1. ## Probability (Bridge Hands)

Hello this is my first posting so I hope I'm doing everything right!

Okay so here is the question
----------------------------
What is the probability that a Bridge Hande (13 cards) contains:
a) 4 spades, 5 hearts, 3 diamonds, and 1 club

c) a void? (at least one suit that is not represented)
-----------------------------

Now I think I answered the first one correctly, here it goes:
a) (13C4 * 13C5 * 13C3 * 13C1)/(52C13)
= 0.00538779

b) now im not sure for this one but (is it correct?)...
(39C13)/(52C13) = 0.01279094804

c) I thought that the answer to this would be the same as b) since one suit was not represented in b)
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Well thanks in advance to anyone who responds with some input, I'm not great at this stats stuff, and our prof is not giving great notes :S.

Edit: woops sorry, I didn't see that I was in the Elementary and middle school section, I will try to repost in the correct section if someone could delete this posting

Hello this is my first posting so I hope I'm doing everything right!

Okay so here is the question
----------------------------
What is the probability that a Bridge Hande (13 cards) contains:
a) 4 spades, 5 hearts, 3 diamonds, and 1 club

c) a void? (at least one suit that is not represented)
-----------------------------

Now I think I answered the first one correctly, here it goes:
a) (13C4 * 13C5 * 13C3 * 13C1)/(52C13)
= 0.00538779

b) now im not sure for this one but (is it correct?)...
(39C13)/(52C13) = 0.01279094804

c) I thought that the answer to this would be the same as b) since one suit was not represented in b)
-------------------------------

Well thanks in advance to anyone who responds with some input, I'm not great at this stats stuff, and our prof is not giving great notes :S.

Edit: woops sorry, I didn't see that I was in the Elementary and middle school section, I will try to repost in the correct section if someone could delete this posting
Next time, just leave it. the moderators can move it to the correct forum

Welcome aboard!

What is the probability that a Bridge Hande (13 cards) contains:

a) 4 spades, 5 hearts, 3 diamonds, and 1 club?

c) a void? (at least one suit that is not represented)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Now I think I answered the first one correctly, here it goes:

$\displaystyle a)\;\;\frac{(_{13}C_4)(_{13}C_5)(_{13}C_3)(_{13}C_ 1)}{_{52}C_{13}} \;=\; 0.00538779$ . . . . correct!

b) now im not sure for this one ... but is it correct? . . . . yes!
. . $\displaystyle \frac{_{39}C_{13}} {_{52}C_{13}} \:= \:0.01279094804$

c) I thought that the answer to this would be the same as b)
since one suit was not represented in b)

In part (b), we have the probability that one particular suit is void.
. . And there are four suits that could be void.

Note that it says: at least one suit is void.
. . This means: one suit is void, or two suits are void, or three suits are void.

We must consider the three cases:

One suit void
There are 4 choices for the suit to be void.
Then there are: $\displaystyle _{39}C_{13}$ ways to select 13 cards from the other 39 cards.
. . $\displaystyle P(\text{1 suit void}) \;=\;\frac{4\cdot(_{39}C_{13})}{_{52}C_{13}}$

Two suits void
There are $\displaystyle _4C_2 = 6$ ways to choose the two suits to be void.
Then there are: $\displaystyle _{26}C_{13}$ ways to choose the 13 cards from the other 26 cards.
. . $\displaystyle P(\text{2 suits void}) \;=\;\frac{6\cdot(_{26}C_{13})}{_{52}C_{13})}$

Three suits void
There are $\displaystyle _4C_3 \,=\,4$ choices for the 3 suits to be void.
If three suits are void, all 13 cards are from the one remaining suit.
There is one way to select 13 cards of one suit.
. . $\displaystyle P(\text{3 suits void}) \;=\;\frac{4\cdot1}{_{52}C_{13}}$

The answer is the sum of these three probabilities.

4. Thank you so much for adding that reply, the detail really helped me understand what was meant. It is greatly appreciated, now hopefully I can help someone else out some day to return the favour

~Randy