Monty Hall problem explained visually(blog.vctr.me)
blog.vctr.me
Monty Hall problem explained visually
http://blog.vctr.me/monty-hall/
83 comments
I do the same thing by expanding it out to more doors, but when talking about it or explaining it to others (great cocktail party discussion btw), I use 100 doors, to make the percentages easy to follow.
Example - there are 100 doors, you pick 1 that has the grand prize. There's a 1% chance you picked the prize on your selection. Monty Hall eliminates 98 of the doors, meaning there are 2 left. There's STILL a 1% chance you picked the door with the prize, but that means there's a 99% chance the only remaining door has the prize. Obviously, you should switch.
Example - there are 100 doors, you pick 1 that has the grand prize. There's a 1% chance you picked the prize on your selection. Monty Hall eliminates 98 of the doors, meaning there are 2 left. There's STILL a 1% chance you picked the door with the prize, but that means there's a 99% chance the only remaining door has the prize. Obviously, you should switch.
Another way is to take a step back and forget about revealing the goats. You have your initial selection (using your numbers) with a 1% chance of picking the prize. Now you're presented the option to leave it shut and open all other doors keeping the prize if it's in any one of them. So if the prize has equal odds of being in any given door, then opening 99 of them gives you a 99% chance of finding the prize versus the 1% chance of only opening one door. This has helped some people I've talked to realize that the odds the prize is in that remaining door (in the proper Monty Hall problem) is actually the sum of the odds it's in all the doors you didn't select.
EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.
EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.
What I find hard to undestand is why I should switch instead of randomly choosing again?
Consider the 3 door case, it's narrowed down to 2 doors. The one you selected has a 1/3 chance of being correct, the one you didn't has a 2/3 chance. If you randomly select between the two (50/50 odds) you come out with a 50% chance of picking the correct door. If you strictly switch, your odds are better, and if you stay the same the odds are worse.
Random choice would be optimal if the host didn't know the location of the car. If the host is blind & car wasn't revealed already, there's 1/2 chance. If the host knows & the car is hidden, then the remaining door has 2/3 chance
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Ah, I used one million because it really hammers the scale of it in, but I like that with 100 you can easily track the percentages.
The way I explain is to expand it but also change the frame of reference to so its clear that the host is an adversary. Imagine we play a game called "who has the Ace of Diamonds"? I deal you one card face down and I deal me 51 cards. I look at my cards and then choose 50 of them to show you, none of which are the Ace of Diamonds. Do you want to keep your card or take the one I have not turned over?
I will never understand how one million doors is easier to understand than 3. Switching doors means you get two doors out of three doors. 66% > 33%. 66/33 illustrates the benefit much better than 1 vs 2 out of a million.
I think what confuses people is that they think a random door is revealed, whereas the problem says a goat is specifically chosen before the offer to switch.
I think what confuses people is that they think a random door is revealed, whereas the problem says a goat is specifically chosen before the offer to switch.
This is the most intuitive description I've found:http://lesswrong.com/lw/2b0/bayes_theorem_illustrated_my_way...
This is a great explanation I hadn't thought of before. Thanks.
Does it matter whether or not the host knows? I don't see how it changes the odds in any way, unless of course he opens the door with the car (very likely with 1 million doors).
The fact that you asked this question shows what's wrong with the supposed "problem". In the usual presentation, it's a "trick question" because the person presenting it omits the crucial information that the offeror is intending to act adversely to the decsion making subject.
Without that information, the interlocutor naturally assumes that the offeror opens doors randomly - in which case the fact that he knows what's behind each door is an irrelevant "red herring" and the correct conclusion is different.
In the random-opening case the intuitive conclusion is right - that the agent (offeree)'s decision does not affect the odds. But if the question-poser revealed that the offeror is trying to maneuver the offeree into the result that's less valuable for the offeree, then the supposed mystery of the whole thing evaporates and the correct conclsion is obvious.
