I was involved in the following situation at the recent New Frontier tournament... We can ignore first place for the purposes of this example, and assume that me and Mike Sherman are battling for second place. He has a small lead on me and has matched my bet since he bets last. (I think the numbers were $180 to $192.50 with both of us betting $90.) After the cards are dealt, here's the situation: Dealer has a ten up. Me.......$180.....$90.......T9......Stand Mike.....$192.....$90.......A25.....??? The question, should Mike hit or stand on his soft 18? At the time, I thought standing was the best play. Afterwards, I kept flip-flopping on the decision. By standing, Mike advances unless the dealer makes 18 or 19. That's a 75.86% shot. (Infinite deck approximations are used throughout this analysis.) If he hits once and only once, it's interesting... There's NO CHANGE in his chances. 3/13 of the time, he's a lock by drawing Ace,2, or 3. (100%) 4/13 of the time, he's in exactly the same situation, drawing a ten. (75.86%) 6/13 of the time, his hand gets worse, and now he loses to a dealer 17 as well, giving him a smaller chance. (63.79%) Average it out, and you get exactly the same answer as for standing! His chance of advancing is still 75.86%! However, hitting once and only once is not optimal for Mike. If he ends up with hard 12 or hard 13, he should hit. That increases his chances marginally. The final numbers: Standing, he advances 75.86% of the time. By hitting to hard 14 or soft 19, he advances 76.07% of the time. That's a pretty close call. No wonder we couldn't decide! FYI, he stood, the dealer was pat with 18, and I advanced as a 3 to 1 underdog. P.S. I'm well aware that using two decimal places of precision is pointless when I'm introducing more error than that by using infinite decks.

shortcut Mike ask my opinion about this case and I remind from my charts of "How to avoid swing" that we must hit 'til hard 14 (opponent 19, dealer T), however I couldn't remember about softs. Around the table it would be possible to find out if one knows that Dealer makes 17 and 18 from T with the same probability. Here's the logic: if we hit only ones with 3/13 (A, 2, 3) we gain extra dealer's 18 and 19, with 6/13 (4,5,6,7,8,9) we lose dealer's 17 and with 4/13 we just get same outcome. Now since D17 = D18 = K by hitting ones we get same that we lose (3 x 2K = 6 x K). As Ken accurately mentioned in his post you can slightly improve your play hitting hard 12 and 13. That biases our decision to HIT. Tirle_bj