Question

Suppose that a deck of 52 cards contains 26 red cards and 26 black cards. Say we use the 52 cards to randomly distribute 13 cards each among two players (2 players receive 13 card each).

a. How many ways are there to pass out 13 cards to each of the two players?

b. What is the probability that player 1 will receive 13 cards of one color and player 2 receive 13 cards of the other color?

Answer 1

Solution

Back-up Theory

Probability of an event E, denoted by P(E) = n/N ……………………………………………(1)

where n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}……(2)

Now to work out the solution,

Part (a)

Vide (2), the first player can be given any 13 cards out of 52 cards in ⁵²C₁₃ ways

= (52!)/{(13!)(39)!}.

Now, only 39 cards are available and hence the second player can be given any 13 cards out of 39 cards in ³⁹C₁₃ ways

= (39!)/{(13!)(26)!}.

Thus, number of ways to pass out 13 cards to each of the two players

= [(52!)/{(13!)(39)!}] x [(39!)/{(13!)(26)!}]

= 5.15785E+21 ANSWER

Part (b)

The first player can 13 of red or 13 of black cards – this can happen in two ways. For each of these, 13 cards out of 26 cards of that colour can be given in ²⁶C₁₃ ways

= {(26!)/{(13!)(13)!}.

Once one colour is given to the first player, the second has to be given only the other colour. So, this happen only in 1 way. Again, 13 cards out of 26 cards of that colour can be given in ²⁶C₁₃ ways

={ (26!)/{(13!)(13)!}.

Thus, number of ways player 1 will receive 13 cards of one color and player 2 receive 13 cards of the other color = 2 x{ (26!)/{(13!)(13)!}²

So, vide (1), n = 2 x{ (26!)/{(13!)(13)!}² = 20801200.

And vide (1), N = 5.15785E+21 [as obtained in Part (a)]

Thus, the required probability = 20801200/5.15785E+21 = 4.03E-15 ANSWER

DONE