Without that information, the interlocutor naturally assumes that the offeror opens doors randomly - in which case the fact that he knows what's behind each door is an irrelevant "red herring" and the correct conclusion is different.
In the random-opening case the intuitive conclusion is right - that the agent (offeree)'s decision does not affect the odds. But if the question-poser revealed that the offeror is trying to maneuver the offeree into the result that's less valuable for the offeree, then the supposed mystery of the whole thing evaporates and the correct conclsion is obvious.
Ha, interesting. Yes, well, back when the problem was invented it was assumed that everyone was aware of the TV show "Let's Make a Deal" and how the scenario worked. The host/offeree (a man named Monty Hall) _never_ picked a door that had the prize behind it, implying that he _always_ knew where the prize was.
The host knows where the car is and so leaves it hidden. If some one shows you this with even just 10 doors, it will be immediately and patently obvious that you must switch.
This is, indeed, the most effective way to explain it, imo.
This is, indeed, the most effective way to explain it, imo.
>Does it matter whether or not the host knows? I don't see how it changes the odds in any way
Of course it matters.
The host uses his knowledge to AVOID opening the door with the car (and thus ending the game prematurely).
So he essentially gives out this information: "the door with the car is one that I, the host, haven't opened".
Of course it matters.
The host uses his knowledge to AVOID opening the door with the car (and thus ending the game prematurely).
So he essentially gives out this information: "the door with the car is one that I, the host, haven't opened".
Let's say that you have 3 doors and you pick door #1.
At that point the host says "Ok, I'm going to open one of the other doors at random and then give you the chance to change your choice".
At this point 2 things could happen. Either he opens the door with the car in which case you will definitely switch to that door and have a 100% chance of winning.
Or he opens the door with a goat. There is a 1/3 chance that you picked correctly in the first place, therefore there is a 2/3 chance that you picked wrongly, so it's safer to assume that you picked wrongly.
You know that he didn't show you the car, so either you picked correctly to being with (1/3) or you picked wrongly (2/3), but if you picked wrongly then it must be behind the door that the host did not open. So you still have better odds from switching even though the host did not know.
At that point the host says "Ok, I'm going to open one of the other doors at random and then give you the chance to change your choice".
At this point 2 things could happen. Either he opens the door with the car in which case you will definitely switch to that door and have a 100% chance of winning.
Or he opens the door with a goat. There is a 1/3 chance that you picked correctly in the first place, therefore there is a 2/3 chance that you picked wrongly, so it's safer to assume that you picked wrongly.
You know that he didn't show you the car, so either you picked correctly to being with (1/3) or you picked wrongly (2/3), but if you picked wrongly then it must be behind the door that the host did not open. So you still have better odds from switching even though the host did not know.
You are correct, as long as the host doesn't reveal the prize the odds of winning by switching remain better than the odds of winning with your initial selection. EDIT: And in the case that he reveals the prize your odds have gone up. I suspect (but haven't calculated) that a version where you can switch to the winning prize when it's revealed would give you much better odds of success and a bankrupt game show would result.
Crunched the numbers, I'm wrong. If the host shows a goat (when the host doesn't know where the car is), switch or stay doesn't matter as you have the same odds either way.
There are 4 outcomes to the selection phase of this game:
1/3 - both pick goats
1/3 - you pick the goat, host picks the car
1/3 - you pick the car, host picks the goat
0 - you pick the car, host picks the car
So given the host picks a goat, switching or staying doesn't matter it's 50/50.
EDIT: An aside - why we shouldn't trust our intuitions, but instead actually work out these problems.
There are 4 outcomes to the selection phase of this game:
1/3 - both pick goats
1/3 - you pick the goat, host picks the car
1/3 - you pick the car, host picks the goat
0 - you pick the car, host picks the car
So given the host picks a goat, switching or staying doesn't matter it's 50/50.
EDIT: An aside - why we shouldn't trust our intuitions, but instead actually work out these problems.
You're right. When doors are opened randomly you can not tell whether the host was "lucky" (to not reveal the car) or whether you picked the right door to begin with since these are both equally likely.
