Some question about probability (i guess using Markov chains) I'm working on the following problem:
"A boy and his father are playing chess, and the game is set to be finished when one of them reaches $v$ victories. In each match, the boy has a $p$ probability of winning, and his father $q$. If $p+q<1$, calculate the probability of the boy's victory" I guess $p+q<1$ basically means that stalemates are possible. I was trying to use Markov chains but having to reach $v$ victories in order to win the game is causing me some trouble. Thank u so much if you read the whole thing :)
 A: To simplify notation, let me call the initial probabilities $p' \; and \;q'$ for boy and father, $p'+q' <1$, and  ignore draws (which don't matter)
P(boy wins a game) $=p=\frac{p'}{p'+q'}$
P(father wins a game) $=q= \frac{q'}{p'+q'}$
but now $p+q=1$
For concreteness, let $v = 7,$ say
Since whoever wins $7$ games wins, it is like a tennis tournament where first to $7$ wins gets the prize, and no recursion is needed
You need a maximum of $13$ games to decide who wins.
The easiest way to compute this is to allow  $13$ games to played even if the match has already been decided, but restricting the loser to a maximum of $6$ wins.
Thus P(boy wins) $= \binom{13}0q^0p^{13} + \binom{13}1q^1p^{12} + \binom{13}2q^2p^{11}+ ... +\binom{13}6q^6p^7$
Restoring for the v wins needed, and making the formula compact,
$$P(boy \;wins)=\large{ \sum_{i=0}^{v-1}}\binom{2v-1}{i}q^ip^{2v-1-i}$$
If this seems counterintuitive, look here where all doubts about the concept have been cleared in detail.
NOTE:
Had the question been that you need to win $v$ more games than your opponent, then it would be a Markov process.
