30 a little harsh but true
How would the game strategy change if you and the dealer swapped who goes first each round or if it was a 50/50 for who goes first each round?
Us C# devs gotta stick together 🤜🤛
Finally watching this
So if there are 3 doors. Door A, B, and C. Me and my buddy can pick any door, even the same door. The host knows which one we each picked, but my buddy and I won't know what the other picked.
I pick door A, my buddy picks B. The host then removes door C, because it did not have a prize. We are then asked if we want to change doors.
We both would have a 60% chance of being right if we changed doors?
Interesting question.
1/3 of the time, your bud will pick the winning door. Given that, 1/3 of the time you will pick the winning door too, 2/3 you pick a losing door. So 1/3 of the time, switching will win 2/3 of the time, overall odds 2/9.
2/3 of the time, your bud will pick a losing door. Given that, 1/3 of the time you will pick the winning door, 1/3 of the time you will pick the same losing door as your bud, 1/3 of the time you will pick a diff losing door than your bud. In the last case, the scenario becomes invalid bc there is no losing door for the host to eliminate, so we remove this as a possibility, meaning there's a 50% chance you pick the winning door and a 50% chance you pick the same losing door as your bud given they picked a losing door. So 2/3 of the time, it's a 50/50 winning or losing by switching, overall chance 1/3.
That means by switching, from our perspective, we have a 2/9 + 1/3 = 5/9 (55%) of winning, and so does our friend from their perspective. Obviously in your scenario, one player will lose and one will win, but given the hidden info, each from their own perspective is slightly more likely to win by switching.
ah yes my favourite houses in hogwarts
the Winningdoor and the Losingdoor
Whenever I hear your name I can’t help but think that you’re the boy kisser meme
thankyou for this video! i was discussing this actually with a friend, she thought it was monty hall - i thought it wasnt. however i couldnt figure out how to prove it wasnt.
You should teach a statistics class
nice my man kissboy
I haven't watched the full video yet, but I remember learning about this years ago! I just forgot what it ends up being... Am I still a clown even though I knew it wasn't a 50/50..?
Love your videos by the way! Very entertaining! And an oddly comforting voice.
Nope, you passed the clown test 👌
Not the Monty Hall Problem, Buckshot Roulette is essentially "Nim 2"
I feel like this game in itself is a Monty Hall problem since every turn you should shoot the dealer or in multiplayer the person ahead of you. Speaking of which when you gonna do an analysis guide for multiplayer?
Soon! Been writing a bunch of code for it, so it's been takin a bit
@delightfulkissboy8926 nice; I can only imagine how brutal it would be for the last player or how items is gonna affect that turnout.
DKB can i get a smooch
💋
the code isn't exactly what you described, as there is a specific rule how to choose the opening door, so if the second door is opened, you know that it is in the first one, because else the first one would be opened.
when the first is opened, it is a 50/50 case. the reason why the first explanation, that you said was wrong doesn't work, is because the 3 had only a half chance of choosing that door, not a full.
True, although the sim doesn't use that knowledge, so the numbers are still accurate. But right, should choose a random door in the case that the picked door is the winner
2:45 why is scenario 1 impossible? wouldnt the host just open door 2 and say the same thing? there's no prize there
You're right, this was an example of how to look at the problem incorrectly and conclude it's still a 50/50. It can seem like scenario 1 is impossible bc the host revealed there's no prize behind the left door in my example
5:34 Is that a Hazbin Hotel reference?
Haven't seen that show, so probably not lol
hey kissboy, how'd you come up with your name?
