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Coin flip variation?

  1. Mar 2, 2014 #1
    Hello. A friend sent me the following problem that she wants to include in an essay:

    There are two epistemic peers whose mental faculties are of equal standing and who have access to all the same relevant evidence.
    The two go to dinner

    Case 1:
    One week later, the first man, A, states that their dinner consisted of Chinese food
    The second, B, disagrees and states that their dinner consisted of Indian food

    Case 2:
    One week later A, states that their dinner consisted of Chinese food
    B is never consulted on the issue

    In each of these cases there is some possibility of one or more misleading events occurring which would cause one or both parties to misremember their meal and to hold an erroneous belief about the subject.
    Is the chance of a misleading having affected A the same in each case?
    If not, how and why do the probabilities differ?


    I've thought about it and have concluded that it is essentially a variation on a (weighted) coin toss problem. The probability of either person being correct is analogous to that of a weighted coin showing (say) heads, since there are only two outcomes (incorrect or correct) that would correspond to the two outcomes of the coin (heads or tails)

    So, let's assume that A and B are both 80% correct, usually. That would correspond to a coin being 80% likely to fall on heads. If this is the case, then in the first case, since it has been revealed that the pair have conflicting answers, we can conclude that one of them is wrong (that one of them has flipped a tails), and they can therefore both not be correct. The crucial point, I think, is that we cannot tell which one is wrong. In the coin flip, the flippers could only be notified that they have both flipped differently, and not the specifics as to who flipped heads and who flipped tails. Since they are both equally likely to be correct/heads (at 80%), then it is impossible to tell which is mistaken and so we must use the base rate of 80%. In other words, it makes no difference, and A is 80% likely to be correct in both cases.

    Is this reasoning sound? Any help is greatly appreciated.
     
  2. jcsd
  3. Mar 2, 2014 #2

    PeroK

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    You have to be careful of your hypotheses in these situations. So, I would clarify your assumption as follows:

    The probability that someone correctly remembers the meal they had is 80%.

    In the second case, all you know is that A recalls Chinese food. So, it's 80% that he is correct.

    In the first case, one of them must be wrong, possibly both. So, you have to look at the options:

    64% of the time they'd both be correct.
    16% of the time A would be correct and B wrong
    16% of the time B would be correct and A wrong
    4% of the time they would both be wrong

    Approximately (* see below), all you know is that they can't both be correct. So, the probability that A is correct is 16/36.

    * There is a twist, as it depends on how many options there are. Suppose there were only Chinese and Indian food, then they can't both be wrong. So, the probability A is correct would be 50%.

    Suppose there are lots of options, then the result is very close to the 16/36 given.

    Somewhere in between, you'd have to calculate the probability that if they are both wrong they choose different options.

    In other words, the 4% will actually vary from 0 to 4% depending on how many answers are possible.
     
  4. Mar 3, 2014 #3

    FactChecker

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    I have to disagree with you.

    Case 1: Knowing that one of the two is wrong will increase the odds that A is wrong. In Case 1, you know that at least one is wrong and so it is over 50/50 that A is wrong (if you knew that only one is wrong, it is 50/50 that A is wrong, but you have to add some probability that both are wrong). For the exact answer Baye's Rule should be used.

    Case 2: There is no reason to think that A is wrong. The 80% is correct.
     
    Last edited: Mar 3, 2014
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