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Guessing coin flip paradox

  1. Sep 10, 2010 #1
    A warm up problem 1

    Somebody flips two coins on table, and then hides them with some two pieces of clothing.

    You observe, when one of the coins is revealed, and it is heads or tails. You are then asked to guess what's the side of the unrevealed coin.

    Which ever answer you give, you are right and wrong with probability 1/2. This is because the unrevealed coin is independent from the revealed one. Right?

    A warm up problem 2

    Two coins are again flipped and they are both hidden with some curtain.

    You are told that "one of the coins is heads." Now you should guess that the other coin is tails. This way you are right with probability 2/3, and wrong with probability 1/3. This is because in the beginning there was probability 1/4 that both are heads, 1/4 that both are tails, and 1/2 that one is head and one is tails. When you receive the information that one the coins is heads, then the corresponding conditional probabilities are going to be 1/3, 0 and 2/3. Right?

    The game

    One hundred pairs of coins are flipped, and each pair is hidden with a curtain, so that we get one hundred "curtain spots". You will go through all of these curtain spots, and you are always told truthfully that "one of the coins is heads" or "one of the coins is tails", and you are supposed to guess what the untold coin is.

    If you guess correctly, you are given a dollar (or euro), and if you guess incorrectly, you lose a dollar (or euro).

    A winning strategy

    We know that if, at every curtains spot, you always guess that the untold coin is different from the told one...

    meaning that if you are told that "one of the coins is head", you answer "the other one is tails", and
    if you are told that "one of the coins is tails", you answer "the other one is head"

    ... then you will always be right with probability 2/3. This means that there is going to be approximately 67 spots where you win, and 33 spots where you lose. You should make approximately 34$ (or 34€) profit with this strategy.

    But the winning strategy doesn't work!

    There is going to be approximately 25 spots where both coins are heads, 25 spots where both coins are tails, and 50 spots where one coin is head and one is tails. So with the winning strategy, there is going to be approximately 50 spots where you win, and 50 spots where you lose, the profit will be approximately 0$ (or 0€).

    How is it possible, that you will only win at approximately 50 spots, out of 100, despite the fact that at every single spot, your winning probability is 2/3?
     
  2. jcsd
  3. Sep 10, 2010 #2

    russ_watters

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    your answer to #2 is wrong: the scenario is identical to #1.
     
  4. Sep 10, 2010 #3
    No! No!

    Suppose we can say that one of the coins is a "coin on the left", and the other one is "coin on the right".

    In the warm up problem 1 we know that for example the coin on the left is heads, and then you are supposed to guess the coin on the right.

    But in the warm up problem 2 you are not told which coin is checked. You are told that "one of the coins is heads", but it could be the coin on the left or the coin on the right. It is not an identical situation.
     
  5. Sep 10, 2010 #4
    Look at it this way.

    The sample space of the new problem

    {HHH, HHT, HTH, TTH, THT, TTT} with the relevant probabilities being
    {1/4, 1/8, 1/8, 1/8, 1/8, 1/4}

    *The first H or T is what you are told.

    When you are unsure of whether you are told heads or tails, the probabilities change.
     
  6. Sep 10, 2010 #5

    Hurkyl

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    Warm-up problem 2 is misleading at best: "one of the coins is heads" is a flat-out lie 1/4 of the time.
     
  7. Sep 10, 2010 #6

    russ_watters

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    Riddle: two coins add up to 30 cents and one of them isn't a nickle......

    .......but the other one is!

    Its just a word game. If one coin is known to be heads it doesn't matter which coin it is. The other only has a 50% chance of being tails. By not saying which one is known you are using a word game to invoke gambler's fallacy.
     
  8. Sep 10, 2010 #7
  9. Sep 10, 2010 #8
    Read the last paragraph of question 2. That is essentially what is happening here.
     
  10. Sep 10, 2010 #9

    russ_watters

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    Could you please clarify whether you are trying to test/trick us here or if you honestly don't understand how this APPARENT paradox is resolved.

    I get very annoyed when I feel like I'm being trolled.
     
  11. Sep 10, 2010 #10

    Office_Shredder

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    Russ, the answer to problem 2 is correct, it just doesn't apply to the third one in the way that it is intended to make you believe it does (as has been noted by a couple poster above).

    You can try flipping two coins a thousand times and try it yourself
     
  12. Sep 10, 2010 #11
    When I wrote the opening post, I didn't feel like understanding the solution myself. So I came here to seek the answer.

    However, as usual, at the same time while I wrote the problem, I started to think about it more. I think I got a good hold of this now.

    Or I guess it would be safer to say that the warm up problem 2 can be interpreted so that it is correct. And IMO it is the most natural interpretation. (Alternative interpretations would be ad hoc and unreasonable.)

