| Take a deck, shuffle it. Cut it in half and focus on one half. There is a 65% chance that there are 2, 3, or 4 aces in this half, conversely a 35% chance that there are 0 or 1 aces. Likewise, there is a 65% chance that the other half has 2, 3, or 4 aces, and a 35% chance that it has 0 or 1.
Let's apply even a little more common sense. Shuffle the deck and cut it into "left" and "right" halves. You then have the following possibilities:
Label | Aces in | Aces in
| Left | Right
------|---------+----------
A | 0 | 4
B | 1 | 3
C | 2 | 2
D | 3 | 1
E | 4 | 0
Since this is the complete list of possibilities, P(A) + P(B) + P(C) + P(D) + P(E) == 1.
Assuming that P(C) > 0, then it follows that P(A) + P(B) + P(D) + P(C) < 1.
Because of the equality of A and E, and of B and D, it further follows that P(A) + P(B) < 0.5, and from that it follows that P(C) + P(D) + P(E) > 0.5 ...
Just coincidentally (not), P(C) + P(D) + P(E) is "the probability of finding at least two aces" in the "left" half of the deck.
Not a complete proof due to the assumption, but unless you want to argue that it's impossible to find exactly two aces, I think we're settled ;) |
