The question is basically to find out the probability of getting odd number of heads when $ n$ biased coins,with $m^{th}$ coin having probability of throwing head equal to $\frac{1}{2m+1}$ ($m=1,2,\cdots,n$) are tossed once.The results for each coin are independent.
If we consider first that only one head turns up.The probability then is equal to $$\sum_{m=1}^{n} [\frac{1}{2m+1} \prod_{k=1,k \neq m}^{n} (1-\frac{1}{2k+1})]$$ which seems very difficult to evaluate.It gets more complicated if we increase the number of heads.I couldnot find out a simple way to do this.Any suggestion would be highly appreciated.Thanks.
Best Answer
We work recursively. Let $P_n$ denote the answer if there are $n$ coins. Of course $P_1=\frac 13$. Considering the toss of the last coin we see that $$P_{n+1}=P_n\times \left(1-\frac 1{2n+3}\right)+(1-P_n)\times \frac 1{2n+3}=P_n\times \left(1-\frac 2{2n+3}\right)+\frac 1{2n+3}$$
This can be solved in closed form. We get $$P_n=\frac n{2n+1}$$
Not a lot of insight to offer on that...I just looked at the table, saw the pattern, then mechanically showed that the expression satisfied the recursion. Of course, the simplicity of the final form suggests that there might be a direct argument for it.