Pick the Door with the Prize!

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Pick the Door with the Prize!
One of my favorite applications of probability theory is the Monty Hall Prize Behind the Door game.
In the most basic version of the game, there are three doors, behind one of which is a prize. You, the player, don’t know the prize door, but your host, Monty, does. He asks you to pick one of the three doors as the prize door. At least one of the two remaining doors has no prize behind it. Monty opens one of these doors with no prize behind it. This leaves two unopened doors: The one you picked, and the one Monty did not open. Then Monty asks you if you want to switch your choice to the unopened door you did not pick in the first place.
Finally, Monty opens the door of your choice, and if the prize is behind it, you win the prize.
Should you make this switch? Justify your answer: For example, does your chance of winning the prize improve or not by making the switch, and why?
If there is enough interest, I’ll cover some generalizations of this game to more than three doors, with Monty opening varying numbers of nonprize doors and you deciding whether to change your initial choice.
In the most basic version of the game, there are three doors, behind one of which is a prize. You, the player, don’t know the prize door, but your host, Monty, does. He asks you to pick one of the three doors as the prize door. At least one of the two remaining doors has no prize behind it. Monty opens one of these doors with no prize behind it. This leaves two unopened doors: The one you picked, and the one Monty did not open. Then Monty asks you if you want to switch your choice to the unopened door you did not pick in the first place.
Finally, Monty opens the door of your choice, and if the prize is behind it, you win the prize.
Should you make this switch? Justify your answer: For example, does your chance of winning the prize improve or not by making the switch, and why?
If there is enough interest, I’ll cover some generalizations of this game to more than three doors, with Monty opening varying numbers of nonprize doors and you deciding whether to change your initial choice.

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Re: Pick the Door with the Prize!
I keep to my original choice simply because I have no reason to change my mind
Of course it might actually be the wrong door but I have no way of knowing it is
Of course it might actually be the wrong door but I have no way of knowing it is
Re: Pick the Door with the Prize!
With 3 doors, it's hard to call. Although your odds are only 1 in 3 of picking the right door in the first place, your intuition may guide you. The more doors, the greater the probability that switching to Monty's remaining switchdoor would be the right way to go, since odds are against you for picking the ONE correct door in the first place, while Monty can eliminate any number of doors to bring you to a choice between 2 doors, giving the illusion that you have a 50/50 chance. Right?
Re: Pick the Door with the Prize!
I have yet to see actual proof, instead of the usual theoretical proof.
 vegetariantaxidermy
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Re: Pick the Door with the Prize!
Absolutely right. As no one ever seems to change their choice then there is no way of proving whether the theory translates into the practical. Logic would tell you that it makes no difference to change, since the prize is still behind the same door that it was in the first place.
It was done on 'Mythbusters' but one artificially changed his 'choice' over and over. That's not the same.