The host doesn't open a door at random. He always shows a goat.
You are discussing some other game show.
You are discussing some other game show.
Let's assume that both of these game shows are showing on different channels and you were invited to participate in one of them. But once you get to the studio you have a case of nerves and forget which show you are on. Should that change how you make your decision?
No, the odds for switching at the real show are 66-33 and the odds at the fake show are 50-50, you might as well switch after the reveal (what you are calling 'better odds' boils down to 'sometimes you can see the car so you switch to it'. When you see a goat, you have equal information for the remaining doors.).
It does. If the host doesn't know, there's a 1/3 probability that the game will end (you won't get a chance to switch) if the host picks the car. If this doesn't happen, it's 50:50 whether you keep your door or switch to the remaining unopened door.
Yes, it matters. The host opens every door other than the one with the car - or one with the goat if you picked the car.
The host is not opening just one door.
The host is not opening just one door.
If the host knows, you should switch. If he doesn't know, it doesn't matter--you have a 50/50 chance of winning either way.
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Nice explanation. Small critique: I think this page would be better if the animations could be played manually rather than on a continuous loop which is quite distracting. Also, maybe add in a bit where the user can play the game themselves to see the logic work out.
Something like this was useful when I was first learning about the Monty Hall problem.
http://www.grand-illusions.com/simulator/montysim.htm
Something like this was useful when I was first learning about the Monty Hall problem.
http://www.grand-illusions.com/simulator/montysim.htm
There's no need for the animations at all. Besides not letting the reader work through the problem at her own pace, the animations hide the previous steps.
The best explanation I've seen of the Monty Hall problem is still Oscar Bonilla's: http://oscarbonilla.com/2009/05/the-monty-hall-problem/
Best explanation I've seen was using statistics: simulate the problem 1000 times. Look at results when you switch and when you don't switch doors.
Or do it in your head: If I switch doors, I will win with 2 out of 3 doors. Otherwise, I'll win only with one door.
I liked the simulation, and it reminds me of a tangentially related probability issue from early in my career.
While still in college, I had a part-time job testing a spreadsheet product (back in the days when there were other commercial choices). One of the things we testers did to keep ourselves entertained was that we each picked a casino game, and implemented that game in spreadsheet macros. My game was blackjack, but another guy did craps.
Just for kicks, the guy who built the craps game re-wired it as an autoplay simulation, and left it playing against itself overnight. When we came back in in the morning, we found that the virtual player was rich, which seemed odd. So the next night we left it running again, and again the following morning, the player had made a bundle.
It turns out that my coworker had accidentally discovered a bug in the random number generator: the distribution of the results was a bit non-uniform.
While still in college, I had a part-time job testing a spreadsheet product (back in the days when there were other commercial choices). One of the things we testers did to keep ourselves entertained was that we each picked a casino game, and implemented that game in spreadsheet macros. My game was blackjack, but another guy did craps.
Just for kicks, the guy who built the craps game re-wired it as an autoplay simulation, and left it playing against itself overnight. When we came back in in the morning, we found that the virtual player was rich, which seemed odd. So the next night we left it running again, and again the following morning, the player had made a bundle.
It turns out that my coworker had accidentally discovered a bug in the random number generator: the distribution of the results was a bit non-uniform.
Very cool way of illustrating it. It looks like you just used iframes for the animations so they were in separate pages with distinct stylesheets, JS resources, etc., why did you do that rather than having the Jquery animations happen in the main document? Was it just to keep things nicely modular?
I've usually found the problem poorly worded (probably intentionally). An assumption that Monty knows which door is which is required but often not stated. If Monty doesn't know which door is which, the odds are 50-50 like you'd imagine.
He just has to not open a door with the car behind, it doesn't matter whether intentional or not. If he forgot his intentionality but takes the exact same actions by chance, the stats don't change for the player.
