Monty Two Face

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong.  The correct answer for the Reverse Monty Hall Problem is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).  Note also that I has already accepted in this article that I had been wrong, but even in that acceptance of being wrong, I was wrong.  But I'm right now ... just not in what is written below.

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When I wrote this article, I thought 1) that it was right and 2) the scenario describedt was analogous with the Reverse Monty Hall Problem.  I was certainly wrong with 1) and if so, I might well be with 2).  I will need to think carefully though to see if it the below is analogous with the Reverse Monty Hall Problem, because if it is (which I have come to doubt) then I am wrong about the Reverse Monty Hall Problem.

Interestingly, I came to believe that I am wrong about what follows via conditional probability which I am told is not applicable to the Monty Hall Problem - so this is an argument in support of 2) being wrong.

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Say that Monty has two coins, one is a standard fair coin with a head and a tail.  The other is a fair coin, in that it is balanced so as to land on either side with equal likelihood, but it has a head on both sides.

Say that Monty selects a coin at random and he tosses it and he records the result.  He repeats this process nineteen times to get the following result (random tosses generated by a spreadsheet):

Table 1

Now, say that we want to look only at the instances in which there might be confusion as to which coin, Monty tossed.  We would have to eliminate all the times that tails was tossed.  So we have:

Table 2

What is the likelihood that, given that Monty tossed a coin and got heads that it was the HH coin?  The answer is 2/3.  In the relative small sample we have here, it looks like this (with HH in red):

Table 3

Heads appeared 15 times, and the HH coin was responsible for that appearance 9 times.  The fact that each of these events were independent and the sample size is small make it quite likely that the HH coin would have been observed somewhere between 8 and 12 times.  With a sample size of 500, the result was still closer to 0.7 than to 0.66.

However, if Monty stopped on the first run, what would have been the likelihood that the HH coin was used, if a heads had resulted?  Sure the result was a heads, but I’m not eliminating the possibility of there having been a tails.  I’m just forcing the scenario to one in which there was a heads.

In this scenario, we only know that Monty selected the coin at random from a choice of two.  We must conclude that the likelihood that he picked that HH coin is 1/2.  This is because the sample space we would need to consider, from multiple iterations is that in Table 1, rather than that in Table 2.

                                                                                                                           

An analogous situation applies in the Reverse Monty Hall Problem.  In the Red Mary scenario (as introduced in Marilyn Gets My Goat), HH is equivalent to the car being behind the Green Door and HT is equivalent to Ava being behind the Green Door.  In the latter case, Monty could have chosen to open the Green Door to reveal a Green Ava, but he didn’t.

This is presented in order to explain how, in a single run, we can get a 1/2 result while getting a 2/3 result over multiple runs.  Those arguing for a 2/3 result in a single iteration, one shot instance version of the game do so by incorrectly dealing with the possibility that Monty could have opened the other door if two goats have been selected by the contestant.

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Basically, I overreached.  I was trying desperately to think of a scenario that could help explain a feature of the Reverse Monty Hall Problem, I thought that I had found one and I burst into print without mulling it over sufficiently.

The likelihood of the coin being a HT coin is given by Pr(HT|H) = Pr(HT ∧ H) / Pr(H) where:

    Pr(HT|H) = the likelihood of the coin being an HT coin given that we see a heads

    Pr(HT ∧ H) = the likelihood of the coin being an HT and our seeing a heads = 1/4

    Pr(H) = the likelihood of seeing a heads = 3/4

    So Pr(HT|H) = 1/3 and Pr(HH|H) = 2/3

We can calculate Pr(HH|H) directly too:

    Pr(HH|H) = the likelihood of the coin being an HH coin given that we see a heads

    Pr(HH ∧ H) = the likelihood of the coin being an HT and our seeing a heads = 2/4

    Pr(H) = the likelihood of seeing a heads = 3/4

And so, Pr(HH|H) = 2/3

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ChalkboardCowboy has a comment below.  Here is a table that is relevant to my response:

Note that MC = Magic Coin (which is simulated to provide Heads with a likelihood of 1/10^12, using an American trillion) and NC = Normal Coin (which is simulated to provide Heads with a likelihood of 1/2).

If I see a heads, I am going to be pretty confident that we have a normal coin and not ChalkboardCowboy's magic coin.

(Note while I was wrong above, ChalkboardCowboy was also being a little silly with his magic coin example.  That said, no harm done, I found my error eventually - largely thanks to "drip" who forced me to go over the workings that I should have done from the start.)

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The dice do actually model something rather close to the Reverse Monty Hall Problem, so not only was I wrong, I was wrong about why I was wrong.  Since I now conform with the majority opinion, if I am now wrong again about being wrong about why I was wrong, I am at least wrong in the same way as almost everyone else.