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Subject : Some numbers -- or check my math, PLEASE! posted by ETFan on Nov-27-2006 05:16am |
| I'll check it myself again tomorrow. Here's how it looks to me today ... You say I should assume 50 rounds per hour. This is very slow. It couldn't be 50 rounds per hour with one player at the table, so I'm going to assume you mean 50 rounds per hour with 4 civilians (I prefer that term to ploppy) at the table. I'm further going to assume, though you didn't specify, that the dealer takes just as long as a civilian to play his hand, and another equal parcel of time to deal plus make the payoffs. So there are 6 actions that take place each round, and each takes an equal amount of time. It's easy to see that each separate action is 12 seconds long. 12 x 6 x 50 = 3600 seconds = 1 hour. You also say that you play about 150% faster than a civilian. That means it takes you 8 seconds to play a hand. You don't say anything about how much faster you can play two hands, so I'll just assume a second hand uses up another 8 seconds. A $10 to $200 spread at a 6D DAS 75% pen spread brings in a little over $0.20 a round overall, playing one spot. I'm assuming H17. A quick sim with PowerSim gives an EV of -$0.11 per hand in negative counts, and $1.03 per hand in positive counts. I'm rounding these numbers off, since this is strictly intended as a "ballpark" calculation. The negative counts occur 72.7% of the time, and the posiives, 27.3%. When you add it all up, it comes to $0.2012 per hand. Now you'll get in a little more than 50 rounds per hour, playing with 3 civilians, since you play a little faster. With the numbers above, it will take 68 seconds for a round, so we have 3600/68 = 52.94 rounds per hour. Total EV playing one spot = $0.20122 x 52.94 = $10.65 an hour. (I wouldn't travel too far to get to this game.) Now we'll see what happens to the EV when the counter plays two spots all the time with the same 3 civilians playing along. As I mentioned in my previous post, it's well established that the counter should bet approximately 73% as much on each of two hands in the positive counts to keep the same risk of ruin (because of covariance). Therefore, on each round in positive counts, the counter will not average $1.03, but $1.03 x 146% = $1.504. In negative counts he still bets the minimum, so he loses 2 x $0.11 = $0.22. The percentage of positive vs. negative rounds stays the same here, but the number of rounds per hour changes. Each round takes 5 x 12 secs + 2 x 8 secs = 76 seconds. Rounds per hour = 3600/76 = 47.368, and total EV per hour = ($1.5038 x 27.3% - $0.22 x 72.7%) x 47.368 = $11.87. Purists will object that I've changed the bet spread for the two hand case. Alright, lets assume the table minimum is actually $5, and the counter is able to juggle his $5 and $10 bets so they average out to $7.30 per hand in negative counts. EV per round in negative counts is then 146% x (-$.11) = -$0.1606. You plug that in place of the -$0.22 in the formula above, you come out with EV per hour = $13.92. A nice little 31% gain over playing one spot at all times, with no increase in risk of ruin. That was the easy part. ;) Now, for playing two spots in positive counts and one spot in negs. To work out the card eating effect, the guiding principle is that the percentage of cards with positive vs. negative counts is the same as that original 27.3% vs. 72.7%. That's all cards to everybody. However, the percentage of cards you actually receive in plus or minus territory will change because you're varying the number of hands. For this ballpark estimate, I'm going to assume that the average number of cards per hand in negative territory is the same as the average number of cards in positive territory, and also that the dealer's average hand has the same number cards as an average player hand. Neither of these assumptions are true, but they're close enough to get a picture of what's happening. So! In positive counts, the counter plays two hands and gets 2/6 of the +EV cards. In negative counts, he plays one hand and gets 1/5 of the -EV cards. Out of every 100 cards dealt, he gets 2/6 x 27.3 = 9.1 plus count cards, and 1/5 x 72.7 = 14.54 minus count cards. The other players and the dealer get the other 76.36 cards. 9.1 + 14.54 + 76.36 = 100. Say a hand consists of C cards. We can see the counter gets 9.1/C hands in positive territory, and 14.54/C hands in negative tundra. Since one round = 2 plus hands, and one round = 1 neg hand, the counter gets 4.55/C plus rounds versus 14.54/C negative rounds. Don't fret about what 0.55 of a card is. These numbers are all averages. I just refuse to say "an average of" X number of cards over and over and over. !Whew! Now, it doesn't really matter what C is (2.8?). What's important is the ratio 4.55/14.54. For this ratio to hold, the percentage of plus/minus count rounds must be: 23.83%/76.17%. So in terms of rounds (as opposed to cards or hands) the percentage of plus counts has gone down. This is a subtle point that aft gangs agley! So we know 23.83 percent of the time counter is playing two hands and the round takes 2 x 8 secs + 5 x 12 secs = 76 seconds. And 76.17% of the time the round takes 1 x 8 secs + 5 x 12 secs = 68 seconds. So your average round takes 68 x 76.17% + 76 x 23.83% = 69.91 seconds, and we're getting in 51.50 rounds per hour. 12.27 of those rounds are positive and 39.23 of them are negative. Again, our EV on each positive round is $1.03 x 146% = $1.5038. In negative counts, if he still bets the $10 minimum, he loses 11 cents. Overall EV per hour = 12.27 x $1.5038 - 39.23 x $0.11 = $14.14. Once again, purists will object. OK, to make the bet spread purely equal, counter must average a $14.6 bet on the negative rounds, and his EV on each such round is -0.1606. EV per hour = 12.27 x $1.5038 - 39.23 x $0.1606 = $12.15. So we have five EVs to compare: 1) Play one hand at all times: $10.65 2) Play two hands at 73% in positives and two hands at 100% ($10 x 2) in negatives: $11.87 3) Play two hands at 73% in positives and two hands at 73% (averaging $7.30 x 2) in negatives: $13.92 Same Risk of Ruin and bet spread as case 1). 4) Play two hands at 73% in positives and one hand at 100% ($10 x 1) in negatives: $14.14. Same RoR as cases 1) and 3) 5) Play two hands at 73% in positives and one hand at 146% in negatives: $12.15. Same bet spread as cases 1) and 3). Make of it what you will. Personally, I like case 3). It's very close to case 4), and case 4) technically has a larger bet spread than case 3), though we can argue about which looks better to the eye in the sky. Also, you mentioned people lurking behind, anxious to steal your extra spot. Also, if you can play the second spot a just weeeee bit faster than the first spot -- say 7 seconds instead of 8 -- number 3) will come out on top. Maybe you can do a one hand signal (a swooping wave) if you want to stand on both hands. I don't believe my simplifying assumptions affected the results substantially, though it's possible 3) and 4) might trade places in a well structured sim. My analysis here doesn't touch on T-Hoppers suggestion to play two hands only at certain negative counts. However I will say that when a TC truly goes in the tank, the best option is ZERO hands. And the more you Wong out on negs, the better the calculation looks for playing two hands on all remaining negs. Still, if you're in a crowded casino where sitting out til the shuffle is not allowed, there may come a point where it makes sense to drop to one hand. Also, if your civilian buddies get fast all of a sudden, I wouldn't rule out 4) -- or even 5) if you're a purist. Finally, interpretation of these numbers really depends on your point of view. If you truly feel that you LIKE to play blackjack, and if the game goes a little slower, why you might just play a little LONGER(!), well then, 4) and 5) gain some ground. Maybe you only care how much you win in an evening, and you believe in limiting exposure before the eye to a certain number of hands, rather than a certain number of minutes. That's a defensible attitude for a red/low green player, but it's inconceivable to a pro. ETF |
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