On FB a while ago, you could put in a custom pronounciation for ur name and a lil robot voice would read it out loud if you clicked it. For some reason I thought it would be funny to set mine to say "Delightful Kissboy" in the silly voice, and it made me laugh, so I kept using it as a username lol
4:39 hey dkb what the fuck 🤨
😫👌
Brr skibidi dom dom dom yes yes
below is a copy paste of my reply to a past comment on 1 of your videos claiming there's Monty Hall:
I can see the analogy you're making, but you have an incomplete understanding of the Monty Hall problem
as per your analogy, let's talk in terms of 1 live 2 blank (1 car 2 sheep), your "choice" is the chamber, and the phone reveals a blank (the host reveals a sheep). it's almost a perfect analogy, but as you yourself said, the phone can tell you where the odd shell out is. you think it doesn't affect the math, but it actually makes all the difference. this is the key detail that makes the Monty Hall problem the infamous Monty Hall problem. more specifically, the host always shows a sheep, and if the contestant chose the car, the host shows 1 of the 2 sheep at random (let's call this 'normal Monty Hall'). this is mathematically different from the host opening 1 of the 2 others doors at random regardless of Car/sheep (let's call this 'random Monty Hall')
to show you the difference, let's first re-examine normal Monty Hall. let's run 6 simulations. because (C)ar/(s)heep placement is random, they go: Css, Css, sCs, sCs, ssC, ssC. without loss of generality, let's say the contestant chooses the leftmost door 1. following the rules of the show, the host opens 1 of the 2 other doors (represented w/ parentheses): C(s)s, Cs(s), sC(s), sC(s), s(s)C, s(s)C. now, the above are just hypothetical simulations. when Monty Hall is simulated in real life, we can observe and use information to speculate on the true nature of each particular simulation. for example, let's say Monty Hall starts airing again in 2025. a contestant chooses the leftmost door 1, and the host proceeds to open the middle door 2 with a sheep behind it. this information proves to us that this current simulation isn't Cs(s) or sC(s), so it can be C(s)s or s(s)C. we knows that simulations go s(s)C twice as often as C(s)s, because half of Css simulations go Cs(s). therefore we speculate there's a 2/3 chance this simulation is s(s)C, so the contestant should switch
now, let's examine random Monty Hall. let's run 6 simulations: Css, Css, sCs, sCs, ssC, ssC. without loss of generality, let's say the contestant chooses the leftmost door 1. the host opens 1 of the 2 other doors at random regardless of Car/sheep: C(s)s, Cs(s), s(C)s, sC(s), s(s)C, ss(C). let's say random Monty Hall starts airing. a contestant chooses the leftmost door 1, and the host proceeds to open the middle door 2 revealing a sheep behind it. this information proves to us that this current simulation isn't Cs(s), s(C)s sC(s), or ss(C), so it can be C(s)s or s(s)C. different from the Monty Hall problem, simulations go either way just as often. half of Css simulations still go Cs(s), but so do half of ssC simulations go ss(C). therefore we speculate there's a 1/2 chance for either, and switching isn't better
I hope you understand now
The physical position of the door isn't what's important to the statistics. it's the simple fact that the host will always open a door that doesn't have the prize. Since there is only a 1/3rd chance you initially picked the correct door, there is a 2/3rd chance the remaining door is correct. Simple as that.
@
we are in complete agreement: physical position is irrelevant and the system is symmetric; I take advantage of this fact to run simulations without loss of generality
way to agree with our dearest delightful kissboy
I initially thought your comment was going to be a failed refutation of kissboys video until I read the entire thing and realized I misread the intro. eh whoops.
@@karlneff
seems a lot of ppl thought my comment was "correcting" him. we are in agreement, but we just use different maths to show it
You really don't want to give us multiplayer, do you... Just say it. Say you don't want to play multiplayer and aren't going to record it for us.
Bruh it ain't that deep, just been more motivated to do coding projects for now. Just doin what I wanna do in the moment 👌
This is why I don't think the general public can use statistics, if someone convinces you using statistics alone step back and rethink. Statistics are nothing but manipulation tools presented as fact when they lack full context.
Simple things like being able to see what is behind 1 door at the start completely differentiates things just like how paying 2$ for a candy and 1,000$ for a TV are both paying 100% of their current price. You're both paying 100% but you're not paying the same amount.
Ok you explained the monty hall problem to me and I dislike it even more now. It sounded illogical before and now i hear the only thing making it work is your initial door not being checked for being a false or true. I could compare it to paying the only hooker not tested for stds.
Here is why it works: if your first guess was incorrect (2/3), then the host will always remove the remaining incorrect door. This leaves only the correct door. In this scenario, switching will give you the win. If your first guess was correct (1/3), then the host will remove a random door, and you will switch to an incorrect one. So what’s essentially happening is that, without switching, you would have a 2/3 chance to be wrong and a 1/3 chance to be right. Switching doors also switches these odds, so now you have a 2/3 chance to be right and a 1/3 chance to be wrong. In other words: if you stay with your choice, you are betting that your choice was correct. If you switch your choice, you are betting that your first choice was INcorrect (which is a 66% chance).
Holy! It’s a Kissboy vid! 🩵