October 15th, 2019 at 8:10:18 AM
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I live in India. A casino in nearby Nepal offers comps worth 1000 USD if I play 750 Black jack hands for USD 55 each. The rules are 3:2 BJ, shufflemaster 126, dealer stands on all 17, double allowed on 9,10,11 only; only one split permitted, no hole card, surrender permitted except against a dealer ace. I am pretty good at using the basic strategy and have been using it for years. What is the probability that in the long run, i will loose more than 1000 USD, 2000 USD and 3000 USD respectively. Also, if a similar offer was available on Baccarat should I prefer it. Baccarat here is non commission with half win paid when banker wins on 6 and six deck as well as continuous shuffle options are available.

October 15th, 2019 at 8:29:30 AM
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DogHand replied to me in another thread and explained this very well, here is the extract (just pick EV of -2% approx. or similar for that game you described - with number of decks someone can tell you the exact EV...):

"According to gordonm888, your Expected Value per hand is -4%, or -0.04. This means that if you play "n" hands, your total EV will be -0.04*(n).

For flat-betting with Basic Strategy, BJ has a Standard Deviation per hand of about 1.15. This means that if you play "n" hands, your total SD will be 1.15*(n)^0.5. Approximately 95% of the time, your actual result will be within 2SD of your EV.

Let's play with some numbers to illustrate.

If you play 100 hands, your EV is -0.04*100 = -4 units. Your SD is 1.15*100^0.5 = 1.15*10 = 11.5, so 2SD's is 23. This means that, with 95% certainty, your actual result will be between (-4-23) and (-4+23), so in the range of -27 to +19 units. If you are flat-betting $10 per hand, that's -$270 to +$190. Thus, you have a reasonable chance to be ahead after 100 hands.

If you play 10,000 hands, your EV is -0.04*10,000 = -400 units. Your SD is 1.15*10,000^0.5 = 1.15*100 = 115, so 2SD's is 230. This means that, with 95% certainty, your actual result will be between (-400-230) and (-400+230), so in the range of -630 to -170 units. If you are flat-betting $10 per hand, that's -$6,300 to -$1,700. Thus, you will almost certainly be losing after 10,000 hands."

"According to gordonm888, your Expected Value per hand is -4%, or -0.04. This means that if you play "n" hands, your total EV will be -0.04*(n).

For flat-betting with Basic Strategy, BJ has a Standard Deviation per hand of about 1.15. This means that if you play "n" hands, your total SD will be 1.15*(n)^0.5. Approximately 95% of the time, your actual result will be within 2SD of your EV.

Let's play with some numbers to illustrate.

If you play 100 hands, your EV is -0.04*100 = -4 units. Your SD is 1.15*100^0.5 = 1.15*10 = 11.5, so 2SD's is 23. This means that, with 95% certainty, your actual result will be between (-4-23) and (-4+23), so in the range of -27 to +19 units. If you are flat-betting $10 per hand, that's -$270 to +$190. Thus, you have a reasonable chance to be ahead after 100 hands.

If you play 10,000 hands, your EV is -0.04*10,000 = -400 units. Your SD is 1.15*10,000^0.5 = 1.15*100 = 115, so 2SD's is 230. This means that, with 95% certainty, your actual result will be between (-400-230) and (-400+230), so in the range of -630 to -170 units. If you are flat-betting $10 per hand, that's -$6,300 to -$1,700. Thus, you will almost certainly be losing after 10,000 hands."

November 8th, 2019 at 11:33:54 PM
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OriginalSD = 1.15 * AvgBet

EV(x hands) = (AvgBet*NumHands)*(HouseEdge)

SD(x hands) = Sqrt(x) * OriginalSD

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Example: $10 flat bet at .5% average blackjack game for 750 hands:

OriginalSD = 1.15*10 = 11.5

EV(750 hands) = (10*750)*(-.005) = -$37.50

SD(750 hands) = Sqrt(750) * 11.5 = ~$315 ... 2SD (95% confidence) = $630 ...3SD (99% confidence) = $945

So pending your level of comfort, if you want to do 2SD, with 95% confidence you will be down $37.50 +/- $630... but again on "average" in the long run you will be down $37.50, so that's what you can base your real value on if you're going to do this over and over and over... that's the number that matters, if the comps are that worth it anyways.

EV(x hands) = (AvgBet*NumHands)*(HouseEdge)

SD(x hands) = Sqrt(x) * OriginalSD

----------------------------------------------------------------------------

Example: $10 flat bet at .5% average blackjack game for 750 hands:

OriginalSD = 1.15*10 = 11.5

EV(750 hands) = (10*750)*(-.005) = -$37.50

SD(750 hands) = Sqrt(750) * 11.5 = ~$315 ... 2SD (95% confidence) = $630 ...3SD (99% confidence) = $945

So pending your level of comfort, if you want to do 2SD, with 95% confidence you will be down $37.50 +/- $630... but again on "average" in the long run you will be down $37.50, so that's what you can base your real value on if you're going to do this over and over and over... that's the number that matters, if the comps are that worth it anyways.

Playing it correctly means you've already won.

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