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Two cards are drawn from a deck of 52 cards without replacement. What is the probability that the two cards have a combined value that is between 10 and 16? Jack, queen, king have a value of 10 each, ace has a value of 11.

I have tried to write all the possibilities that has the sum between 10 and 16. But I am not sure if it is the right way of solving this problem.

Edit : I am also not sure if drawing for example first 4 then 6 and first 6 then 4 are the same cases or not

Mustafa Süve
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3 Answers3

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It's ok to do that, also you can make some arpoximations to the value, if you want to take that way you can take a look here: What is the probability that 5 randomly chosen cards in a deck add up to 40 or greater? But in the end, in your case, calcs are not hard.

Community
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thebooort
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Cases: highest card and lowest card.

  • $11$-value and $2$ to $5$ value.
  • $10$-value and $2$ to $6$ value.
  • $9$ and $2$ to $7$ value.
  • $8$ and $2$ to $7$ value.
  • $7$ and $3$ to $6$ value.
  • $6$ and $4$ to $5$ value.

Remember there are two ways to obtain each of these.

Also consider the case of two cards of same-value-different-suits for each: $8$, $7$, and $6$.

Graham Kemp
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Your question about drawing a $4,6$ or a $6,4$ is a good one.

I'd break it down into two cases: one with the cards the same, the other with cards different.

If the cards are the same, first choose the face value of the card ($13$) then choose the two suits ($_4C_2 = 6$). So there are $13 \cdot 6 = 78$ possible ways to draw a pair. Of these, two $5$'s through two $8$'s satisfy your criterion: $4$ values $\times 6$ suit combinations for each $= 24$ possibilities.

If the cards are different, first choose the two face values ($_{13}C_2 = 78$) then choose the suit for each card ($16$). So there are $78 \cdot 16 = 1248$ ways to draw two different cards.

(Note: $_{52}C_2 = 1326 = 78 + 1248,$ so we've indeed accounted for all possibilities.)

Counting the number of possibilities with the sum between $10$ and $16$ with two different cards is bookkeeping. I'll start with the larger card, and work my way up from the bottom with the smaller cards to go through the possibilities methodically. I'll also not write a higher card second, because that would double-count.

$A2, A3, A4, A5$

$K2, K3, K4, K5, K6$

$Q2, Q3, Q4, Q5, Q6$

$J2, J3, J4, J5, J6$

$T2, T3, T4, T5, T6$

$92, 93, 94, 95, 96, 97$

$82, 83, 84, 85, 86, 87$

$73, 74, 75, 76$

$64, 65.$

I count $42$ combinations for the values. Since either card can be any suit, that's $42\cdot 16 = 672$ possible combinations.

Now we're basically done:

$$P = \frac{672+24}{1326} \approx 0.525.$$

John
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