more fun with coins

[BMAD]

Suppose John tosses a coin 250 times and Eric tosses a coin 251 times, what is the probability that john's coin has more heads than Eric? What is the probability that Eric has more heads than John?

[Rob_G]

has a 50% chance of winning from examining smaller cases. I haven't figured out the other way around yet.

[bonanova]

[spoiler=Looks like]Eric has one more toss than John so he has more heads OR more more tails but not both. That would require at least two tosses more than John. Since there is no bias, the two cases are equally likely, with probabilities of 1/2. The second question has a different answer, since we must rule out all cases where the number of heads are equal. I'm thinking.

[Rob_G]

about a 46.44% chance of winning.

[BMAD]

about a 46.44% chance of winning.

Why?

[Rob_G]

about a 46.44% chance of winning.

Why?

We already know that the probability of Eric winning is .5. I found that the number of ties in a game follows a pattern where if the number of coins used is n and n+1 then the number of ways to tie is the binomial coefficient of (2n+1 n+1). Also we know that the total number of outcomes is 2

²ⁿ⁺¹. Given all that the probability of John winning is

501!

----------------

250! x 251!

.5 - ----------------------------- = .4644 (approx)

2⁵⁰¹

I determined all this by looking at 1v2 through 5v6.

The number of ways for John to win when n and n+1 coins is the number of ways to choose at most n-1 items from a set of 2n+1. The number of ways to tie proved to be an easier formula to use.

[dgreening]

For 250 [or 251] tosses, you would expect some sort of distribution centered
heavily around 50% [or 125 heads].

This is not even close to being a uniform distribution, so most points below
110 or above 140 are zero. [just a guess on the end points]. The point is that
there are really only a small number of points that have to be considered.

Another approach might be to do some sort of convolution of the 2
distributions [which look virtually identical].

Since Ties go to John, the probabilityt will be less than 50%

Rob G may be pretty close [i am not sure].

Another perspective

I suspect that the impact of the extra toss is actually very small and it could go either way.

For 250 [or 251] tosses, you would expect some sort of distribution centered heavily around 50% [or 125 heads].

This is not even close to being a uniform distribution, so most points below 110 or above 140 are zero. [just a guess on the end points]. The point is that there are really only a small number of points that have to be considered.

Another approach might be to do some sort of convolution of the 2 distributions [which look virtually identical].

Since Ties go to John, the probabilityt will be less than 50%

Rob G may be pretty close [i am not sure].

[dgreening]

Another perspective

I suspect that the impact of the extra toss is actually very small and it could go either way.

For 250 [or 251] tosses, you would expect some sort of distribution centered heavily around 50% [or 125 heads].

This is not even close to being a uniform distribution, so most points below 110 or above 140 are zero. [just a guess on the end points]. The point is that there are really only a small number of points that have to be considered.

Another approach might be to do some sort of convolution of the 2 distributions [which look virtually identical].

Since Ties go to John, the probabilityt will be less than 50%

Rob G may be pretty close [i am not sure].

Edited April 23, 2014 by dgreening

[dgreening]

Edited April 23, 2014 by dgreening

[bonanova]

Suppose John tosses a coin 250 times and Eric tosses a coin 251 times, what is the probability that john's coin has more heads than Eric? What is the probability that Eric has more heads than John?

Let

J(x) be the probability that John gets x heads in 250 tries

Let E(y) be the probability that Eric gets y heads in 251 tries.

J(x) = [ 250! / (x!)(250-x)! ] (0.5)²⁵⁰

E(y) = [ 251! / (y!)(251-y)! ] (0.5)²⁵¹

Then P (x > y) = Sum (y=0, 250) { Sum (x=y+1, 251) E(y) J(x) }

The sums are left to the reader as an exercise.