Does Revealing a Real Coin Change Odds in the Modified Monty Hall Problem?

[Dweirdo]

Hello,

I'm sure you are all well familiar with the monty hall problem, so i won't 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!

 

[Boorglar]

I think you are right that you are not supposed to switch. But I believe the probability of having the counterfeit without switching is still 4/5. Here's my reasoning (although I admit I might be wrong):

As you said, after throwing 3 heads in a row, there's 4/5 chances of having taken the counterfeit. So obviously, there is 1/5 chances of NOT having taken the counterfeit. Thus, there is only 1/5 chance of coin B to be counterfeit (assuming you chose A, and assuming there is only one counterfeit coin among A, B, C). So switching coins would give you only 1/5, which is less than 4/5.

The difference with the Monty Hall problem is that in Monty Hall, you don't get that additional information from throwing the coin three times. The announcer simply opens another door which contains no prize. In that case, you know the probability of door A (your 1st choice) is 1/3, and probability of door B is 2/3 (because there is 2/3 chances for you NOT to have chosen the door with a prize).

 

[eigenperson]

I'm going to assume that, just like in the Monty Hall problem, the host's decision to reveal a non-counterfeit coin is not conditioned on your actions, and that if both remaining coins are real, the one revealed by the host is selected at random.

The probability that you have the counterfeit coin is still 4/5, so you should not switch (assuming the point is to find the counterfeit coin). This is because the host's choice clearly gives us no information (he was going to reveal one of the two coins no matter what).

If you're skeptical, use Bayes' theorem again. Let C be the event that the coin you picked was the counterfeit. Let R be the event that the host reveals one particular coin (choose it in advance) of the two remaining ones. Then suppose event R happens; we want to know P(C|R). This is ${P(R|C)P(C) \over P(R|C)P(C) + P(R|\overline{C})P(\overline{C})}$. We already agreed that $P(C) = 0.8$ and it is easy to see that $P(R|C) = P(R|\overline{C}) = 0.5$. We get P(C|R) = 0.8, as claimed.

 

[Dweirdo]

Thank you for answering!

i've redone my calculations and you are right, and it can be shown that no matter what is the probability of your first choice, it won't change