Playing Cards

A pack of cards consists of 52 different cards. A malicious dealer changes one of the cards
for a second copy of another card in the pack and he then deals the cards to four players,
giving thirteen to each. How do I calculate what the probability that one player has two identical cards?

What method would I need for this?

Cheers!

Reply 1

Jonny W

8

Number of possible deals = 52!/(13!^4)

[When we count possibile deals, we pretend that the identical cards are distinguished from one another.]

Number of possible deals in which Player 1 gets two identical cards = 50!/(11!*13!^3)

Number of possible deals in which one of the players gets two identical cards = 4*50!/(11!*13!^3)

Probability = [4*50!/(11!*13!^3)] / [52!/(13!^4)] = 4/17

--

Or ...

Deal as follows. Write the numbers 1, 2, ... 52 on the table. Take the pack (which may be in any order - not necessarily shuffled) and one by one put each card on a randomly selected unoccupied number. Give the cards in positions 1-13 to Player 1, those in positions 14-26 to Player 2, etc.

We may assume that the identical cards are dealt first. There are 51 possible places for the second card. Of these possibilities, 12 result in one player getting two identical cards.

So the probability is 12/51 = 4/17.

Reply 2

Chewwy

17

it's been answered, but a slightly simpler way:

p(a player gets duplicates) = 4 x p(player 1 gets duplicates) = 4 x 13/52 x 12/51 = 4/17

Reply 3

Jonny W

8

chewwy

it's been answered, but a slightly simpler way:

p(a player gets duplicates) = 4 x p(player 1 gets duplicates) = 4 x 13/52 x 12/51 = 4/17

Why is p(player 1 gets duplicates) = 13/52 x 12/51 ?

I know it's true, but I don't understand your method.

Reply 4

Chewwy

17

Jonny W

Why is p(player 1 gets duplicates) = 13/52 x 12/51 ?

I know it's true, but I don't understand your method.

p(he gets one of the duplicates) x p(he gets the other)

Reply 5

Jonny W

8

chewwy

p(he gets one of the duplicates) x p(he gets the other)

Thanks.

Reply 6

Goldenratio

OP

2

Thank you very much for all those answers! :biggrin:

I have to admit that I haven't done much stats for a while and feel quite rusty. If you wouldn't mind, could you explain a bit more how you came to figure out that the probability of one player getting a duplicate card was 13/52?

Reply 7

Chewwy

17

Goldenratio

Thank you very much for all those answers! :biggrin:

I have to admit that I haven't done much stats for a while and feel quite rusty. If you wouldn't mind, could you explain a bit more how you came to figure out that the probability of one player getting a duplicate card was 13/52?

imagine 52 places for the cards laid out in front of you. the first 13 are for player 1, the second 13 for player 2, and so on.

so there are 13/52 places for your first duplicate card to be put into, then 12/51 for the second, assuming the first has been put in the same set.

Reply 8

KAISER_MOLE

7

Is not the player thingy just superfluous, as one person is certain to get one of the duplicates, and 12/51 = 4/17

Anyone know if S4 is interesting or not, I might consider it if I can do the exam in January.

Reply 9

Chewwy

17

KAISER_MOLE

Is not the player thingy just superfluous, as one person is certain to get one of the duplicates, and 12/51 = 4/17

Anyone know if S4 is interesting or not, I might consider it if I can do the exam in January.

ooh yeah.

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