This isn't right though, it makes all the difference whether he acts knowingly or happens to pick a goat; it's a 50/50 chance on changing if he chose a goat by luck (see other posts ITT)
I suspect that much of the contention around this famous problem is based on the original boundaries / rules of the problem being unclear.
The problem as stated sometimes leaves out crucial information that affects the outcome.
Example:
Game show, 3 doors. Contestant chooses one door. The host opens another door, which has a goat. Do you switch?
In this situation, it is not explained whether the host randomly picked one of the two remaining doors to open, without knowledge that one of them would contain a goat. This is the key missing information. If the host ALWAYS reveals the goat, then it is beneficial to switch as 2 out of 3 times the contestant wins by switching.
However, if the host is simply opening another door at random, and the door happens to contain a goat, then there is no advantage to switching, as the odds are 50/50.
edit: it appears that from the original version of the problem stated in Parade magazine, the implication is that the host knowingly reveals a goat.
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
The problem as stated sometimes leaves out crucial information that affects the outcome.
Example:
Game show, 3 doors. Contestant chooses one door. The host opens another door, which has a goat. Do you switch?
In this situation, it is not explained whether the host randomly picked one of the two remaining doors to open, without knowledge that one of them would contain a goat. This is the key missing information. If the host ALWAYS reveals the goat, then it is beneficial to switch as 2 out of 3 times the contestant wins by switching.
However, if the host is simply opening another door at random, and the door happens to contain a goat, then there is no advantage to switching, as the odds are 50/50.
edit: it appears that from the original version of the problem stated in Parade magazine, the implication is that the host knowingly reveals a goat.
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
For this to really work, it must be clarified up front that Monty is going to give you the opportunity to switch every time the game is played.
I've tried to think of the best way to make this obvious. This is what I came up with, and I'd like to hear if it's accurate:
Contestant picks a door out of three. 1/3rd chance he's got the car. Then he is given the choice to either stick with his door, or do a coin-toss where you have to pick heads.
In this case, it seems obvious that option for the coin-toss is the better option, because it gives you a 50% chance of winning.
The confusing part of the problem is that you feel like you're doing one thing, where really it's two independent games where one clearly offers a bigger chance of winning.
That said, some people here have mentioned that it's not actually a 50% chance if you switch doors. So maybe my explanation is incorrect? And if it's incorrect, is it still correct enough to show that switching doors (i.e. opting for the coin-toss) is the better option?
Contestant picks a door out of three. 1/3rd chance he's got the car. Then he is given the choice to either stick with his door, or do a coin-toss where you have to pick heads.
In this case, it seems obvious that option for the coin-toss is the better option, because it gives you a 50% chance of winning.
The confusing part of the problem is that you feel like you're doing one thing, where really it's two independent games where one clearly offers a bigger chance of winning.
That said, some people here have mentioned that it's not actually a 50% chance if you switch doors. So maybe my explanation is incorrect? And if it's incorrect, is it still correct enough to show that switching doors (i.e. opting for the coin-toss) is the better option?
Looking this up on Wikipedia reveals it's relationship with the Three Prisoners Problem[1] as well. The solution presented in that case helped me in reasoning about it.
[1]: http://en.wikipedia.org/wiki/Three_Prisoners_problem
[1]: http://en.wikipedia.org/wiki/Three_Prisoners_problem
Here's how it helps me to look at it. As others have said, you can play with the numbers, but the mechanics are basically this:
First step: you are choosing one option from all available choices. Second step: you are now betting whether your original choice was the correct one.
By consciously narrowing down all the remaining options to one, the host is essentially asking you: "Do you think you guessed right?" As long as there are more than two available options at the start, the odds are greater than you guessed wrong.
As others have said, the important dynamic here is that the host _knows_ where the prize is and proceeds to open all the other doors, leaving only yours and one remaining closed. That remaining door then symbolizes an aggregate of all the doors you didn't select.
First step: you are choosing one option from all available choices. Second step: you are now betting whether your original choice was the correct one.
