The buckshot version is like if you had a really dumb quiz master where he just opens one of the other doors at random and sometimes reveals the cash prize.
What's so dumb about it? If he reveals the car it means the candidate just lost and the house can keep the car. Instead of losing the car with 2/3 probability.
EDIT: I know it said he closes doors, but i obviously meant opens, so can we just... ok? Imagine 100 doors with one prize. You pick a door, the trolling quiz master opens* 98 doors and offers you a 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'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.
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.
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
That "smush" visualization is something a lot more people should use, because that's actually how it works. The "reveal" phase essentially collapses all doors into two sets; Set 1 is the doors you picked (this set has one door). Set 2 has all unpicked doors in it. The 'reveal' phase collapses all of set 2 into one door.
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.
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.
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.
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.
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 👌
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’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!
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.
What helps me to remember how it works is thinking of how the initial selection is 1/3 likely to be correct and 2/3 likely to be incorrect. Even if one of the doors is revealed to be false, the odds of the initial choice never changed, and as such the other unselected door remaining is more likely to be correct.
There's one key difference between Monty Hall and Buckshot. The burner phone could've told us the last shell is a live, had that been the case. Monty Hall *NEVER* opens the winning door.
What made the problem easy to understand for me is thinking of it in term of information. Monty has more information than you, he knows where the correct door is, and he is giving you information because he relies on this information to make the choice of which door to open. If he opened a door at random then switching or not wouldn't matter but you'd end up in a situation that's sometimes unwinnable because the prize door might be opened so you get to switch between two losing door. But because that situation can't happen because the door isn't opened at random it gives you information. Initially you had 1/3 chances to find the price and 2/3 to hit an empty door, if you hit an empty door then monty eliminates the incorrect remaining door for you and you get to switch and get the prze. If you find the correct door then he chooses at random and if you switch you lose. If you think about it like that it's very clear you have 2/3 chances to win if you switch because missing the correct door initially guarantees a win if you switch. so the odds are simply reversed to what they initially were.
*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.
Okay i never understood this entire "problem" to begin with when people used to try to explain it to me, even reading the wikipedia page only made me more confused about it. But you explained it perfectly in like 6 minutes , you're really good at explaining and teaching things.
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.
You are really good at explaining these concepts. You should consider cutting just the Monty hall section for better visibility. Also, it would have been cool to see a simulation of the buckshot version of the “Monty hall” problem.
Genuinely thought about this the first day the multiplayer went into beta. Made a scruffy python code with the assumption of "1 live, 2 blanks, a phone, 2 beers for "switching" rounds" and found that it's a 50/50. Was a very interesting afternoon trying to figure this out in Paint first and then resorted to coding.
Thinking about how it works. It's kinda like this. You have the 3 possible scenarios. You've picked the door on the right initially, and then the host always picks the door on the left, regardless of whether it has the bullet or not. 1 0 0 0 1 0 0 0 1 In the first scenario, you know what to do, so this scenario is discarded since the chances you pick the correct door are irrelevant since you know which door is correct. So it then becomes: 0 1 0 0 0 1 Then if we do the hand smoosh (discarding the doors on the left) like in the video it becomes: 1 0 0 1 It's now a 50-50 as to whether you picked the correct door and it doesn't really matter if we switch or not.
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?
Delightful Kissboy, you are the cool uncle version of 3Blue 1 Brown. I'm dead serious when I say you make educational content that I would be happy to share with highschool students (or interested adults) to trick them into learning about statistics, probability, and programing.
finally someone represents the monty hall problem with code, I've had trouble believing it for what feels like years since it feels like they are seperated events of choosing the 3 doors then the remaining 2.
Basically imagine that the host doesn't open a door himself but just offers you to either stick with the door you chose or open both doors that you didn't. Pretty obvious what you should do in this scenario.
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 just convinced my friend to play buckshot roulette today and now im convincing him to watch your vids on it. gotta infect him with the Hyperfixation :))
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.
