Bet it like beckham

let n be the total number of shoots and r be number of goals scored that you have bet on. Given that there was one hit and one miss, we are left with (n-2) total shots. This will have (n-r-1) misses and (r-1) hits

Let us see the term corresponding to the product of all Hits (all terms from 1 to 124, multiplied), so : (r-1)! In the denominator, we will have all the terms from 2 to 255. so, the denominator will have (n-1)!

Now let us calculate the product of the remaining terms : the (n-r-1) misses. If you write down any particular case, you will notice that all the misses consist of the product of consecutive terms from 2 to 130 . Thus making the term (n-r-1)!.

The question is , how many such cases exist? To answer that question, let us rephrase this subproblem this way - In how many ways can we distribute (n-r-1) identical objects in r boxes ( in and around r - 1 objects)

The number of ways of distributing p identical objects in q different boxes is
${ p + q - 1 \choose q - 1 }$

So we get the total number of cases : ${ n-r-1 + r - 1 \choose r-1}$ = ${n-2 \choose r-1}$

So, the final term is $\dfrac{(r-1)!}{(n-1)!} (n-r-1)! {n-2 \choose r-1}$ = $\dfrac{(r-1)!}{(n-1)!} (n-r-1)! \dfrac{(n-2)!}{(r-1)!(n-r-1)!}$

= $\dfrac{1}{n-1}$

= $\dfrac{1}{255}$

So, no matter what you bet, the probability of winning is always the same(it is independent of r). If n is the total number of shots, then the probability that your guess is correct is always $\dfrac{1}{n-1}$

The problem can be considered as a Polya's Urn , so that the probability of scoring $k$ goals after $n$ trials, given that you performed $\alpha$ goals out of $\alpha+\beta$ initial trials, is

$\displaystyle p(n,k)=\binom{n}{k} \frac{B(k+\alpha,n-k+\beta)}{B(\alpha,\beta)}$

where $B$ is the beta function. In our case $n=256-2=254$ , since two trials have been performed, one of them being a goal, so $\alpha=1$ and $\beta=1$ . Eventually, $k=125-1=124$ . Substituing

$\displaystyle p(254,124)=\binom{254}{125} \frac{B(125+1,254-124+1)}{B(1,1)}=\frac{1}{255}=\frac{a}{b} \ \Longrightarrow \ b^2-a^2=255^2-1=\boxed{65024}$