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:
b) p₂ and p₃, combined probability p₂p₃
c) p₄ and p₅, combined probability p₄p₅
We need to know scenarios (a)OR(b)OR(c)=
So probability of B winning is 57.26%.