50 ping pong balls numbered 1-50 are placed in a vat, how many must be drawn to ensure that you have two pairs each summing 50?
In the worst case, the balls are pulled out in order of their numbers (1,2,3,4,5...), so in order to get one pair summing 50, you need to have pulled out at least 26 balls. At that point you'll have drawn ball 24 and ball 26, summing 50. To get another pair that sums to 50, you just need to draw the 27th ball.
So my answer is 27 balls. I think this is correct, but in class my professor wanted us to use this to solve these types of problems:
If you have $n$ objects to distribute into $k$ boxes, then there is a box with at least ceiling$(n/k)$ objects. How would I apply that here?
There are $24$ pairs of numbers that sum to $50$: $\{1,49\}$, $\{2,48\}$, and so on to $\{24,26\}$; put $25$ and $50$ together in a $25$-th pair. In the worst case you draw the $25$ and the $50$ first, neither of which helps you. After that, however, another $25$ draws will ensure that you draw both members of one of the $24$ pairs summing to $50$. Thus, $27$ draws will ensure that you get at least one pair summing to $50$. If it really does take that long to get your first pair, you must have drawn one ball from each of the remaining $23$ pairs, so one more draw will get you a second pair summing to $50$. Thus, in the worst case it will take you $28$ draws.
To put it a little differently, if you draw only $27$ balls, you might get one from each of the $25$ pairs plus a second one from each of two pairs, and one of those pairs might be the useless $\{25,50\}$ pair. If you draw $28$, however, you are guaranteed to draw both balls from at least three pairs, so even if one of those pairs is $\{25,50\}$, you still get two pairs that sum to $50$.