By consciously narrowing down all the remaining options to one, the host is essentially asking you: "Do you think you guessed right?" As long as there are more than two available options at the start, the odds are greater than you guessed wrong.
As others have said, the important dynamic here is that the host _knows_ where the prize is and proceeds to open all the other doors, leaving only yours and one remaining closed. That remaining door then symbolizes an aggregate of all the doors you didn't select.
Consider the version where Monty does not know what door the car is behind. He randomly opens one of the other two doors, and (per the rules) if he reveals a car, game over too bad. This time, he happened to reveal a goat. Why is the question not "did you guess right originally?"?
I'm quite convinced that it is different if the host knows versus if he does not, but I've never quite been able to put my finger on why in an intuitive way.
I'm quite convinced that it is different if the host knows versus if he does not, but I've never quite been able to put my finger on why in an intuitive way.
Imagine the host doesn't know. If you guess right, he opens a door with a goat behind it and you find yourself in the classic situation.
But if you guess wrong, there's a 50% chance that the host reveals a car by accident.
So the probablities look like:
33%: you guess right, the host reveals a goat 33%: you guess wrong, the host reveals a goat 33%: you guess wrong, the host reveals a car
So, of the situations where the host reveals a goat, the car will be behind your door 50% of the time.
If the host knows and avoids the goat, the probabilities look like this:
33% you were right; the host reveals a goat 33% you were wrong; the host reveals a goat behind door B 33% you were wrong; the host reveals a goat behind door C
So in the situations where the host knows about and avoids the goat, 66% of them have the car behind the other door.
In the case where the host doesn't know about the car, picking the car on your first choice makes it more likely for the host to reveal a goat. That's the difference.
But if you guess wrong, there's a 50% chance that the host reveals a car by accident.
So the probablities look like:
33%: you guess right, the host reveals a goat 33%: you guess wrong, the host reveals a goat 33%: you guess wrong, the host reveals a car
So, of the situations where the host reveals a goat, the car will be behind your door 50% of the time.
If the host knows and avoids the goat, the probabilities look like this:
33% you were right; the host reveals a goat 33% you were wrong; the host reveals a goat behind door B 33% you were wrong; the host reveals a goat behind door C
So in the situations where the host knows about and avoids the goat, 66% of them have the car behind the other door.
In the case where the host doesn't know about the car, picking the car on your first choice makes it more likely for the host to reveal a goat. That's the difference.
That's a great question! I think the person who can articulate that intuitively should win the car!
My stab at it: If he doesn't know, then he's essentially another contestant, and his odds are the same 1 in 3 as yours. You've made selections one after the other, but since your selection has yet to be revealed as either right or wrong, you are essentially just picking different options from the same scenario. There are exactly equal chances that you won the car, he flubs and reveals the car, or he reveals a goat.
My stab at it: If he doesn't know, then he's essentially another contestant, and his odds are the same 1 in 3 as yours. You've made selections one after the other, but since your selection has yet to be revealed as either right or wrong, you are essentially just picking different options from the same scenario. There are exactly equal chances that you won the car, he flubs and reveals the car, or he reveals a goat.
I like frankc's intuitive explanation of the original game given in another comment:
The way I explain is to expand it but also change the frame of reference to so its clear that the host is an adversary. Imagine we play a game called "who has the Ace of Diamonds"? I deal you one card face down and I deal me 51 cards. I look at my cards and then choose 50 of them to show you, none of which are the Ace of Diamonds. Do you want to keep your card or take the one I have not turned over?
Now imagine the same set up, but instead, after dealing the cards, I don't look at them. I then proceed to turn 50 of mine over one after the other and, though unlikely, I happen to not reveal the Ace of Diamonds. Now, is it any more likely the last card I haven't turned over yet is the card, than it is that you have it?
The game is no different in this case to one person lining up all 52 cards and turning them over one by one along the line. If you get to 50 cards and you still haven't found it, that obviously doesn't mean it's more likely to be the end card than the penultimate one.
