Christoffer plays 21 with a friend. In front of him he has three cards: a 4, an 8, and a 2. The total is 14, and he is now wondering how likely it is that he will get 21.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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You have a 4, an 8, and a 2 and you want to know the probability of getting 21:

There are a few different ways to reach 21. We will add the probabilities of each to get the total probability. We assume that you are using 1 standard deck of cards.

(1) The next card could be a 7. The probability of this is 4/49. (There are 4 7's in the deck, and there are 49 cards left after drawing the three initial cards.)

(2) You could draw a 6 and an ace. The probability of drawing a 6 then an ace is 4/49*4/48=1/147. The probability of drawing an ace then a 6 is 4/49*4/48=1/147 so the probability of getting a 6 and an ace is 2/147.

(3) You could get a 5 and a 2, or a 5 and 2 aces. The probability of a 5 then a 2 is 4/49*3/48=1/196 (there are only 3 2's remaining in the deck); the probability of a 2 then a 5 is 1/196 so the total for a 5 and a 2 is 1/98. The probability of a 5 with 2 aces is 6(4/49*4/48*3/47)=6/2303. (There are 6 ways to do this: 5Aa,5aA,A5a,a5A,aA5,Aa5)

6/2303+1/98 is 59/4606.

(4) You could get a 4 and a 3; a 4 and a 2 and an ace; or a 4 and 3 aces.

4&3: 2(3/49*4/48)=1/98

4&2&ace: 6(3/49*3/48*4/47)=9/4606

4&3aces: 24(3/49*4/48*3/47*2/46)=18/52969

(5) You could get 2 3's and an ace; a 3 and 2 2's; a 3, a 2, and 2 aces; a 3 and 4 aces:

2 3's and an ace: 6(4/49*3/48*4/47)=6/2303

3 and 2 2's: 6(4/49*3/48*2/47)=3/2303

3,2, 2aces: 24(4/49*3/48*4/47*3/46)=36/52969

3 4aces: 120(4/49*4/48*3/47*2/46*1/45)=8/158907

The total probability is 4/49+2/147+59/4606+1/98+9/4606+18/52969+6/2303+3/2303+36/52969+8/158907=.12518

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Assuming the only cards drawn from a single standard deck are Christoffers, starting with an initial draw of 2,4,8; his probability of getting 21 is a little more than 1/8.

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