In a tournament Tuesday night the following happened. 4 People left at the table, $100-$5000 bet range. Last Hand. In order from 3rd base BR4 - $12,250 BR3 - $12,525 BR1 - $16,700 BR2 - $12,700 I'm BR2 and BR3 was on the button. Everyone bet $5000. Cards were dealt as follows BR4 - K6 BR3 - Q5 BR1 - J7 BR2 - KJ Dealer - 3 Play was as follows BR3 - Doubled and Broke BR4 - Doubled and Broke BR2 - Split and stayed on 18 and 14 BR1 - Stayed Dealer - Turned over a 9 and drew a 5, giving the victory to BR1 My question is what would've been the best play for me. I can think a couple options. First would be stand on my 20. That would give me the win if the dealer hits 17-20. The other option was split like I did. Once I split and got the 8 and the 4, I have a couple options on the 4. Stay like I did or hit the 14 to at least 17. If I can hit the 14 to 17 or higher that makes the decision by BR1 a little more difficult. From my standpoint, what's the highest percentage play?
No problem with your play Mr. Bill, check your last sitting chart, I think you mixed up the seats on the 3rd chart. I think you played the hand correct, I would have split the 20 on that hand. I think you had a better chance of winning with the split vs. hoping the dealer drew out 17-20. Once you split you couldn't afford hitting either hand and busting. You were in a bad situation, stand and and you would have lost also with the dealers bust. Hopefully it was in the finals and 2nd paid something.
I ordered the 3rd chart the way it played out. BR3 played first, followed by BR4, followed by me at BR2 and fianlly BR1. I did notice though that on the first list I had BR4 and BR3 amounts swapped. I've edited the post to fix that. Unfortunately it was the semi-final table and the only thing second got was $100 worth of action chips. I was able to turn that into $150 though because I bet it all on one hand and got a BJ.
Chances of winning with a 20 This question peaked my interest. I looked up the odds of the dealer making a hand starting with a 3 (www.wizardofodds.com) with the following results: 6 decks – dealer hits a soft 17 (my assumption) Dealer making a 17 = 0.1259 Dealer making a 18 = 0.13193 Dealer making a 19 = 0.126656 Dealer making a 20 = 0.122185 TOTAL = 0.506671 or: Dealer probability of making a 17, 18, 19, or 20 = 50.6671%. Dealer probability of busting is 37.6958%. Assuming BR1 stands on his 17: If the dealer hits to 17 or 20 (approx. 25% probability) you gain a max bet. If the dealer hits to 18 or 19 (approx. 25% probability) you get a swing - you gain 2 max bets. If the dealer hits to 21 or busts (approx. 50% probability) your relative bankrolls remain the same. From this it seems to me that standing on the 20 is the action to take. Again assuming BR1 stands on his 17, you have a 50% probability of staying at BR2 and a 50% probability of becoming BR1. A virtual coin flip.
According to Wong In Wong's book, page 135, paragraph 3: Thus Curt's Revenge means doubling the bet for the best chance to win the match. Use it on the last hand of play if winning a double bet is enough to to win the match, but winning a single bet is not enough. Since winning a single bet is all that is necessary 50% of the time (given the described players' bankrolls), seems to me that splitting is taking on additional risk. Also, all the percentages Wong gives under "Doubling on anything" or "Curt's Revenge" are less than a 50% probability of winning.
Not sure which is highest percentage! Like I said earlier, my play would be the split. Is it the highest percentage play? I am not sure! A single bet would not give him a win, MrBill needed a swing with BR1 for the win. With BR1 catching a hard 17 the chances are they weren't going to be hitting and a dealers bust would still give BR1 the win. MrBill had to get a 2nd bet on the board, with him having a hard 20 a DD was out of the question. So the split was really the only way he could go. If the dealer would have made a hand beating BR1 and Mrbill would then just needed one of his hands to win to of won the table. Personally I like those odds better then standing and hoping for the swing.