The way I explain is to expand it but also change the frame of reference to so its clear that the host is an adversary. Imagine we play a game called "who has the Ace of Diamonds"? I deal you one card face down and I deal me 51 cards. I look at my cards and then choose 50 of them to show you, none of which are the Ace of Diamonds. Do you want to keep your card or take the one I have not turned over?
Now imagine the same set up, but instead, after dealing the cards, I don't look at them. I then proceed to turn 50 of mine over one after the other and, though unlikely, I happen to not reveal the Ace of Diamonds. Now, is it any more likely the last card I haven't turned over yet is the card, than it is that you have it?
The game is no different in this case to one person lining up all 52 cards and turning them over one by one along the line. If you get to 50 cards and you still haven't found it, that obviously doesn't mean it's more likely to be the end card than the penultimate one.
Whether we arrive there by chance or by deliberate action, once we know that the open door has a goat, the odds for winning by switching remain the same. The odds only change for the overall game (where there's now a 1/3 chance we lose/win when the host reveals the car). If the goat is revealed, we're still in the original game: 2/3 for switching, 1/3 for staying.
EDIT: Leaving my wrong answer, but bvk is correct.
EDIT: Leaving my wrong answer, but bvk is correct.
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Like everyone who encountered this in their intro stats class, I love this problem. And it bothers everyone the first time.
Are there other problems like the Monty Hall problem, perhaps in other disciplines?
Are there other problems like the Monty Hall problem, perhaps in other disciplines?
The two envelopes paradox is a good one:
http://consc.net/papers/envelope.html
And another that I've figured out the "right" answer to but think there might be more to it is this:
Suppose I set up a two-player game as follows:
http://consc.net/papers/envelope.html
And another that I've figured out the "right" answer to but think there might be more to it is this:
Suppose I set up a two-player game as follows:
1. One player is the "odd" player, the other is
the "even" player.
2. Each player may choose a single number per turn,
either one or two.
3. The result of each turn is the sum of the two players'
choices. For example, if the even player picks 1 and the
odd player picks 2, the result of that turn is 3.
4. If the result is odd, the even player pays the odd
player that amount, and vice-versa. In our example
result of 3, the even player would pay the odd player 3
dollars.
I'll bite.
I guess that the odd player can always choose 1. In that case he can either win 3 or lose 2. Making that the favorable choice for the odd player.
Even player could always choose 2 in which case he can either win 4 or lose 3. Making this the favorable choice for the even player. However, if both players play their favorable choice, odd player always wins: 2 + 1 = 3.
I'm not sure if I got this right, but that's my thinking so far...
I guess that the odd player can always choose 1. In that case he can either win 3 or lose 2. Making that the favorable choice for the odd player.
Even player could always choose 2 in which case he can either win 4 or lose 3. Making this the favorable choice for the even player. However, if both players play their favorable choice, odd player always wins: 2 + 1 = 3.
I'm not sure if I got this right, but that's my thinking so far...
Here's a hint; is there a strategy one player can play such that no matter what the other player does, the first will win money on average?
I like this question:
Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Have you ever cheated on your spouse?"
Now, what is the probability the person is a cheater given they answered yes?
Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Have you ever cheated on your spouse?"
Now, what is the probability the person is a cheater given they answered yes?
Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly.
Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/
Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/
var runs = 1000,
honest_admissions = 0,
automatic_admissions = 0;
function flip() {
return Math.random() < 0.5 ? 'h' : 't';
}
for (var i = 0; i < runs; i++) {
if (flip() == 't') {
honest_admissions++;
} else if (flip() == 'h') {
automatic_admissions++;
}
}
document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.')
Oops. That's completely wrong. It's the case where every honest answer is yes!
I think this is indeterminate. You have to know the ratio of total yesses to total participants.
The real world use-case I was recalling is simpler: flip a coin and answer honestly if it's heads and "yes" if it's tails. Any individual response is non-incriminating, but if N is large enough where you can assume N/2 got heads, you can know the number of "honest cheaters."
