2013C13

For $1\leq i\leq13$ , we have the probability of getting 13 balls to be $\frac{{{2013-i} \choose 13}}{{{2012} \choose 13}}$ (choosing from the $2013-i$ black balls), while for $2001\leq i\leq2013$ , we have the probability to be $\frac{{{i-1} \choose 13}} {{2012} \choose 13}$ (Choosing from the $i-1$ white balls).

Adding up the probabilities in these ranges gives us:

$\frac{\displaystyle \sum_{i=1}^{13} {{2013-i} \choose 13}+\displaystyle \sum_{i=2001}^{2013}{{i-1} \choose 13}}{{2012} \choose 13}=2*\frac{\displaystyle \sum_{i=1}^{13} {{2013-i} \choose 13}}{{2012} \choose 13}$

(Note that the 2 summations are equivalent).

Note that adding ${{2000}\choose 14}$ to $\displaystyle \sum_{i=1}^{13} {{2013-i} \choose 13}$ gives us ${2013}\choose 14$ , so the above summation is equivalent to $2*\frac{{{2013}\choose 14}-{{2000}\choose 14}}{{{2012} \choose 13}}$

For $14\leq i \leq 2000$ , we have the probability of getting 13 balls to be $\frac{{{2013-i} \choose 13}+{{i-1} \choose 13}}{{2012} \choose 13}$ (choosing from the $2013-i$ black balls or the $i-1$ white balls)

Adding up the probabilities in this range then gives us:

$\frac {\displaystyle \sum_{i=14}^{2000} {{2013-i} \choose 13}+\displaystyle \sum_{i=14}^{2000}{{i-1} \choose 13}} {{{2012} \choose 13}}=2*\frac {\displaystyle \sum_{i=13}^{1999} {{i} \choose 13}} {{{2012} \choose 13}}=2*\frac {{{2000} \choose 14}} {{{2012} \choose 13}}$

(Firstly, the two summations are equivalent and secondly, $\displaystyle \sum_{i=14}^{2000}{{i-1} \choose 13}=\displaystyle \sum_{i=13}^{1999} {{i} \choose 13}$ )

Adding up all the probabilities for $1 \leq i \leq 2013$ , we then get:

$2*\frac{ { {2013}\choose 14} -{ {2000} \choose 14} } {{{2012} \choose 13}} + 2*\frac {{{2000} \choose 14}} {{{2012} \choose 13}}=2*\frac{{2013}\choose 14}{{2012}\choose 13}=2*\frac{2013}{14}=\frac{2013}{7}$

Dividing the total sum of the probabilities above by $2013$ , we have the expected probability that the 13 balls taken are all the same colour is then $\frac{1}{7}$ . The sum of the reduced fraction's numerator and denominator is then $1+7=\fbox{8}$

Here is a general solution. Let $n$ be the number of balls per urn, so that there are $n+1$ urns in total. Let $m$ be the number of balls we select, and $P_{\mbox{white}}$ be the probability of choosing $m$ balls in a row and having them be all white.

For an urn with $j \ge m$ white balls, there are $\binom{n}{m}$ ways of choosing $m$ balls, of which $\binom{j}{m}$ of those feature all balls being white. For urns with fewer white balls there are no ways of picking $m$ white balls.

Averaging this over all urns gives: $\begin{aligned} P_{\mbox{white}} &= {1 \over n+1} \sum_{j=m}^{n} ( \dbinom{j}{m} / \dbinom{n}{m} ) \\ &= {1 \over n+1} {1 \over \binom{n}{m}} \sum_{j=m}^{n} \dbinom{j}{m} \\ &= {1 \over n+1} {1 \over \binom{n}{m}} \dbinom{n+1}{m+1} \\ &= {1 \over n+1} {m!(n-m)! \over n!} {(n+1)! \over (m+1)!(n-m)!} \\ &= {1 \over m+1} \end{aligned}$ where the third step comes from Pascal's Christmas Stocking, and can easily be proven by induction considering that $\binom{n+1}{m+1} = \binom{n}{m} + \binom{n}{m+1}$ , i.e. the sum of the two entries above it in Pascal's Triangle.

Surprisingly the result only depends on $m$ . For this actual problem where $m=13$ , and we want to double the answer since we want the probability of white added to the probability of black, we get $2 \over 14$ , or $\boxed{1 \over 7}$ .

Let $P(i)$ denote the probability that we would get 13 balls of same colour from urn $i$ . We know that we can either select $13$ white balls out of $i - 1$ white balls or we can select $13$ black balls from $2013 - i$ black balls.

$\Rightarrow$ $P(i)$ = $\frac{{i - 1 \choose 13} + {2013 - i \choose 13}}{{2012 \choose 13}}$

By Baye's theorem, required probability = $\frac{1}{2013} \displaystyle \sum_{1 = 1}^{2013} P(i)$

= $\frac{1}{2013}\displaystyle \sum_{1 = 1}^{2013}\frac{{i - 1 \choose 13} + {2013 - i \choose 13}}{{2012 \choose 13}}$ (Note that ${6 \choose 13} = 0$ )

= $\frac{1}{2013} \frac{2{2013 \choose 14}}{{2012 \choose 13}}$

= $\frac{1}{7} \Rightarrow a + b = \fbox{8}$

By experimentation with smaller numbers of urns and balls $(7,3)(9,3)(7,5)(9,5)$ I discovered that the number of urns(2013), doesn't matter to the average as long as it is more than the number of balls picked. Therefore, we can look at the average as if there were only 14 urns. Then for the 1st and 14th urns, the probability would be 1, and for the other urns it would be 0. $2/14=1/7$ $1+7=8$

Let S be the number of ways to pick 13 balls of the same color, and T be the number of ways to choose 13 balls of any color. We are looking for $\frac{S}{T}$ . There are 2013 urns, and ${2012 \choose 13}$ ways to select 13 balls from an urn of ${i-1}+{2013+i}=2012$ balls, so $T=2013{2012 \choose 13}$ . Split S into W, the number of ways to select 13 white balls, and B, the number of ways to select 13 black balls. $S=W+B$ . There are ${13 \choose 13}$ ways to select 13 white balls from urn 14, ${14 \choose 13}$ ways to select 13 white balls from urn 15, all the way up to ${2012 \choose 13}$ ways for urn 2013. We are looking for $\sum_{i=14}^{2013}{{i-1 \choose 13}}$ , which by the hockey stick identity, is ${2013 \choose 14}$ . By the same process, $B={2013 \choose 14}$ as well. Therefore, $S=2{2013 \choose 14}$ , so $\frac{S}{T}=\frac{2{2013 \choose 14}}{2013{2012 \choose 13}}=\frac{2*2013}{2013*14}=\frac{1}{7}$ , so $a+b=8$

Lets write out the first couple terms

Urn 1 = 0 White 2012 Black

Urn 2 = 1 White 2011 Black

Urn 3 = 2 White 2010 Black

Now write the probabilities we get

$\frac{1}{{2012 \choose 13}}$ times

${2012 \choose 13}+...+({1999 \choose 13}+{13\choose 13})+...+({13 \choose 13}+{1999\choose 13})+...+{2012 \choose 13}$

Now we see that the mess on top can be simplified with the hockey stick identity.

It simplifies into $2 \times {2013 \choose 14}$ .

We have a $\frac{1}{2013}$ chance of choosing a specific urn

Bringing everything together we get $\frac{2}{2013}\times\frac{{2013 \choose 14}}{{2012 \choose 13}} = \frac{1}{7}$

$1+7 = \boxed{8}$