Deck of Cards Probability Question

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The discussion centers on calculating the probability of getting at least one four-of-a-kind in a thirteen-card hand from a standard fifty-two card deck. Participants initially estimate the probability to be around 3.43%, with one user calculating it as 0.0753 using combinations. There is confusion regarding the elimination of repetitious combinations when multiple four-of-a-kinds are possible in a hand. Clarifications are made that the focus is on at least one four-of-a-kind, not the exact count. The final consensus acknowledges the probability should be expressed as a decimal, approximately 0.034.
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Homework Statement



What is the probability of gettting four-of-a-kind in a thirteen card hand dealt from a standard fifty two card deck?

Homework Equations





The Attempt at a Solution

 
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Are you asking for the probably at least one four of a kind? Or is having multiple four of a kinds in the hand illegal?
 
I am asking for at least one four of a kind.
 
I got about 3.43%. What have you tried so far?
 
vela said:
I got about 3.43%.

I got this answer as well.
 
I got .0753 = ((13)(52C9))/(52C13) although i suspect that there are some repetitious combinations ie. 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.

Could you explain how you arrived at your answer?
 
Your mistake is in the numerator. After you draw the four of a kind, how many cards are left from which to draw the remaining 9 cards?
 
ahh, i see
Thanks!
I'm still not sure how to eliminate the repetitious combinations though...
 
You haven't eliminated repetitions like getting two or three sets of four of a kind, but that's okay because you said you wanted the probability of at least one four of a kind, not exactly one four of a kind.
 
  • #10
"You haven't eliminated repetitions like getting two or three sets of four of a kind"

In calculating the combinations for any particular four-of-a-kind, for example for the combinations of the four-of-a-kinds for cards 4 and 7, there will be one combination that is counted twice: 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.
 
  • #11
Yeah, you're right. I'll have to ponder this a bit.
 
  • #12
This might help:

\{\texttt{hands with three four-of-a-kinds}\} \subset \{\texttt{hands with at least two four-of-a-kinds}\}
\subset \{\texttt{hands with at least one four-of-a-kind}\}
 
Last edited:
  • #13
vela said:
I got about 3.43%. What have you tried so far?

pbandjay said:
I got this answer as well.

I agree, but remember probabilities are numbers between 0 and 1, not percentages.

So you mean .034 approximately.
 
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