Probability: a specific hand of cards

[mbrmbrg]

Homework Statement

Find the probability of a bridge hand (13 cards) with the AKQJ of spades and no other spades.

Homework Equations

$$P(event)=\frac{\# \ \\favorable \ \\outcomes}{total \ \# \\outcomes}$$

$$nCr=\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}$$

There are 52 cards in a deck, 13 of which are spades.

The Attempt at a Solution

• The total number of possible bridge hands is $$\left(\begin{array}{c}52\\13\end{array}\right)$$
• To get a hands with AKQJ of spades, pick them and then fill your hand with 9 other cards from the remaining 4 cards. So the committees containing AKQJ are $$\left(\begin{array}{c}48\\9\end{array}\right)$$
• But how many of those $$\left(\begin{array}{c}48\\9\end{array}\right)$$ hands don't have spades? Well, let's remove the 9 remaining spades from the deck, and see how many ways we now have to fill our hand: $$\left(\begin{array}{c}48-9\\9\end{array}\right)=\left(\begin{array}{c}39\\9\end{array}\right)$$
• Here's where I'm stuck. The result in the previous bullet tells me how many hands will have four spades. But I want 4 particular spades; and intuitively, I know that I'm less likely to pull a run of four spades than any group of 4 spades. I don't know where to go from here. Should I compute the probability of drawing 9 non-spades and then compute (separately) the probability of drawing a run of spades from the remaining cards? I don't know how to compute the probability for a run, nor do I know how to combine those two probabilities to get the answer to the problem.

Thanks in advance for any help!

 

[statdad]

Think about counting the spades this way:
You want 1 Ace of Spades from the total of _____ Aces of Spades in the deck, and
you want 1 King of Spades from the total of _____ Kings of Spades in the deck, and
you want 1 Queen of Spades from the total of _____ Queens of Spades in the deck, and
you want 1 Jack of Spades from the total of _____ Jacks of Spades in the deck.

Count each separately then _________ because of the "and"s

For the other 9 cards: remember that you do not want any more spades in your hand. There are 47 cards that are not the ones you've mentioned above, but some are the other spades. Take the count of the remaining spades from 47 and select the remaining 9 from them.

Sorry for the verbosity - things get wordy with descriptions and no math.

 

[Architeuthis Dux]

Hello, thank you for your question. Let's break it down step by step.

First, we need to find the total number of possible bridge hands, which is given by the combination formula nCr = (n!)/(r!(n-r)!). In this case, n = 52 (total number of cards) and r = 13 (number of cards in a bridge hand). Therefore, the total number of possible bridge hands is (52!)/(13!(52-13)!) = 635,013,559,600.

Next, we need to find the number of hands that contain the AKQJ of spades. This can be done by selecting these four cards and then filling the remaining 9 cards from the remaining 48 cards. Therefore, the number of hands with the AKQJ of spades is (48!)/(9!(48-9)!) = 1,712,304.

Now, we need to find the number of hands that do not contain any other spades. This can be done by selecting 9 cards from the remaining 39 non-spade cards. Therefore, the number of hands without any other spades is (39!)/(9!(39-9)!) = 8,758,330.

Finally, we can calculate the probability of getting a hand with the AKQJ of spades and no other spades by dividing the number of favorable outcomes by the total number of possible outcomes. Therefore, the probability is 1,712,304/635,013,559,600 = 0.0000027 or approximately 0.0003%.

I hope this helps. Let me know if you have any further questions. Good luck with your studies!