The answer is 8.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Is there a simpler explanation which doesn't require you to deal with such massive calculations? There should be some reason why the answer is so nice.

**
Hint:
**
$\frac{1}{7} = \frac{2}{14}$

Log in to reply

That massive calculation really isn't that bad. That sum is sometimes known as the hockey stick theorem in pascal's triangle.

Gino Pagano
- 7 years, 7 months ago

Log in to reply

Use of hockey stick identity!

A Brilliant Member
- 7 years, 7 months ago

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 $\boxed{1 \over 7}$ .

3 Helpful
0 Interesting
0 Brilliant
0 Confused

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 ${6 \choose 13} = 0$ )

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

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

3 Helpful
0 Interesting
0 Brilliant
0 Confused

1 Helpful
0 Interesting
0 Brilliant
0 Confused

Does anyone have reasoning on why the number of urns doesn't matter?

Daniel Chiu
- 7 years, 7 months ago

Log in to reply

Indeed this was what I was hinting at in my initial comment. Here's another hint:

**
Hint:
**
Use the technique of a distinguished element, to relate this to a problem on 2014 balls.

This is not a proof but it gave me the right answer.

Avi Eisenberg
- 7 years, 7 months ago

Log in to reply

Great insight!

Now think about why it is true, and why the answer is $\frac{2}{b+1}$ .

0 Helpful
0 Interesting
0 Brilliant
0 Confused

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 $\frac{2}{2013}\times\frac{{2013 \choose 14}}{{2012 \choose 13}} = \frac{1}{7}$

$1+7 = \boxed{8}$

0 Helpful
0 Interesting
0 Brilliant
0 Confused

×

Problem Loading...

Note Loading...

Set Loading...

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 $1+7=\fbox{8}$