TL:DR the buckshot version acts as if you have to choose a door, but the host opens one door before you lock in anything. It just either eliminates that door as an option or gives it away as the correct answer, and the fact that it can give away the answer is the difference maker from Monty Hall, 2/3rds of the time it reduces things to a 50/50 and 1/3rd of the time it hands you the answer.
One other difference with the monty hall problem is that with the buckshot roulette thing, instead of having monty open a dud door, if he opens the prize door you automatically win, so you actually do get better odds but for different reasons. By default you have 2/3 chances to hit a blank. But if the live is revealed you can hit the opponent with the live 100% of the time. And if a dud is revealed you have 50/50 to hit the opponent with the live. That means you do hit the opponent 2/3 of the time. But only because in this scenario monty lets you switch to a door that was opened, which isn't in the original problem.
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
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.
But the rest of the cases were eliminated arbitrarily, so it should be equivalent to the naive version. You could instead imagine setting aside 2 boxes at the start before revealing the other 23. It seems obvious here that neither case is more likely to be better than the other, and that switching makes no difference. The difference between this one and Monty Hall is that we chose both cases independently. With Monty Hall the door that is left DOES depend on your initial choice, which is why one door can be "better" than the other. Your reasoning holds if, after selecting a box, you're given a second box and told that one of them holds a million. In 1/25 cases your initial box has a million and switching loses you money, but you gain value in every other scenario.
When I played against my girlfriend and had the "two blank, one live situation" and a phone I immediately decided on firing the first, used the phone, then switched to the one not mentioned. I immediately recognized the similarity to the monty hall problem.
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?
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.
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
Challenge: Write a script/bot that plays the game in the most efficient way possible. It takes into consideration everything - your lives, the dealers lives, all items currently on the table, amount of shells currently in the gun. It should make the best choices possible, it has to know when its better to shoot itself rather than the dealer and vice versa. Basically it should know most of the possible combinations and outcomes in the game. It would definitely be a cool concept
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.
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
The fun thing about the Monty hall problem is that even if you don’t know if the person is specifically choosing a door without the prize, or just chose one of the other doors at random (often times people when posing the question don’t properly state the first is correct), which would mean you don’t know if it’s a 66/33 or a 50/50, it’s better to switch. E.g lets say the host could be either and it’s 50/50 how the host runs it. Then the probability of winning if not switch is 50%*50% + 33%*50% = 42% or so. Conversely, the probability of winning if not switch is 50%*50% + 66%*50% = 58%
That would be true if you knew these are the only two options and will be applied with an equal probability. But in fact, there are infinitely many ways the host could play and if you have no idea whatsoever which is the correct one, then it will be a 50:50.
@@insignificantfool8592 they dont need to be with equal probability, though my logic does assume that it only happens in one of the two ways. Since those are the only two reasonable ways its set up, I'd say its a reasonable assumption. Also, which ways could the host play that would lead to not switching being better than switching? There has to be some way, otherwise it could not counterbalance and would therefore not be 50/50. Also, there are not infinitely many ways for the host to play at all - there are three possibilities for the prize, and three for the door you choose, so 9 possible states, and in each state the host chooses one of the other doors to open, so there could not possible be more than 2^9, or 512 possible ways for the host to play.
@mcpecommander5327 there could be various rules the host could use. You yourself acknowledge that there are two: 1.always open one of the other two doors with a goat and offer the switch (2/3 for switching) 2. open one of the other two doors randomly and offer the switch if it's a goat (1/2 for switching) Just two more would be: 3. Open door and offer switch only if chosen door was car door (0 for switching) 4. Open door and offer switch only if chosen door was goat door (1 for switching) And you could do infinitely many variants like 3 and 4, giving higher or lower likelihoods that you even offer the switch depending on whether a car or goat door was chosen.
@@insignificantfool8592 I assumed that they would always offer switch so 2 would be: open one of the other two doors randomly and offer the switch (no condition) Even with that, there would still only be a finite number of variants, just this time 4^9 possibilities
@mcpecommander5327 that seems weird to me. Does that mean you might be lucky and the car door gets opened and you're allowed to switch? I thought you had the idea that the offer to switch is not forced.
