Probability of A getting 4 spades given that B also gets 4 spades?

Question

What is the probability, given a deck of 52 cards randomly distributed to 4 players each getting 13 cards, person A getting 4 spades while person B also gets 4 spades, and the rest of the spades are distriubted among the remaining two people?

Answer 1

P = (13/52)(12/51)(11/50)(10/49) *
(9/48)(8/47)(7/46)(6/45)
P ≈ 1.7102*10^-6
Odds against = 584,721 to 1

If the a priori condition is "B has 4 spades", then
P = (9/48)(8/47)(7/46)(6/45)
P = 6.4755*10^-4
Odds against = 1,543 to 1

Answer 2

jamie b's intuition is misleading him (her?)
With 13 cards each of 4 suits, the expected number of spades would by 13/4 , which is not terribly different from 4.

There's a slight ambiguity in the statement of your problem.

Did you mean:
1. P(A gets 4 | B gets 4) , the conditional probability
or
2. P( A gets 4 and B gets 4)

Number 2 is easier:

With 13 spades in the deck:
...
I might be back to work on this.

What you'll need to do is get the ratio of the number of 13 cards hands with 4 spades, divide by the number of 13 card hands, 52C13 to get P( A gets 4 spades)

then
multiply by B's probability , but here there are 13-4 spades available because 4 of them are in A's hand. The denominator would be the same.

Then multiply these tow numbers.
(I'm a little unsure of this, but most of it should be correct)

Answer 3

For a bridge hand, find the probability that A gets 4 spades given that B got 4 spades.

Since it is given that B has 4 spades, there are only 9 spades and 48 total cards from which A's hand can be dealt.

P(A = 4 | B = 4) = C(13-4,4)*C(39,9) / C(52-4,13)
= C(9,4)*C(39,9) / C(48,13) = 0.1384

This is about 1 chance in 7.225.

Answer 4

(13!/4!9!)(39!/9!30!)(9!/4!5!)(30!/9!21!)=13!39!/4!4!5!9!9!21! That's the number of combinations. total number is 52!/13!13!26!. So just do the quotient.

Answer 5

not very good.