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A 5 card hand is dealt from a well-shuffled deck of 52 poker cards. If the first two cards are the 10 of diamonds and the 10 of hearts, what is the probability of having been dealt a full?

This is what I did:

Let $A$ be the event the first two cards are the 10 of diamonds and the 10 of hearts. We want to compute

$P(full|A)=\frac{P(A|full)P(full)}{P(A)}$

Yet I don't know how to compute $P(A|full)$. It this the right path? How could I solve this problem?

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By "full" I take it you mean a full hand?

You don't need Baye's rule for this.

The probability of being dealt a full hand when it is given that you have been dealt those two tens is the probability of also being dealt (among the 3 other cards drawn from the remaining 50) either:

  • another ten (1 of 2 suits) and a pair of something else (1 of 12 faces, 2 of 4 suits), or
  • a triple of something else (1 of 12 faces, 3 of 4 suits).

$$\mathsf P(F_{house}\mid 10\diamondsuit,10\heartsuit) = \frac{{2\choose 1}{12\choose 1}{4\choose 2}+{12\choose 1}{4\choose 3}}{{50\choose 3}}$$

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Given that you have already been dealt the two tens, you have two potential scenarios for a full house. The first scenario would be to get one more ten and a pair, and the second scenario would be to get three of a kind. We can compute these the probabilities of these two scenarios separately and sum them.

First, the probability of one more ten and another pair. The probability of the third card being a ten is $\frac{2}{50}$. The probability of the fourth card being something other than a ten is $\frac{48}{49},$ since there is now only one ten left. The probability of the last card being the same as the fourth card is $\frac{3}{48}.$ All together, the probability of the first scenario is $\frac{2}{50}*\frac{48}{49}*\frac{3}{48} \approx 0.00245.$

Now, the probability of three of a kind. The probability of the third card being something other than a ten is $\frac{48}{50}.$ The probabilities of the fourth and fifth cards being the same as the third are $\frac{3}{49}$ and $\frac{2}{48},$ respectively. All together, the probability of the second scenario is $\frac{48}{50}*\frac{3}{49}*\frac{2}{48} \approx 0.00245.$

The total probability is the sum, or 0.0049.

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  • $\begingroup$ In the first scenario, you're missing a factor of $3$ because the ten could come as any of the three cards. $\endgroup$ – joriki Mar 24 '16 at 20:58

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