Three Curtains

There is a 1/3 chance you picked the right answer.There is a 2/3 chance you picked the wrong answer. Therefore there is a 2/3 chance the car is behind #2 OR #3Once you are given the additional information it is not behind #3 the state of affairs remainsThere is a 1/3 chance you picked the right answer and 2/3 chance you picked the wrong answer. It is not #3, therefore there is a 2/3 probability it is behind #2 and a 1/3 probability of being behind #1.Therefore, switching doubles your chances of winning a car to 2 in 3.It is functionally equivalent to being given the choice:You can either open one curtain or two, which do you choose?

I'll show you. We'll do every combo possible, which all occur with equal frequency. You pick 1 every time and you stick.

1 2 3
C
​
1 2
C
​
You win! 1-0 for sticking
​
1 2 3
C
​
1 2
C
​
You lose! 1-1 for sticking
​
1 2 3
C
​
1 3
C
​
You lose 2-1

Alternatively, always switching

1 2 3
C
​
1 2
C
​
You lose! 1-0 to sticking
​
1 2 3
C
​
1 2
C
​
You win! 1-1 to sticking
​
1 2 3
C
​
1 3
C
​
You win! 2-1 to switching!

As I said, you are essentially being given the choice of picking 1 OR picking 2 AND 3. The puzzle is just given window dressing designed to bamboozle the way the human mind naturally estimates probability. Basically you think that because there are two possibilities that means they must be equiprobable, but this assumption is not always true, such as in this case where the host knows which curtain the car is behind and simply reveals the one option you didn't pick that is a loser. In the case where the host always reveals 3 when you always pick 1, then of course you should only switch 1/3 of the time, otherwise it doesn't matter.This is the illustration I like to use to demonstrate how professional poker players are able to make a living - and how the people they make money off are just as confident in their estimations of probability as their sharks. You've probably seen people who think the probability of God existing is 50% because he either does or he doesn't. But then, the chances of winning the lottery are the same by the same reasoning.Edited by Modulous, : No reason given.

Well its the rules that govern the game show host are what determines the probabilities. If it helps, replace the host with a computer program that always picks curtain which does not have a car behind it and was not the one you picked. The OP describes the scenario as you hypothetically perceive it. If you detect some communication, tell me - which curtain is the car behind?Edited by Modulous, : No reason given.

They don't. Replace him with a robot you personally programmed if you prefer. And it is occurring in a lab rather than on TV. Pedigrees, Prizes, and Prisoners: The Misuse of Conditional Probability, Journal of Statistics Education Volume 13, Number 2 (2005), Matthew A. Carlton

quote:

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Let C denote the event that the car is behind Door #2; the a priori probability of C is P(C) = 1/3. Let D denote the event that Monty opens Door #3

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P (C | D) = (P(C) P(D|C)) / P(D) = 2/3

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Obviously, if you take into account some of Hall's variants such as immediately opening a door sometimes when he knew the contestant had guessed wrong, or using money incentives to switch or not to switch including reverse psychology, the probabilities become difficult to predict (but long term analysis of Hall's behaviour may lend to estimating the probabilities taking into account the frequency of the variants etc), and if you take into account that the whole thing could be scripted, or there could be a car behind two of the curtains or fixed in some other way... well then you're right we've left the realms of simple mathematical analysis.Alternatively, we could just look at it like we do 'The Prisoner's Dilemma' without adding 'underworld retribution', or adding quantum physics to 'Zeno's paradox' etc etc. Even as such considerations may be interesting discussions in their own right (such as the 'Iterated Prisoner's Dilemma').

So a serial killer kidnaps three people. Alice, Bob and Eve. The serial killer says he is going to kill two of them and let the other go. He knows already who he is going to kill, but he won't tell them.However, in a moment of desperation, Alice begs that he tell her the identity of at least one of his intended victims. 'If Bob is to go free, tell me Eve. If Eve is to be freed, tell me Bob. If I am to be freed, toss a coin to decide which one you tell me is to be killed'. The serial killer subsequently tells her 'Bob will be killed'.Alice shares the news with the other victims.
Bob is mortified.
Alice and Eve are concerned for Bob, but they both seemed pleased with the news, estimating their chances of survival went up.Assuming the killer told the truth and followed the protocol, whose feelings are justified?