Splitting Tens There are additional ways to advance by splitting. He gets the full 37.7% of the time when the dealer busts. He also gets a portion of the 62.3% of the time that the dealer makes his hand. If the dealer hits to 17, then he needs to at least win one and push the other. 0.1259 * 5/13 * 0 = 0.0000 (Does not hit to at least 17, doesn't matter) 0.1259 * 1/13 * 7/13 = 0.0052 (Hit to 17, needs 18 or better) 0.1259 * 7/13 * 8/13 = 0.0417 (Hit to 18 or better, needs 17 or better) If the dealer hits to 18, then he needs to win either hand or push both hands. 0.13193 * 6/13 * 6/13 = 0.0281 (Does not hit to at least 18, needs 19 or better) 0.13193 * 1/13 * 7/13 = 0.0055 (Hits to 18, needs 18 or better) 0.13193 * 6/13 * 1 = 0.0609 (Hits to 19 or better, don't bust) If the dealer hits to 19, then he needs to win either hand or push both hands. 0.126656 * 7/13 * 5/13 = 0.0262 (Does not hit to at least 19, needs 20 or better) 0.126656 * 1/13 * 6/13 = 0.0045 (Hits to 19, needs 19 or better) 0.126656 * 5/13 * 1 = 0.0487 (Hits to 20 or better, don't bust) If the dealer hits to 20, then he needs to win either hand or push both hands. 0.122185 * 8/13 * 1/13 = 0.0058 (Does not hit to at least 20, needs 21 or better) 0.122185 * 4/13 * 5/13 = 0.0145 (Hits to 20, needs 20 or better) 0.122185 * 1/13 * 1 = 0.0094 (Hits to 21 or better, don't bust) If the dealer hits to 21, then he needs to push both hands. 0.116371 * ??? * ??? = ?.???? (Needs 21 on both hands, very small chance) Total: 62.75%
OK - splitting is the way to go Prospect & TXtourplayer: I concede victory. With BR1 having 17, splitting is definitely the way to go. Prospect: I do have a couple of questions about your calculations if you can spare the time. I took the liberty to number your calculation lines for reference purposes. For the situations where the dealer hits to a 18, 19, or 20, what it boils down to is simply that for BR2 to win he needs to either 1) win just 1 hand out of the split or 2) have a double push. With this in mind: 1) Lines 4&5, 7&8, and 10&11. As these calculations are written, aren't these conditions covered by lines 6, 9, and 12 respectively and therefore shouldn't they be eliminated? (pushes are covered in question 2) 2) To cover a double push, why not simply have a line solly for that purpose? The multiplication would be 1/13 * 1/13 for dealer hitting to 18 or 19 and 4/13 *4/13 for the dealer hitting to 20. 3) Lines 6, 9, 12. Since there are 2 hands to play and if either hand beats the dealer then BR2 wins, shouldn't the multiplier be something greater than "1" instead of "1"? This can get a little complicated since the composition of the "deck" changes as each card is dealt and the probability can never exceed 100%. If any of these questions would result in changing your formulas, the result is still that splitting is the correct move. I'm leaving town soon so I may not be responding to your answers for a week or so. Just don't want you think I'm ignoring this thread.
No reason to concede! Toolman, I never said you were wrong or that I was right. I only posted the way I would have played MrBills hand. Ken's site offers us a chance to post our opinons on the questions asked. Not all of us may have the same opinons and that is what the forum board is all about. Part of the beauty of playing in these tournaments is that win or lose it is your hand to do with as you want. The highest percentage play is not alway the winning play (as most everyone here knows). Bottomline is if your playing MrBills hand you would have 10 - 30 seconds to make your play and you either win or lose the table after the hand. What results would you perfer? Losing using the 52% play or Winning with the 48% play? I have a saying: "I would rather win wrong, then lose right".
Proper Play Ah TX, with you're recent spate at fame, you do my heart good, with no animosity either way, to put in perspective the "numbers." 10 to 30 seconds to make a decision with at least 20 considerations running through your head is a challenge to any but the human computers. And yeah there are a couple. You said it, and I've heard it. "I did everything right and I still lost." Or, "oh hell, i didn't know what I was doing, I just took a stab." I'd rather win wong, than to lose Cerritos.
Learning process TXtourplayer: I guess my choice of the word "concede" was incorrect. It happened to be the word that popped into my head when I was typing my response. Anyway, I learn a little more when anyone posts to this site. That's why I stay with it. Nobody has all the answers (OK, maybe Ken Smith ), but there is such a large pool of knowledgeable people here that one cannot help but learn. I have no problem with people challenging my thoughts. If I'm wrong I want to know about it - helps to keep me from making the same mistake over and over.
Let's look at lines 4-6. Lines 4&5 do not cover line 6. Note that 6/13 + 1/13 + 6/13 = 13/13 or 1. This covers all of the possible cards that the first ten could receive. Note how lines 4-6 can be rewritten like this... (4) 0.13193 * 6/13 * 6/13 = 0.0281 (Does not hit to at least 18, needs 19 or better) (5a) 0.13193 * 1/13 * 1/13 = 0.0008 (Hits to 18, Hits to 18) (5b) 0.13193 * 1/13 * 6/13 = 0.0047 (Hits to 18, needs 19 or better) (6) 0.13193 * 6/13 * 1 = 0.0609 (Hits to 19 or better, don't bust) Or this... (4) 0.13193 * 7/13 * 6/13 = 0.0328 (Does not hit to 19 or better, needs 19 or better) (6) 0.13193 * 6/13 * 1 = 0.0609 (Hits to 19 or better, don't bust) (5) 0.13193 * 1/13 * 1/13 = 0.0008 (Hits to 18, Hits to 18) The last multiplier represents the number of available good cards for the second ten. For example, 6/13 represents the 6 cards (9, T, J, Q, K, A) that will help and 1 represents any card will help and 0 represents that no card will help. So, it can never be greater than 1. Yeah, I assumed infinite deck to make the calculations a lot easier.
