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## Main Question or Discussion Point

Hello,

I'm sure you are all well familiar with the monty hall problem, so i wont restate it.

there is also a similar problem: the Counterfeit coin problem.

A reminder of the problem:

Assume that you’re presented with three coins, two of them fair and the other a counterfeit that always lands heads. If you randomly pick one of the three coins, the probability that it’s the counterfeit is 1 in 3. This is the prior probability of the hypothesis that the coin is counterfeit. Now after picking the coin, you flip it three times and observe that it lands heads each time. Seeing this new evidence that your chosen coin has landed heads three times in a row, you want to know the revised posterior probability that it is the counterfeit. The answer to this question, found using Bayes’s theorem (calculation mercifully omitted), is 4 in 5. You thus revise your probability estimate of the coin’s being counterfeit upward from 1 in 3 to 4 in 5.

it is easy to show the the probability afterwards is 4 to 5, but then i propose a new problem,

an incest between the monty hall problem and this problem.

suppose after you tossed your coin 3 times, the presenter of the coins tells you that C(one of the coins you didn't choose) is a real coin, and proposes you to know switch to the other coin(coin B, supposing you chose coin A).

Now, it is obvious that you're not supposed to switch, but i had an argument with few mathematicians about the probability of your choice.

In my opinion, the probability that i now hold the counterfeit coin is less than it was before the revelation about coin C. And by baye's theorem i get it's 2\3.

However, whatever reasoning i had for my solution was not convincing as others' reasoning, they said that the probability DOES NOT change:

just like monty hall's where the probability that door A has a car is 1\3, and the other two together is 2\3, and one of them is revealed the other one stays with the 2\3, and your doesn't change.

I can see how this reasoning holds, but my nay's theorem and some intuition that this problem is different when the probabilities in the first place aren't equal, I think I'm right.

Any help with this will be appreciated :)

Thanks!

I'm sure you are all well familiar with the monty hall problem, so i wont restate it.

there is also a similar problem: the Counterfeit coin problem.

A reminder of the problem:

Assume that you’re presented with three coins, two of them fair and the other a counterfeit that always lands heads. If you randomly pick one of the three coins, the probability that it’s the counterfeit is 1 in 3. This is the prior probability of the hypothesis that the coin is counterfeit. Now after picking the coin, you flip it three times and observe that it lands heads each time. Seeing this new evidence that your chosen coin has landed heads three times in a row, you want to know the revised posterior probability that it is the counterfeit. The answer to this question, found using Bayes’s theorem (calculation mercifully omitted), is 4 in 5. You thus revise your probability estimate of the coin’s being counterfeit upward from 1 in 3 to 4 in 5.

it is easy to show the the probability afterwards is 4 to 5, but then i propose a new problem,

an incest between the monty hall problem and this problem.

suppose after you tossed your coin 3 times, the presenter of the coins tells you that C(one of the coins you didn't choose) is a real coin, and proposes you to know switch to the other coin(coin B, supposing you chose coin A).

Now, it is obvious that you're not supposed to switch, but i had an argument with few mathematicians about the probability of your choice.

In my opinion, the probability that i now hold the counterfeit coin is less than it was before the revelation about coin C. And by baye's theorem i get it's 2\3.

However, whatever reasoning i had for my solution was not convincing as others' reasoning, they said that the probability DOES NOT change:

just like monty hall's where the probability that door A has a car is 1\3, and the other two together is 2\3, and one of them is revealed the other one stays with the 2\3, and your doesn't change.

I can see how this reasoning holds, but my nay's theorem and some intuition that this problem is different when the probabilities in the first place aren't equal, I think I'm right.

Any help with this will be appreciated :)

Thanks!