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)
In buckshot you don't have a 50% chance, but a 66% of winning whether you are switching or not! in brackets (gun enemy/gun yourself) 1/3 you chose live (win/lost) 1/3 you chose blank and last is live (no choice, always win) 1/3 you chose blank and last is blank (lost/win) So in 2/3 cases you win with any choice
so like because the cash prize has a 2/3 chance of being in a door you have not picked and the host is __going to pick a door without the prize,__ he is turning that 2/3 between the unpicked two doors into 2/3 on one door?
2:16 No, one of those is the door you opened so that scenario doesn't exist anymore. It's a 50/50, unless you misrepresent the situation. An easy example for this is: There are 3 doors before you, 1 is already open, what are your chances for getting the right one.
It is not equivalent to start with one door already opened. In such case, the host would have only had one restriction to fulfill: to not reveal the prize door. Any of the other two would have been free for him. In contrast, in the Monty Hall problem he has two restrictions: 1) to not reveal the door the player previously chose, and 2) to not reveal the prize door. Those two restrictions create a disparity, because if the player's choice is the same that contains the car, the host is still free to reveal any of the other two, making it uncertain which he will take in that case, but if the player's choice is a losing option, the host is limited to only one possible losing option to remove. For example, if you choose #1 and he opens #2, we know that the revelation of #2 was 100% mandatory in case the correct were #3, as the other two would have been prohibited. But if the correct were #1 (yours), it was only 50% likely that he would open #2, not 100% guaranteed as in the other case, because we have to deal with the possibility that he would have preferred to open #3 instead. That's what makes it twice as difficult that he opens #2 in a game that #1 is correct than in a game that #3 is correct (having you selected #1). But if you hadn't picked anything and he just opened #2 before you began, it would have been equally likely that he chose to remove #2 regardless of if the correct were #1 or #3, so no further information about a door over the other; both have the same chances.
minor thing, but I think Let's Make A Deal didn't actually work like that, the Monty Hall problem was invented by a statistician and they just used the TV show as context to set it up in
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.
The buckshot version is like if you had a really dumb quiz master where he just opens one of the other doors at random and sometimes reveals the cash prize.
😂😂 this comment is gold
"Oops, let me just close that... Anyways, you can't pick that door. No."
What's so dumb about it? If he reveals the car it means the candidate just lost and the house can keep the car. Instead of losing the car with 2/3 probability.
@@insignificantfool8592 Except in buckshot, if the phone reveals the live round, you didn't lose, you won.
EDIT: I know it said he closes doors, but i obviously meant opens, so can we just... ok?
Imagine 100 doors with one prize. You pick a door, the trolling quiz master opens* 98 doors and offers you a 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...
@@ninjaguyYT I think the people you hang out with might be slow
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
This stuff is so cool, I knew this was related to stats and prob
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!
This, I struggled to comprehend this concept for so long after being first introduced to it back in high schools
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.
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
That "smush" visualization is something a lot more people should use, because that's actually how it works. The "reveal" phase essentially collapses all doors into two sets; Set 1 is the doors you picked (this set has one door). Set 2 has all unpicked doors in it. The 'reveal' phase collapses all of set 2 into one door.
As a visual learner the smush also helped me understand easier. I appreciated it
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.
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.
Unironically one of the best explanations of the Monty Hall Problem on TH-cam. lol
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.
I blame Zero Escape games for making me obsessed with topics like these, good video
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've known about the monty hall problem for 10 years. i'm studying statistics in college right now. I FINALLY GET IT NOW THANK YOU DELIGHTFUL KISSBOY
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 👌
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 was literally like "that sounds like some cognitive bias, I wish there was a code to check it", and then immediately it was shown lol
Appreciate the new buckshot thought experiment video Sensational SmoochMan 🙏
One of the only youtubers to make powerpoint presentations engaging
Genuinely carrying me through probability this semester thanks dkb
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!
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.
