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Math Help - Conditional Probability brainteaser

  1. #1
    Senior Member sfspitfire23's Avatar
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    Conditional Probability brainteaser

    Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?

    Through cond prob. I think the answer is 1/4, but others are saying its 1/3. There are 4 possible combinations, and knowing that at least one coin turned up heads doesnt change the probability of getting HH in flipping 2 coins.

    we cant just eliminate TT from the equation bc before you flipped the coins, TT has 1/4 chance of turning up.

    But if you said right from the start that the coins are somehow biased that at least 1 coin will always turned up heads, then i would agree that the probability of getting HH is 1/3.

    otherwise if getting HH, HT, TH or TT have equal chances, then it should be 1/4.


    Thoughts?
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  2. #2
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    Re: Conditional Probability brainteaser

    For me the answer is 1/3

    HH HT TH TT have equal chance in general.

    One of the coins is head, this means that only

    HH HT TH

    are possible, all with equal chance.

    -> 1/3
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Re: Conditional Probability brainteaser

    Quote Originally Posted by wnvl View Post
    For me the answer is 1/3

    HH HT TH TT have equal chance in general.

    One of the coins is head, this means that only

    HH HT TH

    are possible, all with equal chance.

    -> 1/3
    I still think that you can't eliminate the proability of TT in calculations just because one is heads after the fact.
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  4. #4
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    Re: Conditional Probability brainteaser

    Yes you can, because the chance for TT is zero after telling that at least one of the coins came up H.
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  5. #5
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    Re: Conditional Probability brainteaser

    Another way of looking at this problem: I have a bag with 1 red, 1 blue, 1 green and 1 black marble. I pull a marble out. At this point the probability are:
    P(red)=0.25
    P(blue)=0.25
    P(green)=0.25
    P(black)=0.25
    P(red)+P(blue)+P(green)+P(black) = 1
    Which makes sense. I have 100% chance of having 1 marble if I pull 1 marble.

    Now I look at it and tell you its NOT black. The chances are now
    P(red)=0.3333
    P(blue)=0.3333
    P(green)=0.3333
    P(black)=0.0
    P(red)+P(blue)+P(green)+P(black) = 1
    Again i have 100% of having at least 1 marble in my hand.

    If I tried to insist the probabilites remained .25 for the colors I'd have this:
    P(red)=0.25
    P(blue)=0.25
    P(green)=0.25
    P(black)=0.0
    P(red)+P(blue)+P(green)+P(black) = .75
    Which is saying after pulling 1 marble out of the bag, looking at it and saying its not black. There is now a 75% chance of me having a marble in my hand. Which is obviously absurd I definitely have a marble in my hand.


    Now you can equate this:
    red = HH
    green = HT
    blue = TH
    black = TT.

    Now telling you that at least one coin is Heads is the same as saying its not black. So the odds of double heads coming up (red) is 1/3.
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