Expected value, unexpected famous fraction

• If at any point of the process we have three heads, the expected value of rounds of tosses at that point is $E_3=0$ since we have achieved the goal. $E_3$ is the "state" of having three heads.

• If at any point we have two heads $(E_2)$ , we can write the following equation: $E_2=1+\frac{1}{2}E_3+\frac{1}{2}E_2 \implies E_2=1+\frac{1}{2}\cdot 0+\frac{1}{2}E_2 \implies E_2=2$ .

Make sure you understand what this last equation means: we toss the coins once more $(1+\ldots)$ and we have a 50% chance of achieving the goal $(E_3)$ or a 50% chance of going back to where we started $(E_2)$ .

• If at a certain point we have one heads, the equation now is: $E_1=1+\frac{1}{4}E_3+\frac{2}{4}E_2+\frac{1}{4}E_3 \implies E_1=1+\frac{1}{4}0+\frac{2}{4}2+\frac{1}{4}E_3\implies E_3=\dfrac{8}{3}$ .

Note that in this case there is one way out of four of reaching the goal (H,H), two ways out of four of jumping to state $E_2$ , (H,T) and (T,H), and one way out of four of remaining in the same state $E_1$ , (T,T), hence the probabilities used.

• Finally, at the start we have no heads, so the equation becomes: $E_0=1+\frac{1}{8}E_3+\frac{3}{8}E_2+\frac{3}{8}E_1+\frac{1}{8}E_0 \implies E_0=1+\frac{1}{8}\cdot 0+\frac{3}{8}\cdot2+\frac{3}{8}\cdot\frac{8}{3}+\frac{1}{8}E_0 \implies E_0=\dfrac{22}{7}$

$A=22, B=7$ and $\boxed{A+B=29}$

The title of the problem refers to the fraction $\dfrac{22}{7}$ , which is the famous approximation of $\pi$ used by Archimedes.