# Conditional Probabilities on Coin Toss

Was having some trouble understanding the last step of the solution where it was determined where the two expressions for $P(F_{n,m})$ were equivalent, specifically:

$P(F_{n,m}) = \frac{1}{6} P(F_{n-1,m}) + \frac{1}{2} P(F_{n,m-1}) + \frac{1}{3} P(F_{n,m})$

$P(F_{n,m}) = \frac{1}{4} P(F_{n-1,m}) + \frac{3}{4} P(F_{n,m-1})$

Apologies for the alignment issues in the two images, if anyone can edit to make it clearer, it would be really helpful.

• Note that you have $P(F_{n,m})$ on both sides of the first equation. It looks like the author just subtracted $1/3 P(F_{n,m})$ from both sides, then multiplied by $3/2$. – Bill the Lizard Oct 6 '15 at 19:50
• I didn't even catch that. That took unnecessarily long to spot. Thanks! – Vakalate Oct 7 '15 at 9:19

$$\begin{array}{lrll} &P(F_{n,m}) &= \tfrac{1}{6} P(F_{n-1,m}) + \tfrac{1}{2} P(F_{n,m-1}) + \tfrac{1}{3} P(F_{n,m}) \\[1ex] \implies & \tfrac{2}{3}P(F_{n,m}) &= \tfrac{1}{6} P(F_{n-1,m}) + \tfrac{1}{2} P(F_{n,m-1}) &(\text{rearranging}) \\[1ex] \implies & P(F_{n,m}) &= \tfrac{1}{4} P(F_{n-1,m}) + \tfrac{3}{4} P(F_{n,m-1}) &(\text{multiply through by }\tfrac{3}{2}) \\ \end{array}$$
Note that the first equation arises from the fact that for two dice, out of the $36$ possible rolls there are $6$ for which the sum is seven, $18$ for which the sum is even, and $12$ for which the sum is neither seven nor even.