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Politics : Formerly About Advanced Micro Devices
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To: jcholewa who wrote (128011)11/10/2000 2:30:00 PM
From: andreas_wonisch  Read Replies (2) of 1570986
 
JC, Re: I believe this is pretty easy. The chances of flipping 1500 heads where all other flips can be any combination of heads and tails should be something like:

0.5 ^ 1500 x 1.0 ^ 500 =
0.5 ^ 1500 =
2.85 x 10^(-452)

No, that's not right. With your solution the chances of getting more than one head would be 0.5, more than two 0.25, more than three 0.125. The problem with your solution is that you compute the probability for getting 1,500 head first and then any combination of heads and tails. You don't consider that heads can be arranged in any order.

The correct approach to this problem is with the help of the binomial distribution. The probability of getting exactly k heads is:

p(X=k) = (n over k) * p^k * q^(n-k)

In this case the probability is p = q = 0.5 since head and tails are equally probable. Further we have n = 2000. But we want to get the probability of getting more than k heads. So we need to sum up:

p(X>=k) = sum_{k=0}^{n} p(k) = (n over k) * p^k * q^(n-k)

It's very difficult (from a computation standpoint) to sum up all these small numbers. But since 2000 is a fairly big n we can approximate the binomial distribution with the gaussian distribution. With the normalized gaussian distribution we have:

P(X>=k) = 1 - Phi[(k - n*p)/(sqrt(n*p*q))]

where Phi(z) is int_-z^z 1/sqrt(2*pi) * e^(-x^2/2) dx.

With k = 1500 this probability is so extremely small that it comes very close to your number. With k = 1000 it's -- like expected -- 0.5. The probablity for k = 1100 would already be below 0.01.

Andreas
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