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[EMAIL PROTECTED] (Bob R) wrote, "Let's say I'm playing blackjack at a table and the dealer's hole card has been accidentally exposed to reveal he has a hard 16. The pit knows the card is up so I don't have to be fearful of making outrageous plays. I have two tens, so I split them. On one of them I get an ace. Do I double? I am guessing yes." You're right to double. Your EV for a 21 is 12/13, and your EV for doubling is 14/13. Below is the derivation. (In fact, there's a similar game called "double exposure" in which both dealer's cards are face up. When the dealer has hard 16 the player should split a pair of tens and double on any resulting soft 21's, if allowed to.) Probablity of dealer getting various totals after hitting hard 16________________________: dealer21: 1/13 dealer20: 1/13 dealer19: 1/13 dealer18: 1/13 dealer17: 1/13 Player's EV=1-2prob(lose)-prob(push): player21: 12/13 player20: 10/13 player19: 8/13 player18: 6/13 player17: 4/13 player<17: 3/13 EV for player doubling soft 21 = 2x(4x12/13+10/13+8/13+6/13+4/13+5x3/13)/13 =14/13
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