Bill has n+1 coins and Ben has n coins.
Both players throw all of their coins simultaneously and observe the number that come up heads.
Assuming all the coins are fair, what is the probability that Bill obtains more heads than Ben?
Originally posted by THUDandBLUNDER Bill has n+1 coins and Ben has n coins.
Both players throw all of their coins simultaneously and observe the number that come up heads.
Assuming all the coins are fair, what is the probability that Bill obtains more heads than Ben?
Yeah, but it's 50% that he'll get MORE heads than him. If they've got an equal number, it doesn't count. I haven't worked out the probability, but I know that much.
Originally posted by ark13 Yeah, but it's 50% that he'll get MORE heads than him. If they've got an equal number, it doesn't count. I haven't worked out the probability, but I know that much.
It's easy. In n coins the expected value of heads is the same, so it's 50% on the last coin.
Originally posted by THUDandBLUNDER What about if they both had n coins.
What would the probability be then?
That is a function of n. For n=1, the probabilty is 1/4, and I believe as n approaches infinity, the probability approaches 1/2 (someone better at math than I will have to confirm that).
Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n.
Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n.[/b]
I don't think the problem is as easy as everyone seems to think it is. I think the answer is 1/2 only if n=0, which is why the above answers are only conditionally correct. Here's what I think it is:
if they both had n coins, then the probability of Bill getting more heads would be n/(2n+2). This is very easy to show.
So my approach is the following: consider they both throw n coins, and the probability of Bill having more heads is given above. Then consider that Bill has one more coin to throw which of course can be heads with 1/2 probability. Then the overall probability that Bill gets more heads should simply be a weighted average of these probabilities:
(n(n/2n+2) + 1/2)/(n+1).
This only equals 1/2 for n=0, which makes sense. for all other possibilities, it depends on n.
yeah, after i look at it, Bowmann is right: the answer is 1/2 regardless of n.
the n/(2n+2) is right for n coins each, but i don't think i can take the weighted average like i did.
the proof that the answer to the original question is 1/2 is straightforward now that i think about it:
bill has n+1 coins, ben has n:
consider that ben has gotten i heads (i = 0,1,2,3,...,n). Then Bill has gotten j heads (j = 0,1,2,...,n+1). Then there are clearly (n+1)(n+2) possibilities (you have to include the zero). But the number of times that Bill has more heads than Ben is only given by SUMMATION(from i=1 to i=n+1) of i. this is equal to (1/2)(n+1)(n+2) (this is easy to show by simple number theory). then the probability we are looking for is (1/2)(n+1)(n+2)/((n+1)(n+2)) = 1/2.
that was probably clear as mud, but it works out...little hard to generalize in clear terms.
03 Apr '05 19:121 edit
Coin Tossing
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