You have a set of n biased coins. The mth coin has probability 1/(2m+1) of landing heads (m = 1, 2, ... , n) and the results for each coin are independent. What is the probability that if each coin is tossed once, you get an odd number of heads?
Solution
Answer: n/(2n+1).
Consider the expansion of (2/3 - 1/3)(4/5 - 1/5)(6/7 - 1/7) ... (2n/((2n+1) - 1/(2n+1)). The negative terms correspond to an odd number of heads. So the product is just prob even - prob odd. But the product telescopes down to 1/(2n+1). Obviously prob even + prob odd = 1, so prob odd = (1 - 1/(2n+1))/2 = n/(2n+1).
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2001