January 30th, 2020 at 12:14:07 PM
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Say you have a 2 percent constant advantage in BJ every hand. If i played 7200 hands with this advantage,what percent of the time would I be not in profit? How many hands would you need to play to bet 98 percent confident level in profit?
January 30th, 2020 at 3:15:54 PM
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The game is statistically similar to a game with p=0.44 and a payout of 1.32 to 1, which gives a standard deviation of 1.15.
96 percent of results will be within 2.05 SDs of the mean, so (100 - 96) / 2 = 2 percent will fall below this range.
The expectation is n * .02 +/- 1.15 * n^.5
You want to know when n * .02 = 2.05 * 1.15 * n^.5. Rearrange and n = (2.05 * 1.15 / .02)^2 = 13,895 hands to be 98% confident of a profit. Call it 14,000 since this is a “back of an envelope” calc.
At 1000 hands per week, that will take over 3 months
96 percent of results will be within 2.05 SDs of the mean, so (100 - 96) / 2 = 2 percent will fall below this range.
The expectation is n * .02 +/- 1.15 * n^.5
You want to know when n * .02 = 2.05 * 1.15 * n^.5. Rearrange and n = (2.05 * 1.15 / .02)^2 = 13,895 hands to be 98% confident of a profit. Call it 14,000 since this is a “back of an envelope” calc.
At 1000 hands per week, that will take over 3 months
It’s all about making that GTA
January 30th, 2020 at 3:27:14 PM
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For 7200 hands your expectation is 144 +/ 98.
144 / 98 is 1.47 SDs, and your result will be a loss (to the left of that grouping) about 7% of the time.
144 / 98 is 1.47 SDs, and your result will be a loss (to the left of that grouping) about 7% of the time.
It’s all about making that GTA
January 30th, 2020 at 6:16:45 PM
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awesome thank you, I guessed about 5 percent, and chalking it up to bad luck. I am actually the one who is the player-dealer bank. and bj only pays 6:5. no double after splits, so I am trying to see if its really bad luck, or perhaps more players are counting vs me, but I dont think this is the case.
January 31st, 2020 at 3:07:16 PM
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Ace2 looks like he gave you a good answer, let's see if my estimate can look at it from a different perspective:
OriginalSD = 1.15*AvgBet
EV(x) = (x*AvgBet)*(Advantage)
SD(x hands) = Sqrt(x) * OriginalSD
EV(7200 hands at 2% edge) = (7200*AvgBet)*(.02)
SD(7200 hands) = Sqrt(7200)*OriginalSD
...you said you're player banking, so I'm going to assume an average bet of $10 to show you the formulas and you can correct them with your real averages that I assume you know or can figure out.
EV = $1440
1SD (68% confidence) = ~$975.. 2SD (95% confidence) = $1950... 3SD (99% confidence) = $2927
So after 7200 hands you're looking at being up $1440 +/- 2927 (3SD, 99% confidence)... so quite possible to be negative.
Let's see how that plays out after 15,000 hands though:
EV = $3000
3SD = $1408
You expect to be up $3000 +/- $1408, which means it should be mathematically impossible (assuming a fare game) for you to be down at this point.
OriginalSD = 1.15*AvgBet
EV(x) = (x*AvgBet)*(Advantage)
SD(x hands) = Sqrt(x) * OriginalSD
EV(7200 hands at 2% edge) = (7200*AvgBet)*(.02)
SD(7200 hands) = Sqrt(7200)*OriginalSD
...you said you're player banking, so I'm going to assume an average bet of $10 to show you the formulas and you can correct them with your real averages that I assume you know or can figure out.
EV = $1440
1SD (68% confidence) = ~$975.. 2SD (95% confidence) = $1950... 3SD (99% confidence) = $2927
So after 7200 hands you're looking at being up $1440 +/- 2927 (3SD, 99% confidence)... so quite possible to be negative.
Let's see how that plays out after 15,000 hands though:
EV = $3000
3SD = $1408
You expect to be up $3000 +/- $1408, which means it should be mathematically impossible (assuming a fare game) for you to be down at this point.
Playing it correctly means you've already won.
February 3rd, 2020 at 5:22:27 AM
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I also pay $3 collection to bank per hand. The average action is $450 per hand (spread out on about 3 hands)