1. ## Probability...

Find the probability that a bridge hand (of 13 cards) is void in at least one suit (i.e. contains no cards of that suit)

So... The ways ive tried to do this filled 9 pages so ill try summarise.

The number of different ways 13 cards can be chosen from a 52 card pack is roughly $\displaystyle 6.35$x$\displaystyle 10^{11}$ (i think, im doing this from memory) so ive tried to figure out the number of hands that involve all 4 suits and also the number of hands that involve only 1,2 and 3 suits. Cant figure out any of these tho...

Couple of other ways that i cant remember, dont have my sheets with me, Can anyone give an answer with working? (by the way the answer is 0.051)

Cheers

Find the probability that a bridge hand (13 cards) is void in at least one suit
There are $\displaystyle {52\choose13}$ possible hands.

Void in one suit

There are 4 choices for the voided suit.
We must choose 13 cards from the other 39 cards.
. . There are $\displaystyle {39\choose13}$ ways.
Hence, there are: .$\displaystyle 4{39\choose13}$ ways to be void in one suit.

Void in two suits.

There are $\displaystyle {4\choose2} = 6$ choices for the two suits.
We must choose 13 cards from the other 26 cards.
. . There are $\displaystyle {26\choose13}$ ways.
Hence, there are: .$\displaystyle 6{26\choose13}$ ways to be void in two suits.

Void in three suits.

There are $\displaystyle {4\choose3} = 4$ choices for the three suits.
We must choose 13 cards from the other 13 cards.
. . There is $\displaystyle {13\choose13} = 1$ ways.
Hence, there are: .$\displaystyle 4(1) = 4$ ways to be void in three suits.

$\displaystyle P(\text{void in at least one suit}) \;=\;\frac{4{39\choose13} + 6{26\choose13} + 4}{{52\choose13}}$

Find the probability that a bridge hand (of 13 cards) is void in at least one suit (i.e. contains no cards of that suit)

So... The ways ive tried to do this filled 9 pages so ill try summarise.

The number of different ways 13 cards can be chosen from a 52 card pack is roughly $\displaystyle 6.35$x$\displaystyle 10^{11}$ (i think, im doing this from memory) so ive tried to figure out the number of hands that involve all 4 suits and also the number of hands that involve only 1,2 and 3 suits. Cant figure out any of these tho...

Couple of other ways that i cant remember, dont have my sheets with me, Can anyone give an answer with working? (by the way the answer is 0.051)

Cheers
The number of hands void in at least one suit is $\displaystyle 4 \, {39 \choose 13} - 6 \, {26 \choose 13} + 4$.

The first term comes from the number of ways of picking a hand void in one suit.

But it counts the number of hands void in two suits twice ...... So you must subtract the number of hands void in two suits. Hence the second term - the number of hands void in two suits. Note: There are six possible pairs of suits, hence the 6.

But wait ..... what about hands void in three suits ....... ? They're counted three times (once for each suit) in the first term, then subtracted due to the second term for each of the three pairs of suits they are void in. This means they haven't been counted at all! Hence the third term is required - the number of hands void in three suits.

Make sense?

Now divide the above number by $\displaystyle {52 \choose 13}$ (the number of ways of picking a hand without restriction) to get the required probability.

4. Originally Posted by Soroban

There are $\displaystyle {52\choose13}$ possible hands.

Void in one suit

There are 4 choices for the voided suit.
We must choose 13 cards from the other 39 cards.
. . There are $\displaystyle {39\choose13}$ ways.
Hence, there are: .$\displaystyle 4{39\choose13}$ ways to be void in one suit.

Void in two suits.

There are $\displaystyle {4\choose2} = 6$ choices for the two suits.
We must choose 13 cards from the other 26 cards.
. . There are $\displaystyle {26\choose13}$ ways.
Hence, there are: .$\displaystyle 6{26\choose13}$ ways to be void in two suits.

Void in three suits.

There are $\displaystyle {4\choose3} = 4$ choices for the three suits.
We must choose 13 cards from the other 13 cards.
. . There is $\displaystyle {13\choose13} = 1$ ways.
Hence, there are: .$\displaystyle 4(1) = 4$ ways to be void in three suits.

$\displaystyle P(\text{void in at least one suit}) \;=\;\frac{4{39\choose13} + 6{26\choose13} + 4}{{52\choose13}}$
Soroban, I think you'll find your total number of hands void in at least one suit has too many hands void in two suits and too many hands void in three suits ....... (My reply contains the reason why).

5. Hey, mr fantastic!

You're absolutely right . . . *blush*

I had a lot of overlapping . . . Thanks for catching it!