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The Probability of the Improbable
- Thread starter jmf314159
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Count Iblis
Junior Member
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- Feb 13, 2007
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- 108
First calculate the probabity of 60 heads in a single test. Call this p. Then thye probability of not 60 heads is 1 - p. The probability of not 60 heads in N tests is thus (1-p)^N which for small p can be written as:jmf314159 said:Let's assume that test T involves tossing a coin 100 times. How many times must I run T before there is a 50-50 chance of registering 60 heads?
(1-p)^N = Exp[N Log(1-p)] = Exp[N(-p -p^2/2 - p^3/3...)]
=approximately Exp[-N p]
Equate this to 1/2 and solve for N.
Sorry, count, but how does one "calculate the probabity of 60 heads in a single test"? Thanks for your patience.Count Iblis said:First calculate the probabity of 60 heads in a single test....
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
You can use the binomial formula, or just derive the result from first principles. In case of coin tosses, all sequences of heads and tails are equally likely. In this case there are 2^(100) possible sequences of heads and tails. The number of sequences that contain 60 heads can be obtained as follows. Clearly given one sequence containing 60 heads you can obtain all others by permuting the elements of the sequence. Interchanging two heads won't give you a new sequience, neither will interchanging two tails. Call the number of sequences X. 'To find X, consider what would happen if the tails and the heads were somehow distinguishable. Then for each of the X sequences there would be 60! times 40! new sequences. The total number of sequences would thus be X 40! 60!, but this would have to be equal to 100!, so we can conclude that X = 100!/(60! 40!)jmf314159 said:Sorry, count, but how does one "calculate the probabity of 60 heads in a single test"? Thanks for your patience.
You can compute X using Stirling's formula:
Log(N!) = N Log N - N + 1/2Log(2 pi N) + 1/(12 N) - 1/(360 N^3) + 1/(1260 N^5) -1/(1680 N^7)+...
Copmpute Log(100!), Log(60!), Log(40!) using this formula, compute Log(X) from these results. We must divide X by 2^100 to obtain the probability, so you subtract Log(2^100) = 100 Log(2) from Log(X) and then you exponentiate to find the result.