#1
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coin tossing and EV
Hi there
Suppose you have a biased coin that is 90% it will come up heads and 10% chance it will come up tails. What is the expected number of heads you expect to get in a row before you get a tail if you start tossing the coin? a formula would be helpful as well. thanks // hazeel |
#2
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Re: coin tossing and EV
P(0 heads) = .1
P(1) = (.9)^1*(.1) P(2) = (.9)^2*(.1) P(x) = (.9)^x*(.1) E(x) = (Sum 0-inf) x*P(x) = (Sum 0-inf) x(.9)^x(.1) = 0(.1) + .9(.1) + 2(.9)^2(.1) + 3(.9)^3(.1) + ... = .9(.1)[1 + 2(.9) + 3(.9)^2 + ...] = .9(.1)(1 / (1-.9)^2) (The sum of the infinite series 1 + 2x + 3x^2 + ... is 1 / (1-x)^2 = .9(.1)/(.1)^2 = 9 -- Homer |
#3
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Re: coin tossing and EV
Incidentally, if the probability of tossing a head is p and the probability of tossing a tail is q, the expected number of heads before throwing a tail is equal to p divided by q.
For example, if P(H) = .72 and P(T) = .28, E(X) = .72/.28, where X is the expected number of heads thrown before throwing a tail. -- Homer |
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