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**