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Five cards are dealt from a standard 52 card deck. What is the probability that the sum of the faces on the five cards is 48 or more?

Attempt: Five cards can be selected out of $52$ cards in $\binom{52}{5}$ ways. Then There will be 47 cards remaining. Let $A_1 =$ number on first card, $\ldots, A_5 =$ number on fifth card. Then we need to find $P(A_1 \cap A_2 \cap \ldots \cap A_5) = P(A_1) + P(A_2) + \ldots + P(A_5)$.

I am not sure if this is a way to approach this problem. Please can someone help me start this problem? Thank you very much.

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    $\begingroup$ How are we giving value to the face cards? Are they J 11, Q 12, K 13? Are we doing it the blackjack way, where all the face cards are worth 10? $\endgroup$ – Dan Uznanski Sep 29 '14 at 2:08
  • $\begingroup$ Nothing else is stated, just the above words. $\endgroup$ – user20877 Sep 29 '14 at 2:11
  • $\begingroup$ @user58756496 well, without that information, we cant tell you the exact answer with 100% certainty, because we dont know what 48 or more means. $\endgroup$ – Asimov Sep 29 '14 at 2:13
  • $\begingroup$ oh oh, or it could be the face cards are of value 0, because they don't have any numbers on them. To adjust David's answer below, you'd need to use C(4,k) for tens instead of of C(16, k). $\endgroup$ – Dan Uznanski Sep 29 '14 at 2:24
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If face cards count as $11,12,13$ there are going to be lots of possible sums (up to $64$) and I don't see any quick way of doing the problem.

Assuming that face cards count as $10$, there are four ways you can get a sum of $48$ or more from $5$ cards:

  • $10+10+10+10+10$;
  • $10+10+10+10+9$;
  • $10+10+10+10+8$;
  • $10+10+10+9+9$.

You need to find the probability of each case and then add them. I'll do the last one as an example. To choose a hand of this kind,

  • choose three $10$-cards from $16$ possibilities. . . . . $C(16,3)$ ways;
  • choose two $9$s. . . . . $C(4,2)$ ways.

So the probability for this bit is $$\frac{C(16,3)C(4,2)}{C(52,5)}\ .$$ See how you go with the rest.

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  • $\begingroup$ Why choose 10 cards from 16 possibilities? $\endgroup$ – user20877 Sep 29 '14 at 2:17
  • $\begingroup$ Because there are $16$ of them, assuming face cards count as $10$. $\endgroup$ – David Sep 29 '14 at 2:18
  • $\begingroup$ So for the second one it would be: Choose four 10-cards from 16 possibilities...C(16,4). And choose one 9-card from 3 possibilities. Thus C(16,4)*C(9,3)/[C(52,5)]? Also aren't there only 3 possibilities for 9? $\endgroup$ – user20877 Sep 29 '14 at 2:27
  • $\begingroup$ If it's an ordinary pack of cards there are four of every value I think. $\endgroup$ – David Sep 29 '14 at 2:40
  • $\begingroup$ Then how can there be 16 possibilities for a three 10 cards? Then would it not be four too? $\endgroup$ – user20877 Sep 29 '14 at 2:44
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If the cards range from 1 to 10, take the complement which will range from 0 to 9.

With the complements you need to reach 1 or less. There are more than 5 zeros. So the answer is clear.

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