You're in a tournament in which once each round you're allowed to replace a card you don't like immediately after receiving it. No surrender, max bet 500. It's the last hand and one of you advances. BR2 2000 bet 500 cards 4,4,T You 2010 bet 500 cards 4,2,T ? Dealer 7 1. What is your best decision now that you've gotten the ten? Why? 2. What is your hard standing number with the dealer upcard 7 and your opponent hand 18 or higher?
I'll give it a shot 1. If you stand on 16 then your opponent advances if the dealer makes 17 or 18. Using infinite deck figures, that is 36.86% + 13.78% = 50.64%, which makes you a small underdog. Since your total is currently the worst possible stiff total, replacing the ten can only improve things, or at worst leave things the same if you get another ten. (If you get a smaller stiff total than 16, your chances of hitting it and avoiding going bust are at least improved.) 2. I've arrived at some answers, but I'm not very confident about my method. So, I'll just post the results and if by some miracle they are correct I'll reveal how I arrived at them. Opp. Hand 18 - Stand on h17. Opp. Hand 19 - Stand on h17. Opp. Hand 20 - Stand on h18. Opp. Hand 21 - Stand on h18.
I'd change the ten Monkeysystem, since it is the last hand I'd change out the last card (the ten) and re-hit the 4/2 and hope I catch a small card. Should I catch a 6 - King, (making another bustable hand) I'd stand on it and not give away the dealers bust. I just wouldn't give up the dealers bust even should I catch a 6 for 12. This is how I'd play it anyway, just my opinion.
1. Replace the 10; 2. Stand on 17 or higher 1. Replace the 10. As already pointed out, you can do no worse than hard 16, so a replacement for the 10 will be at least as good as the 10. 2. With a dealer 7 up, you should stand on hard 17 or higher if your opponent has stood with with an 18 or higher. If your opponent stands with a 17, you should stand with a 16 or higher. If your opponent has stood stiff, you should stand with any stiff. The reasoning behind this is that this is what Wong says on p. 133 of CTS David
Some Math With card replacement, if you hit hard 17 you have a 52% chance to improve your hand. If you improve your hand you have a minimum 63% probability of beating the dealer, so you have a minimum 32.8% probability to advance if you hit hard 17 and your opponent has 21. That compares favorably to the 26% chance of the dealer busting, which is what you need if you stand on 17 or below. The most extreme case for testing the hard standing total is if you have hard 17 and your opponent hard 18. Any improvement of your hand locks it for you and you have a 52% probability of doing that. If you stand, then the dealer bust, 19, 20, or 21 wins it for you and you're a 49% underdog. You'll never hit hard 18, even if your opponent has 21. With 18 you are a 63% favorite to advance if your oppenent has 21 and even better if your opponent has 19 or 20. The 52% figure for improving your hand by hitting hard 17 with card replacement comes like this: Busting twice in a row is a failure. Any other outcome is a success. The probability of busting when you hit hard 17 is 9/13. Therefore, the probability of busting twice in a row is (9/13)^2. The probability of not busting twice in a row is 1 - (9/13)^2. That's 1 - (81/169) = 88/169 = 52%. With card replacement, your hard standing number is 18 against dealer 7 with opponent total >/= 18. If you hit the hard 16 and get an ace you'll hit it again. The odds of success by doing that are 52% * 1/13 = 4%. Your odds of success busting and replacing with an ace are 8/13 * 1/13 * 26% = 1%. The odds of getting 18 or higher are 52%. Added together hitting the hard 16 gives you a 57% probability of winning. What if you replaced the ten? Your hard standing number reverts to 16 as in games without card replacement. We can calculate your probability of winning from a starting total of six by taking numbers from Ken's tables for a hard standing number of 17 and then assuming you'll do a little better with a hard standing number of 16. Your probability of winning is your probability of drawing 6 out to 18 or higher from the H17 table, added to your probability of winning if you draw out to 17. It turns out to be 44.56% + (11.48% * 26.23%) = 47.57%. Your hard standing number is 16, not 17, so you're probably a little over 50% but not really that close to the 57% chance you get if you hit the 16. The best play is to hit and hit it again if you get an ace. If you bust and replace with an ace, stand.
