I was playing in a qualifying round for a tournament on December 21st. I'm just curious as to what my chances were the way things unfolded. The only thing at stake was first place(100 points) vs second place(95 points). Going into the last hand in betting and playing order here are the bankrolls and bets BR3 - $450 - $50 BR4 - $100 - $100 BR2 - $2400 - $1200 BR1(me) - $2785 - $1000 I now realize I should've bet $1200 but that's not the point. Dealer had a 9 up. Don't remember what BR3 had or did. BR4 had a 16 and stood on it. BR2 had a 20 and stood on it. I had a 13 and hit it, getting a 5 for 18. I stood there. Given that with a dealer 17, 21 or bust I win. 18, 19 or 20 and I lose. I have 2 questions. 1) What is my chance of winning given the dealer 9 up? 2) Dealer had a 3 down for 12. Now what are my chance of winning? Dealer ended up with a 7 for 19 so I ended up 2nd. I'm just curious what my chances were.
Based on infinite deck figures, you were approximately a 41% underdog to begin with, and a 69% favourite once the dealer's hole card was revealed.
Your chances were slim to none. Anyway after screwing up on my initial bet I would have doubled down on my 13 to get to 18 In that case the other player might be forced to split depending on how he was playing but atleast giving me a chance to catch up. Not sure if any one would have done the same
Doubling would be the wrong play since BR2 played before BR1 and BR2 stood on his 20. A single win by BR1 would have guaranteed BR1 1st place. Also, a double results in a reduction of BR1's probability of winning because if the dealer drew to 21 and BR1's double down card did not result in a 21 then BR2 wins. Without the double, BR1 would win if the dealer drew to 21 irregardless of BR1's total or even if BR1 busted.
