A game of In-between

For any 3 cards chosen, with the numbers $a$ , $b$ , $c$ on them, since they are different, there are only 2 ways out of the $3!=6$ ways to draw them from the deck, such that $a<c<b$ , namely $(a,b,c)$ or $(b,a,c)$ . Note that the order of the first 2 cards, $a$ and $b$ , can be swapped around because they are drawn together in the question and the value of $b$ is automatically the larger number of the two. Thus, $p(x)=\frac{2}{6}=\frac{1}{3}$ for any value of $x \geq 3$ .

Hence, the value of

$\displaystyle \sum_{k=3}^{2015} {p(k)} = (2015-3+1) \times \frac{1}{3} = 2013 \times \frac{1}{3} = \boxed{671}$

In each "game," you are drawing three cards from the deck. The probability $p(x)$ is effectively the probability that the "middle card" is the last of the three that you draw, which is $\frac{2!}{3!}=\frac{1}{3}$ . Note that this is the case regardless of the value of x .

So the value of the sum is

$\sum_{k=3}^{2015}\frac{1}{3}=\frac{2013}{3}=\boxed{671}.$

Firstly, let us calculate $p(x)$ for generic x : If $b-a = n+1$ then there are $x-n-1$ possibilities for the pair $(a,b)$ and $n$ possibilities for c such that $a <c <b$ . The total number of possibilities $(a,b,c)$ without the condition that c is necessarily between a and b is $\dbinom{x}{2} \times (x-2) = \dfrac{x(x-1)(x-2)}{2}$ Therefore: $p(x) = \dfrac{\displaystyle\sum_{n=1}^{x-2} n(x-n-1)}{\dfrac{x(x-1)(x-2)}{2}}$ Splitting the summand down into its component parts and from there evaluating each part using known results, we get $p(x) = \dfrac{\dfrac{(x-1)(x-1)(x-2)}{2} - \dfrac{(x-2)(x-1)(2x-3)}{6}}{\dfrac{x(x-1)(x-2)}{2}}$ In fact pretty much all of these terms cancel out after a bit of simplifying, to leave $p(x) = \dfrac{1}{3}$ irrespective of x . Therefore $\displaystyle\sum_{k=3}^{2015} p(k) = 2013 \times \dfrac{1}{3}$ so the answer is $\fbox{671}$