- 03 Apr '05 19:12 / 1 edit

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).*Originally posted by THUDandBLUNDER***What about if they***both*had n coins.

What would the probability be then?

Not sure what that has to do with your original problem, since the answer is always 1/2 there, regardless of n. - 04 Apr '05 01:35
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.

THUD, is this right? - 04 Apr '05 02:12

Actually, Bowmann was right for n+1 coins but he gave no reasoning.*Originally posted by davegage*

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.

THUD, is this right?

When they both have n coins the calculation is a bit more complicated. - 04 Apr '05 02:57yeah, 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.

good problem.