I think this is indeterminate. You have to know the ratio of total yesses to total participants.
The real world use-case I was recalling is simpler: flip a coin and answer honestly if it's heads and "yes" if it's tails. Any individual response is non-incriminating, but if N is large enough where you can assume N/2 got heads, you can know the number of "honest cheaters."
[deleted]
Wait, is there an interesting answer to this?
I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse.
If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%).
If I flip tails, I answer 2Y (X) or 2N (1-X).
So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X)
And combined, Y = 0.25+0.5X
What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)
I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse.
If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%).
If I flip tails, I answer 2Y (X) or 2N (1-X).
So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X)
And combined, Y = 0.25+0.5X
What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)
This could be the most bothersome problem of all:
http://www.airplaneonatreadmill.com/
http://www.airplaneonatreadmill.com/
I know of one other mathematical fact that matches the Monty Hall problem in the loudness of its opponents: 0.999(...) exactly equals one. Despite a number of proofs existing, you can still find ten large pages of arguments on Wikipedia's talk page.
http://en.wikipedia.org/wiki/Talk:0.999.../Arguments
http://en.wikipedia.org/wiki/Talk:0.999.../Arguments
This problem is known as "Restricted Choice" in the card game bridge. It occurs when deciding whether to play for the drop of a missing honor card or finesse. See http://en.wikipedia.org/wiki/Principle_of_restricted_choice for more.
It bothers everyone because the teacher never _stresses_ the fact that the host doesn't open a door at random. It's purposely deceptive.
Given two choices (equal probability; redundant?) - which is the case once you have two doors - switching cannot change the odds. Have someone flip a coin. Call it. But before the flipper looks at the coin - (whether he KNOWS the result or not) he asks you to consider switching. Explain how the option of switching changes the probability of a correct or incorrect call. Inference?
The best explanation I've seen was in series 3, episode 3 of Jame May's Man Lab, where they repeatedly iterated the problem and showed that when repeated enough times, the frequency of winning/losing tends to a different value with a change of door than without (by having the presenters open booby-trapped beer cans and either get splashed with beer or not)
I can't see how your generalization of the problem to more doors is correct.
In the classical problem with 3 doors, if you pick a door with a car, the host will only open 1 other door.
In your demo, if I use 40 doors and pick one, the host opens all other doors but one. Why?
If the one of those 40 I pick has a car, will the host open all remaining doors? That doesn't seem to be the case for the 3-door variant.
In the classical problem with 3 doors, if you pick a door with a car, the host will only open 1 other door.
In your demo, if I use 40 doors and pick one, the host opens all other doors but one. Why?
If the one of those 40 I pick has a car, will the host open all remaining doors? That doesn't seem to be the case for the 3-door variant.
In the Monty Hall problem the door with the car is never revealed until the end. In the 3 door version if you've selected the prize door, the goat revealed is selected at random. If you've selected a goat, the other goat is revealed and the car remains hidden behind the remaining (at this point unselected) door. So with 40 doors, after your selection 38 goats would be revealed. It may be that you selected the car by chance (1/40 chance), in which case 38 of the 39 remaining doors would be selected at random (as it doesn't matter). If you selected a goat only the other 38 goat doors would be opened (but the contestant can't tell the difference between these circumstances).
Still, why? Why open 38 doors? I don't think this emphasizes the difference in probability, it just creates artificial bias.
Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.
With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.
Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.
IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.
In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.
So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.
In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.
Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.
With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.
Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.
IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.
In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.
So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.
In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.
(c), I feel, matches the original problem best when taken to an increasing number of doors, but makes for an absurd game show. It also most closely resembles the odds of the original game show (win by switching (n-1) of n times, or 2 of 3 with 3 doors). Anything else becomes a different game (you don't have a single choice, you have a multitude of choices). And it doesn't turn it into a different problem, it exagerates the odds of the original problem to better illustrate the statistics to people whose intuition is wrong. The case of 100 doors where switching results in winning 99% of the time gets passed the questioners intuition that it should be even odds (switching or staying).
