Conditional Probability and Drawing Cards

[Sizwe]

Homework Statement

A standard deck of 52 cards of 4 suits, each with 13 denominations, is well shuffled and dealt out to four players, N, S, E and W, who each receive 13 cards. If N and S have exactly ten cards of a specified suit between them, show that the probability that three remaining cards of the suit are in one player's hand (either E or W) is 0.22

Homework Equations

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

The Attempt at a Solution

I completed this question a few months ago with the solution:

$\frac{2\binom{23}{10}}{\binom{26}{13}\binom{13}{13}} = 0.22$

Problem is, I have no idea how I got to that solution. I try now but end up with probabilities greater than 1, or very very small probabilities. Some clarification on how I got my solution, or how anyone would solve this, would be appreciated :)

 

[LCKurtz]

Homework Statement

A standard deck of 52 cards of 4 suits, each with 13 denominations, is well shuffled and dealt out to four players, N, S, E and W, who each receive 13 cards. If N and S have exactly ten cards of a specified suit between them, show that the probability that three remaining cards of the suit are in one player's hand (either E or W) is 0.22

Homework Equations

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

The Attempt at a Solution

I completed this question a few months ago with the solution:

$\frac{2\binom{23}{10}}{\binom{26}{13}\binom{13}{13}} = 0.22$

Problem is, I have no idea how I got to that solution. I try now but end up with probabilities greater than 1, or very very small probabilities. Some clarification on how I got my solution, or how anyone would solve this, would be appreciated :)

Once N and S have their cards, there are 26 cards left for E and W, 3 of which are of the particular suit. For E to get all three, he must choose 3 from those 3 then 10 more from the remaining 23. Does that help?

 

[Ray Vickson]

Once N and S have their cards, there are 26 cards left for E and W, 3 of which are of the particular suit. For E to get all three, he must choose 3 from those 3 then 10 more from the remaining 23. Does that help?

Your formula is weird. It looks wrong, but happens by accident to deliver a correct result---essentially because some of the factors or divisors are 1.0, and so do not affect anything. I suggest you try to use a formula similar to yours on the problem of computing the probability that E gets 2 of the "special" cards (instead of 3). Then you really will see a difference between your formula and the correct one.

RGV

 

[Sizwe]

Once N and S have their cards, there are 26 cards left for E and W, 3 of which are of the particular suit. For E to get all three, he must choose 3 from those 3 then 10 more from the remaining 23. Does that help?

Yes :) Thank you! I see it now as give N and S their cards including ten from the suit. Then give the three remaining to either E or W (hence I mutliply by 2), then all that remains is for E or W to choose 10 cards from the 23 left over. Any other choices made cancel through division of the number of ways N and S can get 10 of a suit between them (the denominator). I think the lesson learned here is to show all my steps :P

Thanks again guys :)