Bridge Probability

**mistykz** · January 30th 2009, 02:02 PM

I have absolutely no idea how bridge works, although I don't think it really applies to this problem...but anyway.

a) what is the probability that your bridge partner has exactly two aces, given that she has at least one ace?

b) what is the probability that your bridge partner has exactly two aces, given that she has the ace of spades?

I know that Pr(2 aces given >= 1 ace) = P(2A and >= 1A)/P(>=1A), but I don't know what these values are. What would be the probability of having at least 1 ace? I'm not sure how the cards are dealt and/or how many are held at a time

**awkward** · January 30th 2009, 03:51 PM

Originally Posted by mistykz

I have absolutely no idea how bridge works, although I don't think it really applies to this problem...but anyway.

a) what is the probability that your bridge partner has exactly two aces, given that she has at least one ace?

b) what is the probability that your bridge partner has exactly two aces, given that she has the ace of spades?

I know that Pr(2 aces given >= 1 ace) = P(2A and >= 1A)/P(>=1A), but I don't know what these values are. What would be the probability of having at least 1 ace? I'm not sure how the cards are dealt and/or how many are held at a time

A bridge hand is 13 cards dealt from the standard 52-card deck.

So, for example, the probability of having at least one ace in your hand is

$1 -\Pr(\text{no aces}) = 1 - \frac{\binom{48}{13}} {\binom{52}{13}}$

Is this enough to get you started?

**mistykz** · January 30th 2009, 06:33 PM

I think so... so the probability of having no aces would be 48/52? Did I read that correctly?

How would I find the probability of having exactly 2?

**awkward** · January 31st 2009, 07:48 AM

Originally Posted by mistykz

I think so... so the probability of having no aces would be 48/52? Did I read that correctly?

No, you're misunderstanding the notation.

$\binom{48}{13} = \frac{48!}{13! 35!}$ is the number of combinations of 48 objects taken 13 at a time (also know as a "binomial coefficient").

The probability of having no aces is
$\frac{\binom{48}{13}}{\binom{52}{13}}$
$= \frac{48!}{13! 35!} \cdot \frac{13! 39!}{52!}$
$= \frac {36 \times 37 \times 38 \times 39} {49 \times 50 \times 51 \times 52}$
$= 0.3038$

How would I find the probability of having exactly 2?

There are $\binom{52}{13}$ possible hands, each of which is equally likely. If you have exactly 2 aces then there are $\binom{4}{2}$ ways to pick the aces and $\binom{48}{11}$ ways to pick the remaining 11 cards, so the probability is

$\frac{\binom{4}{2} \binom{48}{11}}{\binom{52}{13}}$ .

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