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Author | Topic: Three Curtains | |||||||||||||||||||||||
Capt Stormfield Member Posts: 429 From: Vancouver Island Joined: |
If the rules allow the host to choose, then we introduce factors which are not knowable to us. Are they over budget this week? Has the show run out of time? All fascinating gaming I'm sure, but not amenable to probabilistic analysis.
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Dr Adequate Member (Idle past 314 days) Posts: 16113 Joined: |
The original question is meaningless unless we know the structure of the game. We do.
If the host has knowledge, and gets to use it in the choice of which unchosen curtains are opened ... He does, that's how the game works. He knows where the car is, and will never reveal it. Hence if you didn't pick the curtain with the car behind it, then the one curtain which you didn't pick and which he didn't draw back must have the car behind it. Edited by Dr Adequate, : No reason given.
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PaulK Member Posts: 17828 Joined: Member Rating: 2.3 |
Yes, the strategy followed by the host matters.
In the standard problem it is assumed that the host ALWAYS reveals one of the prizes, and never reveals the car. In that case the solution is as has been stated. Observing the show is the only way to work out if there is a strategy, and what it might be.
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Capt Stormfield Member Posts: 429 From: Vancouver Island Joined: |
You are correct. I had not internalized the concept that the host was not randomly choosing which curtain to open.
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Modulous Member Posts: 7801 From: Manchester, UK Joined: |
If the rules allow the host to choose, then we introduce factors which are not knowable to us. They don't. Replace him with a robot you personally programmed if you prefer.
Are they over budget this week? Has the show run out of time? And it is occurring in a lab rather than on TV.
All fascinating gaming I'm sure, but not amenable to probabilistic analysis. Pedigrees, Prizes, and Prisoners: The Misuse of Conditional Probability, Journal of Statistics Education Volume 13, Number 2 (2005), Matthew A. Carlton
quote: 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').
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Minnemooseus Member Posts: 3945 From: Duluth, Minnesota, U.S. (West end of Lake Superior) Joined: Member Rating: 10.0 |
He does, that's how the game works. He knows where the car is, and will never reveal it. Hence if you didn't pick the curtain with the car behind it, then the one curtain which you didn't pick and which he didn't draw back must have the car behind it. Monty knows where the car is, but he isn't obligated to show a non-car selection ("window"). I have no idea how often Monty did not do a reveal. I imagine that someone must have gone through all or at least a bunch of "Price is Right" episodes, and compiled the curtain game results. It would be interesting to see the statistics of how Monty chose or didn't choose a window to reveal. etc. By the way, the Monty Hall thing is included in "Car Talk's" list of hardest puzzlers:http://www.cartalk.com/...-stuff-car-talks-toughest-puzzlers How they presented the problem:http://www.cartalk.com/content/monty-hall-puzzler-12?ques... And how they presented the solution:http://www.cartalk.com/content/monty-hall-puzzler-12?answer quote: I haven't now reviewed upthread, but my recollection is that the decision was that the odds improve from 1-in-3 to 1-in-2 (not 2-in-3). The mathematical proof stuff Car Talk presents:http://www.cartalk.com/content/monty-hall-puzzler-0 Moose
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NoNukes Inactive Member |
I'm not sure, but I bet Michael Larson could have figured it out. Perhaps he could have. Michael Larson does not appear to have had any special mathematical talent. But perhaps it is possible to figure out the odds by watching a few dozen episodes of the show and simply tallying the success rates of people who changed their choices versus those of people who stick. In any event, the solution to the problem is well known by now, and this is at least the second time we have discussed the Monty Hall problem in these forums. Dr. Adequate's explanation using the 100 doors is the clearest, most succinct one I've ever read.Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846) I have never met a man so ignorant that I couldn't learn something from him. Galileo Galilei If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass
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PaulK Member Posts: 17828 Joined: Member Rating: 2.3 |
For analysing the problem, I'd say that the more interesting questions are
"is the car ever revealed?" and "what relationship, if any, is there between the choice to reveal and the contestant choosing correctly" If the presenter has a strategy, it can be exploited.
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rstrats Member (Idle past 132 days) Posts: 138 Joined: |
I don't know why Monty Hall is being brought into the conversation. The OP is nothing like the Monty Hall game. That game uses doors, whereas the OP game uses curtains. Also, the Monty Hall game has goats or some other undesirable "prize" behind two of the doors, whereas the OP game has nothing behind two of the curtains. The OP game is a one time occurance. It says that the host knows where the car is located so as not to end the game prematurely. From that it should be obvious that he will not open that curtain and so will not be choosing at random.
Edited by rstrats, : No reason given. Edited by rstrats, : No reason given.
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Modulous Member Posts: 7801 From: Manchester, UK Joined: |
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?
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PaulK Member Posts: 17828 Joined: Member Rating: 2.3 |
Curtains or doors, goats or nothing - those are details which don't change the mathematics of the problem, and looking at the general problem will help understand this one.
If this is a one-time occurrence then we do need to know if the host is required to offer the choice to switch, and if not what criteria he uses to decide to offer the choice or not. Without that the problem is not solvable.
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rstrats Member (Idle past 132 days) Posts: 138 Joined: |
PaulK,
re: "If this is a one-time occurrence then we do need to know if the host is required to offer the choice to switch..." Required or not is irrelevant to the OP. In this specific game, the OP says that the host does offer the chance to switch. The question is in regard to this specific game and not to any future ones.
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PaulK Member Posts: 17828 Joined: Member Rating: 2.3 |
Then the question is unanswerable. Switching could be advantageous, it could be disadvantageous, There is no way to tell.
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Modulous Member Posts: 7801 From: Manchester, UK Joined: |
If the presenter has a strategy, it can be exploited. 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}
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PaulK Member Posts: 17828 Joined: Member Rating: 2.3 |
I was speaking loosely, but this is a rather different game. Switching has to be better or worse or the same as not switching. If it's the same (worst case for the contestant) then it's still a 50-50 guess. And even that's better than the 1/3 odds before revealing.
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