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Re: Pick the Door with the Prize!
Thanks to surreptitious57, Lacewing, HexHammer, and vegetariantaxidermy, for your interest! I'll try to make your gracious attention to my post both interesting and rewarding.
Practical applications of probability  games of chance such as card games and other casino games of chance, for example, are amenable to exact calculation of odds, so the gambling house will know its longrun profit. Luckily for the nonmathematical types, such applications are also amenable to simulation, to observe the longrun odds.
In this case (the three doors), if you don't want (or can't stand) to go through the math calculation, take three cards from a deck, one red suit and two black. The red suit represents the prize door. Have a friend keep track of which card is red. Mix the cards face down, spread them out, and point to one you guess is red. Have your friend turn up a black card. Two cards remain face down.
1. To simulate not switching: Stay with your first choice and flip the cards up. After, say, 50 trials of this, you'll find that you guessed red only about a third of the time. Not so surprising.
2. To simulate switching: Run another 50 trials, but this time, for each trial, switch to the other card that is still face down. What do you see happening?
I hope these simulations may be more enjoyable, at least for awhile, than, say, playing solitaire. If your friend isn't available, you can do the simulations by yourself and pretend you don't know which card is red.
Now, I hope we'll all be able to calculate, or observe in a simulation, the odds of winning by switching in the threedoor game!
Future fun(?)  Ten doors, Monty opens less than 8 nonprize doors. Do you want to switch your choice to one of the other doors remaining unopened? How? You can use a random device. For example, if there are 7 doors left, yours and six others, throw a die to select one of the other six doors. Does this improve your odds of winning?
Practical applications of probability  games of chance such as card games and other casino games of chance, for example, are amenable to exact calculation of odds, so the gambling house will know its longrun profit. Luckily for the nonmathematical types, such applications are also amenable to simulation, to observe the longrun odds.
In this case (the three doors), if you don't want (or can't stand) to go through the math calculation, take three cards from a deck, one red suit and two black. The red suit represents the prize door. Have a friend keep track of which card is red. Mix the cards face down, spread them out, and point to one you guess is red. Have your friend turn up a black card. Two cards remain face down.
1. To simulate not switching: Stay with your first choice and flip the cards up. After, say, 50 trials of this, you'll find that you guessed red only about a third of the time. Not so surprising.
2. To simulate switching: Run another 50 trials, but this time, for each trial, switch to the other card that is still face down. What do you see happening?
I hope these simulations may be more enjoyable, at least for awhile, than, say, playing solitaire. If your friend isn't available, you can do the simulations by yourself and pretend you don't know which card is red.
surreptitious57, you have one chance in three of guessing the prize door (two chances in three of having the wrong door) at the outset. You're right, no way of knowing it, but please think in terms of odds, or chances. It's still one in three to win, if you don't switch. You'll find out that you do have a reason to switch! Not that you'll know for sure, in any case, until the game is over, but you'll actually improve your chances by switching.surreptitious57 wrote: ↑Tue Jul 03, 2018 4:00 am I keep to my original choice simply because I have no reason to change my mind
Of course it might actually be the wrong door but I have no way of knowing it is
Good thought, Lacewing, increasing the number of doors  but better (or worse) than 50/50, if you switch (don't switch). Say there are ten doors. If Monty opened eight nonprize doors, leaving two doors  your first pick and one other door  maybe you can see that switching is a good idea. There is a 9 in 10 chance that your first pick is wrong, because you only have a 1 in 10 chance of guessing correctly the first time. This means there is a 9 in 10 chance that if you switch, you'll get the prize! Switching is good even with the minimum three doors.Lacewing wrote: ↑Tue Jul 03, 2018 4:11 am The more doors, the greater the probability that switching to Monty's remaining switchdoor would be the right way to go, since odds are against you for picking the ONE correct door in the first place, while Monty can eliminate any number of doors to bring you to a choice between 2 doors, giving the illusion that you have a 50/50 chance. Right?
HexHammer, I hope you try the simulation described above. But my reply to Lacewing should give you a good clue to the proof in the threedoor case.
vegetarian ..., there is a proof that changing (switching your choice) is the way to go, even with just three doors. Check out the tendoor situation above in my reply to HexHammer, which dramatizes the virtues of switching. The mathematical theory of probability provides such proof, and casinos depend on such proofs and the corresponding calculations, for their illgotten gains.vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 7:08 am Absolutely right. As no one ever seems to change their choice then there is no way of proving whether the theory translates into the practical. Logic would tell you that it makes no difference to change, since the prize is still behind the same door that it was in the first place.
Now, I hope we'll all be able to calculate, or observe in a simulation, the odds of winning by switching in the threedoor game!
Future fun(?)  Ten doors, Monty opens less than 8 nonprize doors. Do you want to switch your choice to one of the other doors remaining unopened? How? You can use a random device. For example, if there are 7 doors left, yours and six others, throw a die to select one of the other six doors. Does this improve your odds of winning?
 vegetariantaxidermy
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Re: Pick the Door with the Prize!
Strange. My posts keep disappearing. I mentioned on my deleted post the Mythbusters experiment. They found the maths to be correct, and the one who changed his choice won the predicted number of times more than the one who didn't, but as I pointed out, in the real situation no one changes (or extremely rarely. That experiment has been done too, and they couldn't get a useful result because of it).