I'm not sure what analysis you have done on this game, or seen done, but that's not true as a general rule.Suppose we're playing a card game. The game is I draw a card from a 42 card deck (I've removed an 8 and a 9 and some random other cards that are not relevant). If I get a 5, 6, 7, 8 or 9 I win, otherwise I lose. I'm an underdog in this game right? I have only 18 winning cards and 24 losing ones. Let's say we both bet $100 before the card is drawn and I get to elect to bet after drawing the card, but only I get to see what the card is until after the betting.If I bet $100 every time, and you called every time - you'd make money. You win $200 24 times and lose $200 18 times.
If I bet when I'm winning, and you fold when I bet - then you'd make money from the bets made pre-draw and wouldn't lose any money post draw. You win when I don't bet 24 times and lose when I do 18 times.But what if I adopt a mixed strategy?Suppose I bet when I pick up a winning card and I also bet if I get the Queen of Hearts.You should still fold when I bet because the odds of my bluffing are 18-1, but this immediately improves things for me as I win 19 times instead of 18.But let's say I pick 5 random cards to also bluff on (All the Queens and the Jack of clubs, say)I now bet 23 times, 18 times with the best. 5 times as a bluff. I could even tell you what I am doing and you'd still be in trouble.The odds of me bluffing are 18 to 5 or 3.6 to 1. There are $200 in pre-draw bets and my $100 bet, totalling a potential win of $300 for you. But you have to bet $100 which means you are being given odds of 3 to 1 that I am bluffing. But the odds of my bluffing are nearly 4 to 1. Though I went in at a probabilistic disadvantage, I adopted a strategy that you cannot defeat within the confines of the game.{Example adopted from David Sklansky, The Theory of Poker}

Actually depending on the strategy in play, switching can be worse than 50/50 for the contestant. At the extreme and boring end, imagine the host only ever offers the switch in the cases where you picked the car. Now switching is a losing proposition every time.If the host is randomly opening one of the three doors, then the strategy becomes:
1/9 host opens the door you picked AND its a winner (stick)
2/9 he opens your door and its a loser (switch)
1/9 he opens door 2 AND it's a winner (switch)
2/9 he opens door 2 AND it's a loser (odds go up, but switching is irrelevant.)
1/9 he opens 3 AND winner (switch)
2/9 he opens 3 and loser (doesnt matter)here, you win 1/9 (x3) of the time = 1/3 of the time outright. And you win 50% of the time the remaining. Your general odds going in are better than 50/50.If he randomly opens the two non picked doors
1/3 he opens a door and it is a winner (1/6 for each door)
2/3 he opens a door and it is a loser.So in this example your odds go up because 1/3 of the time you win outright and 2/3 you have a 50/50 shot, meaning your odds are better than 50/50And we know if he deliberately selects one of the remaining losing doors each time - you win 2/3 of the time by switching 100% of the time.Therefore, if you adopt the correct strategy you do a bit better than 50/50 if any switch is offered.Unless the host can choose to not offer a switch. In the cases where he does this, you lose 2/3 of the time if he chooses at random. If he selects to do this at whim he can make you lose 2/3 of the time or less (ie the times he offers a switch he improves your odds).We also note, that in complete random selections - switching is sometimes worse than 50/50 (1/9 of possibilities where he opens your door and it is a winner).

I'm also trying to put the different strategies out there and in numbers, just to make it more interesting than a standard Monty Hall discussion.1/3 of the time you pick right first time.
But Monty offers you to switch 2/3 of the time (2/9) in this case and you auto win the rest (1/9) of the time2/3 of the time you pick wrong. Monty offers to switch 2/9 of these times, but not the other 4/9ths.Switch option: 2/9 + 2/9 = 4/9ths.
No switch: 5/9ths.If you always switch then 50% of the time you picked right first time and lose.
If you never switch then 50% of the time you picked wrong first time and you lose.That is to say, switching makes no difference,
you win 2/9ths
you lose 2/9thsWhen you don't get a choice, you win 1/9, and lose 4/9ths.Thus you win 3/9ths (1/3), and you lose 6/9ths (2/3).I think that's the best strategy for the host which involves occasional revelation of non-chosen losing options coupled with the choice to switch(which is no different than just picking a random curtain/door) .