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So the Monty Hall Problem itself is widely known and understood. Nonetheless, a friend of mine and I were wondering whether the the same strategy could affectively be applied by a participant of Who wants to be a Millionaire? when using the 50/50 Joker.

Let's imagine the following scenario: The participant P has no clue about the correct answer $ x \in \{A,B,C,D\} $ and wants to use the 50/50 Joker (eliminating two wrong answers). But instead of immediately going for it he first "preselects" one of the answers in his mind. There is no need to tell Quizmaster Q about his "imaginary preselection". Now P tells Q that he wants to use his joker and Q lets the computer eliminate two wrong answers.

(1) In case the answer P had preselected is eliminated he has no choice but to choose between the remaining two answers, effectively leaving him with a 50% chance of success - no magic happening here. (2) But what about the other case when the answer P had preselected survives the elimination? According to the Monty Hall Problemit seems as if changing the selection (i.e. choosing the other remaining option P had not preseleted) seems to give him a 0.75 chance of success.

Nevertheless, I find it hard to believe that this actually holds true, since the so called 50/50 (!) Joker would then not be p(success) = 0.5 after all. Additionally it seems unlikely that making an "imaginary preselection", no one else is told about, actually increases your odds.

I know this problem is not exactly the same as Monty Hall since the the quizmaster does not always eliminate answers only from the ones the participant had not "preselected", meaning that the preselection itself could be eliminated, too, as it happens in (1). Still the second case seems to actually be a just variation of it.

So are we right and making a preselection and then going for the other remaining option is a valid strategy that increases the participant's odds of winning? If not, please help us understand our misconception.

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3 Answers

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Monty Hall does not give you information about your preselection. Therefore the probability that your first choice is right given that it is still available after the intervention, is not changed. The 50/50 Joker does give information about it (esp. when it gets eliminated). Note that many of the misunderstandings of the Monty Hall problem arise from the missing assumption that the host always willfully opens a goat-door other than the preselected door. If you modify the Monti Hall problem so that the host opens any not preselected door at random, the general misconception that both remaining doors are "equal" becomes correct.

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If the contestant has no clue about the correct answer, then the "preselection" can change nothing. On the other hand, if the contestant has information about the probabilities of the four answers, then depending upon which answers are shown the player may gain considerable information.

For example, suppose the player estimates the probabilities at 10%, 20%, 30%, and 40%. If after two answers are eliminated, those that remain are the ones the contestant thought was most likely (the worst case), the more likely of the remaining answers will be correct 4/7 of the time. If the remaining answers are those the contestant thought were the most likely and the least likely (the best case), the more likely of the remaining answers will be correct 4/5 of the time. Other combinations of eliminated answers will yield intermediate probabilities.

BTW, from watching the show, I suspect the best strategy, if one can do it convincingly, would be to pretend that one was dithering between two answers which one believed to be most likely. It appears that the "50/50" doesn't select randomly, but tries to include the wrong answer the contestant favors most highly. Thus, a contestant who convincingly claimed to believe that a low-probability answer was correct might be able to coax the 50/50 into offering up one of the most favorable scenarios.

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No. I don't believe applying the Monty Hall strategy to this problem should increase the probability of winning.

Let's assume that the computer chooses an incorrect answer to eliminate at random (i.e. each wrong answer is equally likely to survive with probability 1/3).

By switching, there are 2 cases in which you can win:

• Your choice is wrong AND survives the elimination. (E1)
• Your choice is wrong AND it doesn't survive AND you guess correctly. (E2)

$$P(win) = P(E1) + P(E2)$$ $$= P(wrong \, \cap \, survives) + P(wrong \, \cap \, eliminated \, \cap correct \, guess)$$ $$=P(wrong)*P(survives \, | \, wrong) + P(wrong)*P(eliminated \,| \,wrong)*P(correct \, guess | eliminated \, \cap \, wrong)$$ $$=\frac{3}{4}\frac{1}{3} + \frac{3}{4}\frac{2}{3}\frac{1}{2} $$ $$=\frac{1}{4} + \frac{1}{4} = 0.5 $$

Thus, making a choice in your head then switching if it survives the elimination will NOT increase your chances of winning.

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