The Student Room Group
Hmmm, Binomial Distribution is out of the question so I'm not sure, it's an odd question too as the way the cards are dealt will have an effect on the probability too I think.
Reply 2
I got really confused on probability questions. don't like them at all
Reply 3
This is the same question as:

There are 52 marbles in a bag. Fifty are red, and two are blue. If I take thirteen marbles at random, what is the probability that the selection includes both blue marbles?
Reply 4
There are 52 cards in a pack. The probability of getting a card of a certain number and certain suit is going to be 1/52 yes? The probability of getting a certain number but not a certain suit will be 4/52.

Now... the probability of getting two of the same number (specific) is the same as 4/52 x 4/52 (=16/2704). There are 13 different cards in the pack so multiply the numerator by 13 (=208/2704 OR 1/13). Notice anything?

Anyway... that was the basic stuff behind the probability of the cards before the dealer changed anything. Bear in mind, he changes only 2 NUMBERS. He removes 1 NUMBER and replaces it with ANOTHER. So he changes the one he takes away and changes the one he adds a copy of.

So we'll do the same as above. 13-2=11. 11 numbers are still exactly the same as before. So

11 x 16/2704 = 176/2704

Keep that fraction (or probability) in mind.

Now, the probability of getting two of the card he removes is 3/52 x 3/52 = 9/2704.
Likewise the probability of getting two of the card he adds a copy of is 5/52 x 5/52 = 25/2704.

Kk, now before looking at the spoiler I want you to make sure you understand everything lol.

Spoiler



I tried to explain it... someone correct me if I'm wrong plz (I'm sure someone will :smile:)
I know that the answer is 12/51