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The problem states: $5$ cards are dealt from a standard $52$ card deck. What is the probability that the sum of the values on the five cards is $48$ or more?

It is assumed of course that the value of face cards is $10$ and that of aces $11$. I know I am looking for the ratio between the number of possible outcomes with sum of values at least $48$, and the total number of possible outcomes, but I am having trouble finding the former quantity.

Any help is appreciated, thank you!

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  • $\begingroup$ I think the only solution is to casework. You will have to consider cases where the sum is equal to $48, 49, 50, 51, 52, 53$ or $54$. $\endgroup$ – Michael Wang Feb 26 '17 at 22:38
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    $\begingroup$ This seems like a job for brute force case-by-case based on the smallest card. If you have a $4$, all of the remaining cards must be aces. If you have a $5$ all of the remaining cards must be aces or three aces and a face card. If you have a $6$ it could be $69AAA,~6FFAA,~6FAAA,~6AAAA$ etc... You can count how many hands fall into the $6FFAA$ category for example by picking which $6$, which two face cards, and which two aces for a total of $4\cdot \binom{12}{2}\binom{4}{2}$. This will be tedious, but it will eventually yield a solution. $\endgroup$ – JMoravitz Feb 26 '17 at 22:38
  • $\begingroup$ @JMoravitz it's actually even worse than that as you also have to count all possible combinations for the higher sums as well, since the questions asks for all combinations whose sum is at least 48. $\endgroup$ – Michael Wang Feb 26 '17 at 22:45
  • $\begingroup$ @MichaelWang I already accounted for that in my comment, note that $6AAAA$ has a different total than $69AAA$. Whether you break into cases based on the sum or based on the lowest card, they both accomplish the same thing of beginning to partition the valid arrangements into smaller sets. $\endgroup$ – JMoravitz Feb 26 '17 at 22:45
  • $\begingroup$ @JMoravitz Oh right, sorry. I should have taken more time to read, oops. However, you also have to account for cases when the lowest card is a $7$, $8$, $9$, etc. $\endgroup$ – Michael Wang Feb 26 '17 at 22:50
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These are the ways to do it.

A) 4 aces and a card between 4 and 10.

B) 3 aces and two cards whose sum is 15 or higher and whose values are between 5 and 10.

C) 2 aces and three cards whose sum is 26 or higher and whose values are between 6 and 10.

D) 1 ace and 4 cards whose sum is 37 or higher and whose values are between 7 and 10.

E) no aces and 5 cards whose sum is 48 or higher and whose values are between 8 and 10.

So

A) there is one way to have 4 aces. And there are 40 cards between $4$ and $10$. (Ten ranks, 4,5,6,7,8,9,10,J,Q,K and 4 suits. So A) is 40.

B) There are 4 ways to have 3 aces. And the ways to have two cards add to more than 15 are

B1) two 10s. There are $12*11/2$ ways to do this.

B2) a ten and a card between 5 and 9. There are $12$ tens and $5*4$ cards between 5 and 9. So there are $240$ ways to do this.

B3) 2 9s. There are $4*3/2=6$ ways to do this.

B4) A 9 and a card betweem 6 and 8. There are $4*(3*4) = 48$ ways to do this.

B5) two 8s. There are $4*3/2=6$ ways to do this.

B6) an 8 and a 7. There are $4*4=16$ ways to do this.

So B1) = $4*12*11/2$;B2) $4*240$; B3) $4*6$; B4) =$4* 48$ B5) $4*6$; B6) = $4*16$

C) there are $3*4/2$ ways to pick two aces. Two pick three cards that are 26 or higher between $6$ and $10$ are

C1) Three 10s. There are $12*11*10/3!$ ways to do this.

C2) Two 10s. and a card between $6$ and $9$. There are $12*11/2$ times $4*4$ ways to do this.

C3) One 10. and two 9s. There are $12$ times $4*3/2$ ways to do this.

C4) A 10, a 9, and a 7 or an 8. There are $4*4$ times $2*4$ ways to do this.

C5) A 10, and two 8s. There are $4$ times $4*3/2$ ways to do this.

C6) three 9s. There are $4$ ways to do this.

C7) two 9s and an 8. There are $4*3/2$ times $4$ ways to do this.

D) there are $4$ ways to have one ace. The ways to have 4 carsd whose sum is 37 or higher for $7$ through $10$ is:

D1) four tens: $12*11*10*9/4!$ ways to do that.

D2) three tens: and a $7$, $8$ or $9$. $4$ times $3*4$ ways to do that

D3) two 10s, two nines. there are $4*3/2$ times $4*3/2$ ways to do that.

D4) two 10s, a nine and an eight. There are $4*3/2$ times $4*4$ ways to do that.

D5) one 10, 3 nines. There are $4$ times $4*3/2$ ways to do that.

E) the ways to do this are

E1) five 10s. There are $12*11*10*9*8/5!$ ways to do that.

E2) four 10s and an 8 or 9. There are $12*11*10*9/4!$ times $8$ ways to do that.

E3) three 10s and two 9s. There are ${12\choose 3}$ times ${4 \choose 3}$ ways to do that.

Multiply and add them up and divide by ${52 \choose 5}$.

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  • $\begingroup$ This would make for a decent recursive program. By hand though... difficult. $\endgroup$ – fleablood Feb 26 '17 at 23:32
  • $\begingroup$ Wouldn't you multiply B1-B6 by 4, since there are 4 ways to get 3 aces? $\endgroup$ – varcharvi Feb 27 '17 at 3:57
  • $\begingroup$ Oops. Typo disconnect somewhere along the line. It was four when I started typing. Somehow by the time I got to the end it became a three in my mind (because I was thinking three aces; who knows. It was an error.) $\endgroup$ – fleablood Feb 27 '17 at 7:30
  • $\begingroup$ Also a lot of them I should have written as ${12 \choose 4}$ rather than $12*11*10*9/4!$ but somehow I started typing and ... that came out. $\endgroup$ – fleablood Feb 27 '17 at 7:33

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