You face an urn with 5555 cards in it, each has a non-zero integer written on it. Nothing is said about the distribution of those numbers. You are told to draw randomly a card, copy the number, return it back, shuffle and draw randomly a card, then write down the sum of both numbers, say S.
(i) Prove: The probability of S being an even number is higher than S being odd.
(ii) Is it true for any initial number of cards? Comment.
Let the number of even cards = E
the number of odd cards = O
the difference = D = O-E
Then the total number of cards in the urn is N = 2E + D
A selected card is even with probability E/(2E+D) and odd with probability (E+D)/(2E+D)
Then, S is even if both selected cards are even or if both selected cards are odd. This occurs with probability
(E/(2E+D))^2 + ((E+D)/(2E+D))^2
which simplifies to
1/2 + (1/2)(D/N)^2
This probability is obviously greater than 1/2, unless D = 0.
If E + O is odd, then D cannot equal 0.
In particular, 5555 is odd, so the probability is greater than 1/2
Edited on February 15, 2017, 5:12 pm
Argument (with math)
