Originally Posted by
PinkChip
But in the second case (dealer gets an Ace, with probability of 1/13):
when dealer has 15, his probability of busting by the next card is not 6/13 anymore.
If he has Hard 15, there are now 7 cards which would him bust: 7,8,9,10,J,Q,K.
So you must compute 1/13 * 7/13 in the second case.
Analogously, 1/13 * 8/13 in the third case, since if the dealer has 16,
he busts with 8 cards: 6,7,8,9,10,J,Q,K.
It gets further complicated, since in the second case, if the dealer has 15,
he could draw another Ace (1/13) and then get into the third case.
In my opinion, the result of these overall four cases is
6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)
= 6/13 + 7/169 + 8/169 + 8/2197
= 1014/2197 + 91/2197 + 104/2197 + 8/2197
= 1217/2197
= 55.4 %
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