(a) demonstrates that the questioners intuition about the probabilities is off, but (again, IMO) doesn't really offer the clarity of (c).
Re: Summary - I agree, it's not an accident. It was chosen because it was a good number for a game show, and leaving the participant with 2 options (switch or stay) ensures a certain swiftness of decision. While (a) (say they used 4 or 5 doors) offers only a slight improvement on the odds (contestants may lose too often) and (c) (again with 4 or 5 doors) offers a perhaps too great improvement of the odds (contestants win too much if they know the trick).
(a) demonstrates that the questioners intuition about the probabilities is off, but (again, IMO) doesn't really offer the clarity of (c).
Re: Summary - I agree, it's not an accident. It was chosen because it was a good number for a game show, and leaving the participant with 2 options (switch or stay) ensures a certain swiftness of decision. While (a) (say they used 4 or 5 doors) offers only a slight improvement on the odds (contestants may lose too often) and (c) (again with 4 or 5 doors) offers a perhaps too great improvement of the odds (contestants win too much if they know the trick).
I have a much easier explanation than this: By offering you the switch, the host has given you two doors to open - it's irrelevant that he opened one for you - so, it comes down to picking one door (the one you're on) or two doors (of which he has opened one for you, again, irrelevant - the effect is the same) ergo, it's 1/3 or 2/3.
That's a good looking demo, quite smooth. Implementing this was actually a recent DailyProgrammer challenge (http://www.reddit.com/r/dailyprogrammer/comments/1qdw40/1111...)
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I always explain this problem with stars: I'm thinking of a star in the night's sky... guess which one.
Alright, good guess, but would you consider changing? You can either keep your original guess, or switch to this one other star I'm pointing to now.
Everyone switches.
Alright, good guess, but would you consider changing? You can either keep your original guess, or switch to this one other star I'm pointing to now.
Everyone switches.
That doesn't explain anything except that people will be influenced because you pointed to another star.
Heh, Monte Carlo simulation of Monty Hall problem. Mildly amusing... if only we could work a Monty Python reference in there somewhere!
Watching this play out really kills any remaining gut response that would say "it's the same either way!"
Watching this play out really kills any remaining gut response that would say "it's the same either way!"
Just make it show alive/dead parrots, or spam, instead of goats. ;)
"Our table has spam, spam, spam, spam, eggs, spam, spam, and spam. You've chosen your dish, now let's see what else was on the table...."
;)
"Our table has spam, spam, spam, spam, eggs, spam, spam, and spam. You've chosen your dish, now let's see what else was on the table...."
;)
some background:
http://en.wikipedia.org/wiki/Monty_Hall_problem
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong
(per above post At least this made it to stats intro class)
(per above post At least this made it to stats intro class)
The rules of the problem aren't explained very well. most of the contention probably is based on unclear boundaries of the game.
I like the simulation at the bottom. Sometimes it's best to just say, "The math gets confusing, let's run this a couple times and see what happens."
In another 50 years, no one is going to know who "Monty Hall" is.
Except for this "problem" which people will still be arguing about.
Except for this "problem" which people will still be arguing about.
To get people thinking I generally have them agree there are 9 possible games to play, so the final win-chance cannot be 50%
No one thinks the final win chance is 50%. They mistakenly think it is 1/3 with either strategy.
got into a huge fight with a family member over this. i even programmed a simulation and showed it run over 1000 tries and he still refused to budge from the 50% answer.
Play a cash game with cards for a while. Dollar bets, dollar return.
Play 50 games every time you see him until he gets it.
Play 50 games every time you see him until he gets it.
You're allowed to pick one door, out of a million. Then, the host opens 999,998 of the other doors which he knows have a goat. This leaves the closed door you picked and one remaining closed door.
So, there's now two possibilities. Out of one million doors you somehow just happened to have guessed right and picked the right one. Or, you're wrong. If you're wrong then the correct door would be the other closed one.