Probability: a specific hand of cards

In summary, the probability of getting a bridge hand with the AKQJ of spades and no other spades is equal to the number of favorable outcomes over the total number of possible outcomes. To calculate this, we first need to determine the total number of possible bridge hands, which is equal to 52 choose 13. To get a hand with the specified spades, we need to choose them and then fill the remaining 9 cards from the remaining 48 cards. However, we also need to account for the possibility of getting other spades in our hand, so we subtract the number of ways to get 9 spades from the remaining 39 cards. Finally, we need to consider the probability of getting the specific AKQ
  • #1
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Homework Statement



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

Homework Equations



[tex]P(event)=\frac{\# \ \\favorable \ \\outcomes}{total \ \# \\outcomes}[/tex]

[tex]nCr=\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}[/tex]

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 [tex]\left(\begin{array}{c}52\\13\end{array}\right)[/tex]
  • 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 [tex]\left(\begin{array}{c}48\\9\end{array}\right)[/tex]
  • But how many of those [tex]\left(\begin{array}{c}48\\9\end{array}\right)[/tex] 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: [tex]\left(\begin{array}{c}48-9\\9\end{array}\right)=\left(\begin{array}{c}39\\9\end{array}\right)[/tex]
  • 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!
 
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  • #2
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.
 
  • #3


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!
 

What is probability?

Probability is a measure of the likelihood of an event occurring. It is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

How is probability calculated?

The probability of a specific hand of cards can be calculated by dividing the number of ways that hand can occur by the total number of possible hands in the deck. For example, the probability of getting a royal flush in a standard 52-card deck is 4 divided by 2,598,960, which is approximately 0.00000154.

What is a specific hand of cards?

A specific hand of cards refers to a combination of cards that a player holds in a card game. This could be a poker hand such as a royal flush, or a blackjack hand such as a 10 and an ace.

How does probability affect card games?

Probability plays a crucial role in card games as it determines the likelihood of a certain hand occurring. Players can use probability calculations to make strategic decisions and improve their chances of winning.

Can probability be influenced or controlled in card games?

No, probability cannot be influenced or controlled in card games. The outcome of each hand is completely random and cannot be predicted or manipulated.

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