Unknown Coin Flipping

Start with an integer N that goes to infinity. The possible probabilities of heads are $\frac{1}{N}$ , $\frac{2}{N}$ , ... $\frac{N}{N}$ . Now, by conditional probability, the probability of a given $\frac{M}{N}$ for heads is $\frac{\frac{M}{N}}{\frac{N(N+1)}{2}}$ , or $\frac{2M}{N(N+1)}$ Given that it already landed heads, the probability that the coin lands heads again with a particular $\frac{M}{N}$ bias is $\frac{2M}{N(N+1)}\times \frac{M}{N}$ = $\frac{2M^{2}}{N^{2}(N+1)}$

Taking into account all values of M≤N, we find $\frac{2(1^{2}+2^{2}+3^{2}+ ... N^{2})}{N^{2}(N+1)}$

The sum of squares from 1 to N is given by the formula $\frac{N(N+1)(2N+1)}{6}$

Plugging the formula into the previous obtains $\frac{2(\frac{N(N+1)(2N+1)}{6})}{N^{2}(N+1)}$

Simplifying and then expanding yields $\frac{2N+1}{3N}$ = $\frac{2}{3}$ + $\frac{1}{3N}$

As N goes to infinity, the fraction tends to $\frac{2}{3}$ , So A=2 and B=3

A+B= $\boxed{5}$