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Re: Pick the Door with the Prize!
Thanks for hanging in there, vegetariantaxidermy! I hope you get a chance to see my reply to you and three other brave souls. You can do a simulation with three playing cards  maybe better than solitaire or Go Fish.vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 9:33 am Strange. My posts keep disappearing. I mentioned on my deleted post the Mythbusters experiment. They found the maths to be correct, and the one who changed his choice won the predicted number of times more than the one who didn't, but as I pointed out, in the real situation almost no one changes (that experiment has been done too. They couldn't get a useful result because of it).
 vegetariantaxidermy
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Re: Pick the Door with the Prize!
I'm sorry that you don't appear to understand what I mean. You could always ask.Mike Strand wrote: ↑Tue Jul 03, 2018 9:37 amThanks for hanging in there, vegetariantaxidermy! I hope you get a chance to see my reply to you and three other brave souls. You can do a simulation with three playing cards  maybe better than solitaire or Go Fish.vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 9:33 am Strange. My posts keep disappearing. I mentioned on my deleted post the Mythbusters experiment. They found the maths to be correct, and the one who changed his choice won the predicted number of times more than the one who didn't, but as I pointed out, in the real situation almost no one changes (that experiment has been done too. They couldn't get a useful result because of it).

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Re: Pick the Door with the Prize!
I apologize, vegetariantaxidermy, for not addressing your main point here, regarding human behavior, I think. In practice, the probabilities hold only for consistent behavior  always switching or always staying with the original choice. If a person sometimes switches, sometimes not, the probability of winning falls somewhere between the two extremes. Only if the person's behavior is predictable (say, switch 30% of the time, not switch 70%) can the new longrun probability of winning be calculated.vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 9:33 am I mentioned on my deleted post the Mythbusters experiment. They found the maths to be correct, and the one who changed his choice won the predicted number of times more than the one who didn't, but as I pointed out, in the real situation no one changes (or extremely rarely. That experiment has been done too, and they couldn't get a useful result because of it).
We can be thankful, I guess, for the surprises that come from humans being "spur of the moment" decisionmakers. Or can we?

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Re: Pick the Door with the Prize!
Related problem: If a gambling casino lets customers play the threedoor problem, what should they charge the customer to play each game? (The casino pays for the prize, but wants to make money). Keep in mind human whimsy  sometimes switch, sometimes don't switch, and unpredictably.
 vegetariantaxidermy
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Re: Pick the Door with the Prize!
I see what they are getting at now. When you choose from the three doors, you have a 2 out of 3 chance of choosing a goat. Not particularly great odds. So after one of the goats has gone you now have 50/50 chance so mathematically you have a better chance of winning if you 'start again' and switch doors. It goes against the grain and our sense of what is logically the case though, hence the fact that pretty much everyone sticks with their original choice. It's a very interesting problem and lesson in what we consider to be reality despite the fact that it can't be proven in a real, practical sense.Mike Strand wrote: ↑Tue Jul 03, 2018 12:57 pmI apologize, vegetariantaxidermy, for not addressing your main point here, regarding human behavior, I think. In practice, the probabilities hold only for consistent behavior  always switching or always staying with the original choice. If a person sometimes switches, sometimes not, the probability of winning falls somewhere between the two extremes. Only if the person's behavior is predictable (say, switch 30% of the time, not switch 70%) can the new longrun probability of winning be calculated.vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 9:33 am I mentioned on my deleted post the Mythbusters experiment. They found the maths to be correct, and the one who changed his choice won the predicted number of times more than the one who didn't, but as I pointed out, in the real situation no one changes (or extremely rarely. That experiment has been done too, and they couldn't get a useful result because of it).
We can be thankful, I guess, for the surprises that come from humans being "spur of the moment" decisionmakers. Or can we?
I still get the sense that something is missing here though.

