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On a game show, there are three doors, behind one of which is a prize. I choose a door and the host opens one of the other doors that has no prize behind it. I get to switch my door choice if I wish.

Now suppose we have three positive numbers $p_1$, $p_2$, $p_3$ such that $p_1+p_2+p_3=1$ and the prize is behind door $i$ with probability $p_i$. By labeling the doors suitably we can assume $p_1>p_2>p_3$. Assume that you know the probabilities $p_1$, $p_2$, $p_3$ associated to each door. What is the strategy that maximizes my chances of winning the prize?

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Teta K
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    When Monty has a choice between two doors to open, how does he decide which one to open? The answer to your question depends on the answer to mine. – Gerry Myerson Feb 07 '19 at 10:13
  • I assume he won't open the door with the prize. – Teta K Feb 07 '19 at 10:29
  • I think, the best strategy depends on the concrete probabilities. If $p_3$ is very small, for instance, a very good chance is to choose that door and switch. If $p_1$ is very large, a good idea (not sure whether always the best idea) is to choose number $2$ and to switch only if $3$ is opened. – Peter Feb 07 '19 at 10:45
  • This question is well known and the solution while counter intuitive to most has been covered here on this site: What's wrong with this equal probability solution for Monty Hall Problem? and many other places on the internet. – Warren Hill Feb 07 '19 at 12:33
  • Possible duplicate of What's wrong with this equal probability solution for Monty Hall Problem? – Warren Hill Feb 07 '19 at 12:33
  • The point, Teta, is that if you have picked the door with the prize, then Monty has to decide which of the other two doors he will open. How does he make that choice? – Gerry Myerson Feb 07 '19 at 22:53
  • @Warren, please convince me that the current question, where the a priori probabilities for prize locations are assumed not to be equal, is a duplicate of that earlier question. – Gerry Myerson Feb 07 '19 at 22:56
  • So, Teta, I repeat: when Monty has a choice to make, how does he choose? – Gerry Myerson Feb 09 '19 at 02:19
  • @GerryMyerson It's true that the probabilities of Monty's actions can affect the answers to question such as, "If I chose door 1 and Monty opened door 2, what's the probability the car is behind door 3?" But I have been unable to come up with any circumstances in which this allows the existence of a better strategy than "choose the least likely door, always switch." Do you have an example in mind? – David K Feb 09 '19 at 23:03
  • @David, no, not a better strategy, just different odds. If Monty always opens the higher-numbered door when he has a choice, and you choose door 1, and Monty opens door 2, that means he couldn't open door 3, so you win 100% of the time by switching. – Gerry Myerson Feb 09 '19 at 23:10

2 Answers2

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The usual form of the "Monty Hall" question asks for the probabilities of the car being behind each of the doors, conditioned on the fact that the contestant initially chose a particular door and host opened a particular door. Those probabilities cannot be computed unless we know (or make some assumptions about) the probability that the host will open a particular door when two doors are eligible to be opened.

Since the question asked here does not say that any doors have been opened yet, I take it that the question is only about the strategy that gives the best chance to win under the condition that the game is about to start, and that the strategy can specify both how to pick the door to choose initially and how to react (switch or stay) depending on which door the host opens.

Start by choosing door 3. After the host opens a door, always switch.

David K
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The usual interpretation on this problem is that Monty knows where the car is.

You pick a door. The probability that the car is behind it is $\dfrac{1}{3}$.

The probability that it is behind one of the other two is $\dfrac{2}{3}$. Since there are two goats and Monty knows where the car is he will always show you a goat. The probability that the car is behind the remaining door is $\dfrac{2}{3}$. You are twice as likely to get the car if you swap.

Warren Hill
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