DannyOcean, great screen name!! Plus 1
No, that's not true. And, of course, it depends on how many other players are at the table. If you were alone, betting one hand of $10, let's say you could play 210 hands per hour. Now, you switch to two hands of $10, but you can play only 2/3 the number of hands per hour. In the first case, your e.v. operates on $2,100 of action. In the second case, it's on $20 x 140 = $2,800 of action. Your e.v. has increased by 33.3%, not 50%.Originally Posted by ZeeBabar
Now, put five other people at the table with you, such that , when you play your two hands, all seven spots are used. I'll leave it as an exercise for you, or others, to now see what happens to e.v.
Don
No, sorry, you're missing the point. The idea is not to state that, with more players at the table, your e.v. drops. First, you hourly win drops, which isn't the same thing. As I previously stated, the e.v. changes very little, if at all.
The idea is to challenge your original statement that, when switching from one hand of $10 to two hands of $10, the effect is as if you were now playing one hand of $15, which just isn't the case. After showing you what the equivalent would be if you were playing alone, I challenged you to do the math for going from one hand of $10 with five other players at the table (total of 7 hands, including the dealer's) to two hands of $10 with the same five other players (total of eight hands, including the dealer's). But, you didn't try to do it (or couldn't do it).
Do you want to see the answer, or do you want to try to work it out and understand for yourself?
Don
Here is my attempt at explaining this:
I think the basic assumption in Don's example is that the dealer manages 420 hands per hour. If a heads-up player uses one spot, he gets half of these managed hands, thus 210 hands per hour. If the wager per hand is 10 dollars, then the player has a throughput of 2100 dollars per hour. When playing two spots heads-up, the player gets 2/3 of the managed hands, thus 2/3 * 420 = 280 managed hands per hour, for a throughput of 2800 dollars, an increase of one third since 2800/2100 = 4/3 = 1 1/3.
Now assume 5 other players at the table, so 7 persons (6 players plus the dealer). A counter playing one spot now gets only 1/7 of the 420 managed hands, thus 60 hands, and his throughput is 600 dollars only. If he plays two spots, he gets 2/8 = 1/4 of the 420 managed hands, thus 105 hands, yielding a throughput of 1050 dollars, which still is only half of his throughput when playing one spot but heads-up. At least, spreading to two spots increased the throughput by 75 percent in this case, since 1050/600 = (7*150)/(4*150) = 7/4 = 1 3/4 = 1.75.
The EV / advantage / edge (e.g. about 1 percent when card counting) is nearly independent of the number of spots and other players, but the win rate is the product of EV and throughput (e.g. 0.01 * 2100 dollars per hour = 21 dollars per hour). Since throughput depends on the circumstances mentioned above, so does the win rate.
Are these considerations correct?
Last edited by PinkChip; 08-19-2019 at 07:18 PM.
Thanks Don for the "review" ;-) I guess some similar computations are contained in BJA3 but I read it years ago together with many other books, so don't remember every detail, e.g. how exactly the SCORE is defined. But will look up this in conjunction with having a look into my newly bought CVCX in order to get a better understanding of optimal bet ramps.
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