What helps me to remember how it works is thinking of how the initial selection is 1/3 likely to be correct and 2/3 likely to be incorrect. Even if one of the doors is revealed to be false, the odds of the initial choice never changed, and as such the other unselected door remaining is more likely to be correct.
There's one key difference between Monty Hall and Buckshot.
The burner phone could've told us the last shell is a live, had that been the case.
Monty Hall *NEVER* opens the winning door.
I literally wondered this when I first played, glad to see this exact thing being addressed, thank you!
Delightful Vid Boy!
smart
What made the problem easy to understand for me is thinking of it in term of information. Monty has more information than you, he knows where the correct door is, and he is giving you information because he relies on this information to make the choice of which door to open.
If he opened a door at random then switching or not wouldn't matter but you'd end up in a situation that's sometimes unwinnable because the prize door might be opened so you get to switch between two losing door.
But because that situation can't happen because the door isn't opened at random it gives you information.
Initially you had 1/3 chances to find the price and 2/3 to hit an empty door, if you hit an empty door then monty eliminates the incorrect remaining door for you and you get to switch and get the prze. If you find the correct door then he chooses at random and if you switch you lose. If you think about it like that it's very clear you have 2/3 chances to win if you switch because missing the correct door initially guarantees a win if you switch. so the odds are simply reversed to what they initially were.
penn jillette explains monty hall problem and talks about video games
This was super satisfying mathematically thank you, it was quite delightful ❤
*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.
Okay i never understood this entire "problem" to begin with when people used to try to explain it to me, even reading the wikipedia page only made me more confused about it.
But you explained it perfectly in like 6 minutes , you're really good at explaining and teaching things.
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.
This actually made the problem make way more sense to me. I appreciate it! You're just a great teacher
That guy looks like a creepypasta Version of one of the guest faces from roller coaster tycoon
😂
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.
You are really good at explaining these concepts. You should consider cutting just the Monty hall section for better visibility. Also, it would have been cool to see a simulation of the buckshot version of the “Monty hall” problem.
How havent you blown up already??? your content is amazing, keep it up G
Genuinely thought about this the first day the multiplayer went into beta. Made a scruffy python code with the assumption of "1 live, 2 blanks, a phone, 2 beers for "switching" rounds" and found that it's a 50/50. Was a very interesting afternoon trying to figure this out in Paint first and then resorted to coding.
Thinking about how it works. It's kinda like this. You have the 3 possible scenarios. You've picked the door on the right initially, and then the host always picks the door on the left, regardless of whether it has the bullet or not.
1 0 0
0 1 0
0 0 1
In the first scenario, you know what to do, so this scenario is discarded since the chances you pick the correct door are irrelevant since you know which door is correct. So it then becomes:
0 1 0
0 0 1
Then if we do the hand smoosh (discarding the doors on the left) like in the video it becomes:
1 0
0 1
It's now a 50-50 as to whether you picked the correct door and it doesn't really matter if we switch or not.
Babe, wake up. Delightful Kissboy uploaded a new Buckshot video...
The fact that I did all the probability tasks in my final math exam DOES NOT mean I still don't hate probability theory...
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
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?
Delightful Kissboy, you are the cool uncle version of 3Blue 1 Brown. I'm dead serious when I say you make educational content that I would be happy to share with highschool students (or interested adults) to trick them into learning about statistics, probability, and programing.
this video was an interesting date night
finally someone represents the monty hall problem with code, I've had trouble believing it for what feels like years since it feels like they are seperated events of choosing the 3 doors then the remaining 2.
Basically imagine that the host doesn't open a door himself but just offers you to either stick with the door you chose or open both doors that you didn't. Pretty obvious what you should do in this scenario.
Another delightful video from our kissboy ❤
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 love your discussions about statistics but it would be rude of me not to mention that this video's thumbnail is art
i was just watching the powerball video and this pops up after i finish it 💜
In the process of learning Godot and GDscript just to comprehend how stupid the Dealer is🔥
i just convinced my friend to play buckshot roulette today and now im convincing him to watch your vids on it. gotta infect him with the Hyperfixation :))
Hell yeah 🙌
your videos are so good i always forget you're still a "small" TH-camr, keep that shit up DKB
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.
