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I have what seems like a simple question, but it's been a while since I've done any P/S. So i come to SE for help!

Two player pool/billiards: P1 has probability p of sinking a ball on any shot and has N balls remaining, while P2 has prob q and M balls remaining.

Question: What is the probability of the first player winning?

I can type out my reasoning (and will in an edit - will post before i reason it out though), but my answer has come down to:

$$\sum_{j=0}^{M-1} [p^N q^j \sum_{i=0}^\infty [(1-p)^i (1-q)^i]]$$

Is this correct or close?

Reasoning:

not really theoretical reasoning, but extrapolating the simple cases outwards:

(hit = h, miss = m, with probability p = w/p)

1 ball each: possible victory paths -

P1 wins w/p,

P1 m w/ (1-p), P2 m w/ (1-q), P1 wins w/ p

... etc - $p \sum_{i=0}^\infty (1-p)^i (1-q)^i$

2 and above balls each:

At some point, all the P1 hits must occur - $p^N$

All possible amounts of P2 hits must be accounted for - $\sum_{j=0}^{M-1} q^j$

Every miss variation is accounted for - * <-- This is where i think I am wrong. Is it actually a double sum in and of itself? IE $\sum_{i=0}^\infty \sum_{k=0}^\infty (1-p)^i (1-q)^k$ ?

EDIT: Some wolfram alpha shows me that $\sum_{i=0}^\infty (1-p)^i = \frac 1 p$, so I guess my final final equation can be simplified to

$$\sum_{j=0}^{M-1} \frac{p^N q^j}{pq} $$ ??

etc.

share|improve this question
    
Formula seems wrong as $N=0$, $p$ is anything, does not always lead to a probability of 1. –  Calvin Lin Jan 4 '13 at 18:36
    
hmm. good point –  im so confused Jan 4 '13 at 18:37
    
well, lmao that final thing can't be right, that's > 1 for one scenario hahahaha oops –  im so confused Jan 4 '13 at 18:42
1  
The problem neglects my usual way of winning: the opponent scratching on the 8 ball. –  user7530 Jan 4 '13 at 20:14
    
hahaha you must usually play against me then –  im so confused Jan 4 '13 at 20:15

1 Answer 1

up vote 2 down vote accepted

Let $P(k)$ denote the probability that $P_1$ sinks all $N$ balls by $k^{th}$ turn (i.e., with exactly $k-1$ misses). The final shot has to be a hit. Before that there were $N-1$ hits and $k-1$ misses, which can be arranged in any order. So

$$ P(k) = \binom{N+k-2}{N-1}p^{N}(1-p)^{k-1} $$

Similarly if $Q(k)$ denotes the probability that $P_2$ sinks all $M$ balls by $k^{th}$ turn, then

$$ Q(k) = \binom{M+k-2}{M-1}q^{M}(1-q)^{k-1} $$

Now, for $P_1$ to win, the number of turns he takes to finish must not be more than that of $P_2$. So, required probability is

$$ \sum_{i\leq j} P(i)Q(j) $$

This might not give a quick idea about the probability value, but compuationally, I don't think it is difficult as computing $P(k)$ and $Q(k)$ can be done very fast by recursion and even though it is an infinite sum, it converges quickly. I am not sure if this is the best we can get though.

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It looks like $P(k)$ and $Q(k)$ can be greater than $1$. Also, wouldn't sinking $N$ balls in $k$ turns mean you missed $k-N$ times, not $k-1$? –  b0uma Jan 6 '13 at 5:47
    
@compadee Thanks for pointing that out. $p$ and $1-p$ terms should actually be in numerator. Now, we can show that $P(k)$ and $Q(k)$ are always $\leq 1$. By a turn, I mean a series of hits without any miss ($P_2$ gets his turn only when $P_1$ misses). –  polkjh Jan 6 '13 at 7:06
    
This is excellent, thank you! It also looks pretty much like the simplest form you can take it too. Nice work –  im so confused Jan 10 '13 at 16:30

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