Posted 11 Apr 2014 · Report post 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? 0 Share this post Link to post Share on other sites

0 Posted 27 Apr 2014 · Report post 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. 0 Share this post Link to post Share on other sites

0 Posted 11 Apr 2014 · Report post has a 50% chance of winning from examining smaller cases. I haven't figured out the other way around yet. 0 Share this post Link to post Share on other sites

0 Posted 13 Apr 2014 · Report post [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. 0 Share this post Link to post Share on other sites

0 Posted 14 Apr 2014 · Report post about a 46.44% chance of winning. 0 Share this post Link to post Share on other sites

0 Posted 15 Apr 2014 · Report post about a 46.44% chance of winning. Why? 0 Share this post Link to post Share on other sites

0 Posted 15 Apr 2014 · Report post 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) 2^{501} 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. 0 Share this post Link to post Share on other sites

0 Posted 23 Apr 2014 · Report post 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]. 0 Share this post Link to post Share on other sites

0 Posted 23 Apr 2014 (edited) · Report post 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 23 Apr 2014 by dgreening 0 Share this post Link to post Share on other sites

0 Posted 23 Apr 2014 (edited) · Report post Edited 23 Apr 2014 by dgreening 0 Share this post Link to post Share on other sites

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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?

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