A certain card game involves trying to get as close to 21 points without going over. The ace is worth 1 point and each face card is worth 10 points.

Suppose Player A draws a nine and a five and Player B draws a three and a ten. If Player A next draws a six, what is the probability of Player B winning the round?
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Player A has drawn a 9, 5 and 6 making 20, and so needs an ace to win. Any other card would make A go “bust” and B would win. B has drawn 2 cards so that makes 5 cards already drawn from a standard 52-card pack, leaving 47 cards. There are 4 aces left in the pack and all 12 face cards left. B must draw an 8 or less on the next draw to win or avoid going bust, and A must draw an ace to win.

Let’s see what’s left in the pack of 47 cards:

4 aces, 4 twos, 3 threes, 4 fours, 3 fives, 3 sixes, 4 sevens, 4 eights, 3 nines, 3 tens, 12 face cards. 29 cards have face values of 8 or less; 18 cards have face values of 9 or more—B would lose with these. B would win outright if 8 were to be drawn next (probability p₁=4/47).

B would win by default if:

(1) B drew a face value between 2 and 7 (probability p₂=21/47); AND 

(2) A drew anything except an ace from the remaining 46 cards (probability p₃=42/46);


(3) B drew an ace (probability p₄=4/47); AND

(4) A drew anything except an ace from the remaining 46 cards (probability p₅=43/46).

So there are 3 scenarios involving probabilities:

a) p₁

b) p₂ and p₃, combined probability p₂p₃

c) p₄ and p₅, combined probability p₄p₅

We need to know scenarios (a)OR(b)OR(c)=

p₁+p₂p₃+p₄p₅=0.0851+0.4080+0.0796=0.5726 approx.

So probability of B winning is 57.26%.

by Top Rated User (840k points)

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