    The problem is that when there is a person who tells you that "one of them is heads", you cannot really know what rules the person is obeying. It would be clearer to state the "somehow you obtain the information that at least one of them is heads".

    I mean that if somebody criticizes the warm up problem 2 by saying that it is vague, that criticism would be justified. IMO it would not be justified to say that the problem 2 would be plain wrong.
     
  13. Sep 10, 2010 #12
    The real deception is here:

    There is no such thing as "the" game, in the opening post!

    Version 1 of the game:

    The game host will follow the following rules:

    If he is at a spot where there are two heads, he will say "one of them is heads" with probability 1 (so he doesn't lie).

    If he is at a spot where there are two tails, he will say "one of them is tails" with probability 1.

    If he is at a spot where there are one heads and one tails, he will say "one of them is head" with probability 1/2, and he will say "one of them is tails" with probability 1/2. (So he might use a third coin flip to decide what he says.)

    Version 2 of the game:

    The game host will follow the following rules:

    If he is at a spot where there are two heads, he will say "one of them is heads" with probability 1.

    If he is at a spot where there are two tails, he will say "one of them is tails" with probability 1.

    If he is at a spot where there are one heads and one tails, he will say "one of them is heads" with probability 1.

    ---

    Other versions can be invented too.

    So if we "define" the game by stating that

    then "the" game is not going to be defined! There are several different games which all satisfy this description.

    In the version 1 there does not exist a winning strategy, but in the version 2 it does exist. In the version 2, simply by always guessing the other coin to be tails will give approximately 50$ (or 50€) profit.

    If we don't know the rules that the game host is applying, then IMO the right answer is that the probabilities don't exist. You can come up with subjective conditional probabilities if you guess possible rules and decide some prior probabilities for them, but they cannot be considered as correct answers alone.
     
    Last edited: Sep 10, 2010
  14. Sep 10, 2010 #13

    russ_watters

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    No, it isn't correct, unless we are being deceived about what is going on. A single coin flip has a 50/50 probability of being heads or tails. Period. The scenario is just confusingly enough worded to mask the fact that that's what is available to you.

    The OP is saying that there are three possible outcomes of a two-coin flip:

    2 heads - (25% chance)
    2 tails - (25% chance)
    1 heads and 1 tails (50% chance)

    What is key here is that this description masks the fact that you actually have four possible events, not three - he's just clustered two of the events together.

    So when you know that one coin is heads, that doesn't leave the first and third possibilities, it leaves the first and half of the third.

    A more correct description is that you have these four scenarios:
    HH
    TT
    HT
    TH

    By knowing one of the coins is heads, you've eliminated the second and fourth, leaving only the first and third. And don't fool yourself into thinking the position of the coin matters (ie, that if you don't know which it is, it can be either): it doesn't. After you learn one is heads, you have these remaining:
    HH
    HT
    or you have
    HH
    TH

    But you don't have:
    HH
    TH
    HT

    You might think that by not knowing which is which you still have to deal with the fact that there are two coins sitting on the table, but you don't. The two coins are no longer the coin on the left and the coin on the right: mathematically, the two coins are now the one that is known to be heads and the one that isn't known at all.
    I haven't even read past scenario #2 yet.
     
    Last edited: Sep 10, 2010
  15. Sep 10, 2010 #14

    russ_watters

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    There is really only one relevant way to know that one of them is heads: by looking at one or both of them.

    You can't even use a computer to randomly check one or the other or both coins: by doing that, you simply make the second person a duplicate of the first and replace the second person with the computer. The bottom line is that each coin has two possible states and when one coin's state is known, it becomes irrelevant and the other coin has two possible states with a 50/50 chance of either being its actual state.

    Someone or something has to physically check one or both of the coins and once it does, you no longer are dealing with the coin on the left and the coin on the right but rather the coin that was checked and the coin that wasn't. It doesn't matter that you don't know which was checked - in fact, depending on the outcome, you may never know which was checked unless the person who checks tells you. It is completely irrelevant to the probability of guessing the combination.
    This is very much like the airplane on a treadmill game that went around the net a couple of years ago. There are essentially two possibilities: interpreting it in a way that is physically impossible in order to make it right and interpreting it in a way that is physically meaningful and therefore making it wrong. If you want to make it physically impossible, you can say literally anything and declare it "correct": maybe the second coin is a double-headed coin...?!

    The caveat being that I don't think there is but one way to read the question. Hurkyl read it another way, but I don't see that being a possible interpretation. Either way, the point stands: if that interpretation is there, it is wrong.
     
    Last edited: Sep 10, 2010
  16. Sep 10, 2010 #15
    You are making a big mistake there.
     
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