name |
message |
date |
| Midnite |
okohiored- I am glad you came out all right in Vegas. As Grandpa used to say, A wink is as good as a nod, to a blind horse. Don't really matter "how" you won, just that you did win..... |
2003-07-30 22:19:20 |
| Midnite |
Philtx- An 11 doubled when the dealer has a 6, will win 67 times and lose 33 times. A BJ will win 100% of the time and for 150% of your bet. Not hard to see why it would be a Very bad play, to DD with a BJ. |
2003-07-30 21:59:30 |
| okohiored |
Bankroll can stand it and have lost 12 in a row before or 4000.My point is agree full martingale is foolish but rather than progressive bets take advantage of odds of not losing 5 in a row to get your self ahead or bust.If i start 4-5 units ahead this helps wipe-out the house advantage |
2003-07-30 20:50:46 |
| philtx |
Interesting stuff ... very intersting indeed. How about a quick change of topics. A few days (weeks) back someone somewhat facetiously mentioned doubling down on a blackjack. In all seriousness, if this were allowed, is it truly a bonehead thing to do, or are there times that it might seriously be an adviseable move - like against dealers 6. |
2003-07-30 20:11:11 |
| BuGhOu§eMASTER |
okohiored: You should not risk it though, unless your bankroll can seriously sustain such massive damages. I've seen someone lose about 10 in a row. It can happen. Although rare, I would just play solidly and not rampantly. |
2003-07-30 20:10:38 |
| okohiored |
stillwould like to start with little martingale at atart and take advantage of midnites odds he quoted of 81-1 against losing 5 in a row so here goes 1-2-4-8-16 units but quit whenever you win except first hand.Now come back to2-1-2-3-4-5.Does this make any sense as a progression.Of course if you lose the 16 units in semi-martingale just go play the horses.Bet the gray one! |
2003-07-30 19:54:26 |
| ~kinkette |
I'm learning the 2:1:2:3:4:5 betting strategy, but what do you do if you lose the first hand? Do you continue betting 2 units until you win..then reduce to 1? |
2003-07-30 19:28:00 |
| Midnite |
Desert Dog- My chart only shows three hands that are profitable against the dealers Ace. They are A-10, 10-10 and A-9. |
2003-07-30 16:46:42 |
| Midnite |
Zebra and others : I feel we are very fortunate to have three sharp posters, sharing their knowlage, on this board. Mr. Thomason, Mr. Renezy and Mr. Grifter. Not that there are not some very good posts, from several other posters, as well. And Zebra, you are quite welcome and good luck in Vegas. |
2003-07-30 16:33:32 |
| Desert Dog |
Midnite -- So would you say doubling 11 vs Ace if table rule is H17 (dealer Hits soft 17) is a rare instance of H17 being more favorable to the player than S17 (dealer Stands on soft 17)? |
2003-07-30 16:29:15 |
| zebra |
very smart - thanks for helping me think this through midnite. |
2003-07-30 16:14:04 |
| Midnite |
Zebra- I am sure Mr. Renezy can answer this better than I, but I will take a shot at it. I think the answer is the H17. As almost half of your DD wins will come from the dealer breaking, doubling an 11 into an Ace, where the dealer will only break about 20 % of the time, is not a good move. If it is a S17 game, the dealer will never break with A-6. If it is H17, he will break (at least sometimes) with the A-6. |
2003-07-30 15:54:11 |
| zebra |
also -- in a double deck / s17 / double any / das / late surrender game -- basic strategy says surrender fifteens to tens but not to aces -- this is counterintuitive to me as well. as you can see, i'm brushing up on my variations -- heading to vegas in a week or so and want to stay sharp when the rules change on me from location to location. |
2003-07-30 15:52:26 |
| zebra |
another question -- when playing double deck / h17 / double any / das / late surrender -- basic strategy says to double elevens against aces. why is this correct in this game but not in other games (even other two deck games)? |
2003-07-30 14:26:59 |
| zebra |
renzey -- i think i just caught up -- the 340 hands represents the total number of hands (each consisting of two or more cards) that you can finish with (without busting) -- and not the total number of hands (consisting of two cards) that you can be initially dealt. |
2003-07-30 14:24:18 |
| Renzey |
To Zebra; It's not a dense question at all, because 6/7 has indeed already been counted as a 13. It's just that a multi-card 21 is unique in that it has a different value than a B/J. Also in that same category as a multi-card 21 is a two card 21 formed by splitting a pair of Aces (or splitting two 10's, or doubling with 11 and catching a 10). Those kinds of 21's are just functionally different hands than a B/J. |
2003-07-30 14:19:47 |
| zebra |
renzey -- thank your for the reply. additional question for you though -- why do you include 3 card 21's in the 340 possible hands that you may be dealt? using your example of 6;7;8, and assuming 6;7 is dealt -- isn't that hand already considered among the hard thirteens (and therefore should not constitute another hand)? i apologize if i'm being dense here -- i'm just not sure where i should file this data (340 possible hands) into my overall comprehension (growing daily) of the game. |
2003-07-30 11:27:36 |
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