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Re: Pick the Door with the Prize!
Hi again, vegetariantaxidermy, and thanks for your continuing interest! You're on the right path, seeing that it looks better to switch doors. Try the simulation with cards, where you always switch. You'll see the odds of winning are better than 50/50. It's interesting, though, that if you flip a coin to decide whether to switch or not, your odds are indeed 50/50. The key is: Always switch!vegetariantaxidermy wrote: ↑Tue Jul 03, 2018 1:15 pm When you choose from the three doors, you have a 2 out of 3 chance of choosing a goat. Not particularly great odds. So after one of the goats has gone you now have 50/50 chance so mathematically you have a better chance of winning if you 'start again' and switch doors. It goes against the grain and our sense of what is logically the case though, hence the fact that pretty much everyone sticks with their original choice. It's a very interesting problem and lesson in what we consider to be reality despite the fact that it can't be proven in a real, practical sense.
I still get the sense that something is missing here though.
Another good thing about doing the simulations: They help train your intuition to understand the actual odds. You start seeing that one of the two cards that are left  one your first choice, the other not  has got to be the red one (cover the prize). But your card only has a 1/3 chance of being the red one. Thus the other card has a chance of ..... ? You may be sorry you switched of course, but we're talking odds and longrun average outcomes here, not the outcome of any single trial of the game.
Last edited by Mike Strand on Tue Jul 03, 2018 4:52 pm, edited 2 times in total.

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Re: Pick the Door with the Prize!
Back to the question of how much a gambling casino, which buys the prize, should charge a player of the threedoor problem to play each game:
vegetariantaxidermy pointed out the inconsistency of human players (if I'm not mistaken), which makes the odds hard to predict.
The casino accounts for human whimsy, however, by assuming the worst case for the casino (best case for the player): This is that player will always choose to switch choice of door. Now you can figure out how much the casino should charge the player in order to make a profit. Obviously, the casino still wants to attract players, so can't charge a huge amount  enough to cover the cost of the prize (on the average, since sometimes the player wins), and some amount of overhead. The casino is only interested in making money over many, many games.
By the way, is anyone ready and willing to state the odds (in the 3door game) of winning, if the player always switches doors?
vegetariantaxidermy pointed out the inconsistency of human players (if I'm not mistaken), which makes the odds hard to predict.
The casino accounts for human whimsy, however, by assuming the worst case for the casino (best case for the player): This is that player will always choose to switch choice of door. Now you can figure out how much the casino should charge the player in order to make a profit. Obviously, the casino still wants to attract players, so can't charge a huge amount  enough to cover the cost of the prize (on the average, since sometimes the player wins), and some amount of overhead. The casino is only interested in making money over many, many games.
By the way, is anyone ready and willing to state the odds (in the 3door game) of winning, if the player always switches doors?
Re: Pick the Door with the Prize!
The problem becomes clear once we explicitly state the assumptions.
* Monty knows what's behind each door.
* Monty must reveal exactly one door.
* Monty must reveal a door concealing a goat.
Now the puzzle is clear.
The player makes a choice.
1/3 of the time the player chooses correctly and the other two doors conceal goats. Monty reveals one of the goats. Switching picks the other goat and the player loses.
The other 2/3 of the time, the player has initially chosen a goat. There are two doors left, one concealing the car and the other concealing a goat. Monty must reveal the door concealing the goat. The remaining door contains the car. Switching wins the car.
So 1/3 of the time switching loses; and 2/3 of the time switching wins.
Let me repeat this. 2/3 of the time the player's initial guess is a door with a goat. That leaves one door with a car and one with a goat. Monty is required by the rules of the game to reveal the door with the goat. That means that switching guarantees we get the car. This case happens 2 times out of 3.
1 time out of 3 the player's initial guess is the door with the car. Switching loses.
So switching is the winning strategy in 2 out of 3 equally likely cases.
* Monty knows what's behind each door.
* Monty must reveal exactly one door.
* Monty must reveal a door concealing a goat.
Now the puzzle is clear.
The player makes a choice.
1/3 of the time the player chooses correctly and the other two doors conceal goats. Monty reveals one of the goats. Switching picks the other goat and the player loses.
The other 2/3 of the time, the player has initially chosen a goat. There are two doors left, one concealing the car and the other concealing a goat. Monty must reveal the door concealing the goat. The remaining door contains the car. Switching wins the car.
So 1/3 of the time switching loses; and 2/3 of the time switching wins.
Let me repeat this. 2/3 of the time the player's initial guess is a door with a goat. That leaves one door with a car and one with a goat. Monty is required by the rules of the game to reveal the door with the goat. That means that switching guarantees we get the car. This case happens 2 times out of 3.
1 time out of 3 the player's initial guess is the door with the car. Switching loses.
So switching is the winning strategy in 2 out of 3 equally likely cases.