Another day, another vid from DKB reminding me not to kms because statistics o7
Professor Kissboy! Thank you for today's lecture! I'm studying up I promise
tbf, in buckshot roulette you get so much items that you rarely have to rely on probabilities as long as you keep note of the shells
now we need a monty hall item
Magnifying glass. Perfect for this
TL:DR the buckshot version acts as if you have to choose a door, but the host opens one door before you lock in anything. It just either eliminates that door as an option or gives it away as the correct answer, and the fact that it can give away the answer is the difference maker from Monty Hall, 2/3rds of the time it reduces things to a 50/50 and 1/3rd of the time it hands you the answer.
One other difference with the monty hall problem is that with the buckshot roulette thing, instead of having monty open a dud door, if he opens the prize door you automatically win, so you actually do get better odds but for different reasons.
By default you have 2/3 chances to hit a blank.
But if the live is revealed you can hit the opponent with the live 100% of the time.
And if a dud is revealed you have 50/50 to hit the opponent with the live.
That means you do hit the opponent 2/3 of the time.
But only because in this scenario monty lets you switch to a door that was opened, which isn't in the original problem.
love this guy
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
"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
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.
But the rest of the cases were eliminated arbitrarily, so it should be equivalent to the naive version. You could instead imagine setting aside 2 boxes at the start before revealing the other 23. It seems obvious here that neither case is more likely to be better than the other, and that switching makes no difference.
The difference between this one and Monty Hall is that we chose both cases independently. With Monty Hall the door that is left DOES depend on your initial choice, which is why one door can be "better" than the other.
Your reasoning holds if, after selecting a box, you're given a second box and told that one of them holds a million. In 1/25 cases your initial box has a million and switching loses you money, but you gain value in every other scenario.
When I played against my girlfriend and had the "two blank, one live situation" and a phone I immediately decided on firing the first, used the phone, then switched to the one not mentioned. I immediately recognized the similarity to the monty hall problem.
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?
this is so good, I'm telling everyone about this
4:40 gave me the same, slightly unhinged feeling as the Click
so essentially, because the burner phone already has the chance of giving you the right answer, you cant increase the chance by switching.
Youre the only person who's made me believe that the Monty Hall problem is real
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.
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
Accurate username as always sir
I was wondering about this exact question
Challenge: Write a script/bot that plays the game in the most efficient way possible. It takes into consideration everything - your lives, the dealers lives, all items currently on the table, amount of shells currently in the gun. It should make the best choices possible, it has to know when its better to shoot itself rather than the dealer and vice versa. Basically it should know most of the possible combinations and outcomes in the game. It would definitely be a cool concept
Check out my last vid; it's set up for 1v1 multiplayer, but planning to adapt it to be able to annihilate the dealer specifically
Thanks for this, cave Johnson
This is my new favorite channel
I missed hearing delightful kissboys voice
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 👌
3:30 a little harsh but true
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
It's far more intuitive in buckshot roulette than with doors
Thank you delightful kissboy:)
Heeey DKB it's been a while, love your content cheers
The fun thing about the Monty hall problem is that even if you don’t know if the person is specifically choosing a door without the prize, or just chose one of the other doors at random (often times people when posing the question don’t properly state the first is correct), which would mean you don’t know if it’s a 66/33 or a 50/50, it’s better to switch. E.g lets say the host could be either and it’s 50/50 how the host runs it. Then the probability of winning if not switch is 50%*50% + 33%*50% = 42% or so. Conversely, the probability of winning if not switch is 50%*50% + 66%*50% = 58%
That would be true if you knew these are the only two options and will be applied with an equal probability.
But in fact, there are infinitely many ways the host could play and if you have no idea whatsoever which is the correct one, then it will be a 50:50.