Good point. I forgot I had that resource to refer to. I'll draw a veil over my own attempt. However, I notice that Ken posted a version which shows h16 as the standing number when your opponent has 18, with a dealer 7 - https://www.blackjacktournaments.com/posts/18029 Maybe that discrepancy is what this question was hinting at? [Monkeysystem added his math as I was composing the above. It may take me a while to digest it. ]
Humbly, I stand corrected. Monkeysystem's analysis is absolutely correct, and mine is absolutely flawed. Shows what I know about playing with card replacement David
I didn't really understand the question. Although it now seems obvious, it never occurred to me that by not replacing the ten you leave yourself the option to replace the next card, even if it bust you.
Power Chip Just spent 90 minutes trying to work it out before I went back into the message board and saw the last posts. I can honestly admit that I will never be the next S.Yama on here when it comes down to the finer points of strategic play (London Colin will verify that….lol). However, it really did get me thinking and made me look at some other aspects of my tournament play. Thanks to Monkeysystem for putting that teaser on the site. Andy
Just to confirm the strategy if you do replace the ten: Starting with (4,2), you should hit to soft 18 or hard 16. The success rate of this strategy is 60.6%. Monkeysystem's second post in this thread comes up with a lower success rate, but I believe it is a misapplication of the numbers he is using. The success of this player is not solely dependent on improving the (4,2) hand. Instead, he also wins whenever the dealer makes 19,20 or 21, no matter how his hitting efforts turned out. Does hitting the 42T hand instead of replacing the ten yield a better result than 60.6%? First let's evaluate BR1's situation for each possible BR1 ending hand total: Bust: He still wins with dealer 19+. (23.13%) 17: He wins unless the dealer makes 17 or 18. (49.36%) 18,19,20,21: He's a lock. (100%) ------------------- OK, he hits the 42T hand once. The results: Bust: 8/13 17: 1/13 18+: 4/13 The times he busts, he uses the power chip and hits again. And we reuse the probabilities, but multiply them by 8/13. Final results, assuming he's willing to stand on 17 and not hit again. Bust: 8/13 * 8/13 = 37.9% 17: 1/13 + (8/13 * 1/13) = 12.4% 18+: 4/13 + (8/13 * 4/13) = 49.7% Now we have to use those probabilities to weight the outcomes in bold above. ---------- Bust: 37.9% * 23.13% = 8.76% 17: 12.4% * 49.36% = 6.12% 18+: 49.7% * 100% = 49.7% ---- Add all those up to get 64.58%. So, hitting the 42T instead of replacing the ten is already better, even if we choose to stand after taking any non-busting card. I was going to look at the numbers if you hit 17 with the power chip available, but I'm out of time before the bounty tournament begins.
Unfortunately my bounty rounds in Tulsa went badly, so I'm able to finish the analysis. Looking at how to play 42TA if you still have the power chip: If you stand with 17, you advance 49.36% of the time (from the bold section in the previous post). If you hit trying to improve it, you have a 52.1% of not double-busting, and that makes you a lock. (Monkey shows this in his post as well.) The other 47.9% of the time, you'll bust twice and then you have a 23.13% chance of advancing. 52.1% + (47.9% * 23.13%) = 63.18% overall chance of succeeding when hitting the 17, which is much improved from the standing 49.36%. Revising the final hand probabilities of BR1 with the new strategy of hitting 17 if we still have the power chip: Bust: 41.56% (8/13 * 8/13) [bust 16 twice] + (1/13 * 9/13 * 9/13) [make 17 then bust it twice] 17: 4.73% (8/13 * 1/13) [bust the 16, then draw an Ace and stand] 18+: 53.71% (4/13) [first draw] + (8/13 * 4/13) [ bust, then hit to 18+] + (1/13 * 4/13) [hit to 17, then one more successful card] + [1/13 * 9/13 * 4/13) [hit to 17, bust, then success] Now we again apply the advance probabilities to those numbers. (41.56% * 23.13%) + (4.73% * 49.36%) + (53.71%) = 65.66% Recapping, all three strategies for the overall problem: 1) Use the power chip now to replace the Ten. Hit to h16s18. You have a 60.6% chance. 2) Save the power chip and just hit the 42T. Stand on any non-bust result. You have a 64.58% chance. 3) Save the power chip and hit the 42T. Also hit 42TA if you still have the power chip. You have a 65.66% chance. Interesting question, and MonkeySystem's strategy is indeed the best.