@@insignificantfool8592 they dont need to be with equal probability, though my logic does assume that it only happens in one of the two ways. Since those are the only two reasonable ways its set up, I'd say its a reasonable assumption. Also, which ways could the host play that would lead to not switching being better than switching? There has to be some way, otherwise it could not counterbalance and would therefore not be 50/50. Also, there are not infinitely many ways for the host to play at all - there are three possibilities for the prize, and three for the door you choose, so 9 possible states, and in each state the host chooses one of the other doors to open, so there could not possible be more than 2^9, or 512 possible ways for the host to play.
@mcpecommander5327 there could be various rules the host could use. You yourself acknowledge that there are two:
1.always open one of the other two doors with a goat and offer the switch (2/3 for switching)
2. open one of the other two doors randomly and offer the switch if it's a goat (1/2 for switching)
Just two more would be:
3. Open door and offer switch only if chosen door was car door (0 for switching)
4. Open door and offer switch only if chosen door was goat door (1 for switching)
And you could do infinitely many variants like 3 and 4, giving higher or lower likelihoods that you even offer the switch depending on whether a car or goat door was chosen.
@@insignificantfool8592 I assumed that they would always offer switch
so 2 would be:
open one of the other two doors randomly and offer the switch (no condition)
Even with that, there would still only be a finite number of variants, just this time 4^9 possibilities
@mcpecommander5327 that seems weird to me. Does that mean you might be lucky and the car door gets opened and you're allowed to switch?
I thought you had the idea that the offer to switch is not forced.
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)
In buckshot you don't have a 50% chance, but a 66% of winning whether you are switching or not!
in brackets (gun enemy/gun yourself)
1/3 you chose live (win/lost)
1/3 you chose blank and last is live (no choice, always win)
1/3 you chose blank and last is blank (lost/win)
So in 2/3 cases you win with any choice
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?
Wait, it still works without the goats and the car (as per the numberphile video)?! I believed those to be crucial...
so like because the cash prize has a 2/3 chance of being in a door you have not picked and the host is __going to pick a door without the prize,__ he is turning that 2/3 between the unpicked two doors into 2/3 on one door?
ah yes my favourite houses in hogwarts
the Winningdoor and the Losingdoor
2:16 No, one of those is the door you opened so that scenario doesn't exist anymore. It's a 50/50, unless you misrepresent the situation.
An easy example for this is: There are 3 doors before you, 1 is already open, what are your chances for getting the right one.
Imagine there are 5 doors. You pick 1. The host opens 3 doors with no prize. Do you switch or do you stay?
It is not equivalent to start with one door already opened. In such case, the host would have only had one restriction to fulfill: to not reveal the prize door. Any of the other two would have been free for him. In contrast, in the Monty Hall problem he has two restrictions: 1) to not reveal the door the player previously chose, and 2) to not reveal the prize door.
Those two restrictions create a disparity, because if the player's choice is the same that contains the car, the host is still free to reveal any of the other two, making it uncertain which he will take in that case, but if the player's choice is a losing option, the host is limited to only one possible losing option to remove.
For example, if you choose #1 and he opens #2, we know that the revelation of #2 was 100% mandatory in case the correct were #3, as the other two would have been prohibited. But if the correct were #1 (yours), it was only 50% likely that he would open #2, not 100% guaranteed as in the other case, because we have to deal with the possibility that he would have preferred to open #3 instead.
That's what makes it twice as difficult that he opens #2 in a game that #1 is correct than in a game that #3 is correct (having you selected #1).
But if you hadn't picked anything and he just opened #2 before you began, it would have been equally likely that he chose to remove #2 regardless of if the correct were #1 or #3, so no further information about a door over the other; both have the same chances.
good video
minor thing, but I think Let's Make A Deal didn't actually work like that, the Monty Hall problem was invented by a statistician and they just used the TV show as context to set it up in
Finally watching this
yes.
hey man, ever thought about making videos about Balatro?
Whenever I hear your name I can’t help but think that you’re the boy kisser meme
Us C# devs gotta stick together 🤜🤛
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.