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## Session Bankroll Question

Asking for a friend since I realize I have no clue how to utilize the qfit calculator tools.

I am planning on having \$75 as my average bet.
I am only playing for 4 hours which I estimate to be 320 hands.
The winrate is likely -\$0.50 per hand.
I believe the standard deviation is \$1529 ( 1.14 x 75 x sqrt(320) ).

My question is how do I calculate the odds of tapping out given x where x is the amount of money I bring.

I also have a very hard time conveying why \$1529 does not mean a 16% ROR ( (100-68) /2 ) where \$1529 is the given bankroll. The person asking keeps insisting bringing \$1529 means not tapping out 16% of the time. Can someone clarify?

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Originally Posted by FishBear
I am planning on having \$75 as my average bet.
I am only playing for 4 hours which I estimate to be 320 hands.
The win rate is likely -\$0.50 per hand.
So, you're not counting? You have a negative win rate? Are you varying your bet? Are you just playing BS? You haven't been very clear.

Originally Posted by FishBear
I believe the standard deviation is \$1529 ( 1.14 x 75 x sqrt(320) ).
Only if you flat bet. If you are varying your bet, you can't use the average bet to calculate the s.d. The s.d. will be larger than what you're stating. You need to be more clear as to what you're doing. Not much of what you've written makes sense.

You're implying a negative edge of 0.67%, although you haven't given us any rules or anything. So, you're supposed to lose \$160, and the total s.d. is (maybe) \$1,529. If you bring \$1,529, your ROR is much greater than 16%, because you can't just calculate the end-point value of a one-s.d. loss (which would be \$1,689, in any event). You have to realize that you could lose the entire \$1,529 BEFORE you finish playing the 320 hands.

In any event, try to be a bit clearer as to what you're doing.

Don

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Thanks Don.

I'm asking for a recreational player who is not card counting. I understand this is theoretically a losing game. The question is how do I calculate the ROR given a 320 hand constraint. I'm assuming near perfect basic strategy and the house edge to be around -0.67%. If I had to guess, the game is probably a standard shoe game, so: h17, das, da2, rsa, no surr.

He is using a betting system where I gauged \$75 to be a rough estimate as the average bet. Based on your tone, it seems like I cannot just use 1.14 unit as stdev.

The betting system to be precise is to start with 2 units and increase the bet for every previous win. The maximum to be risked is 5 units. If he loses a bet then he'll return to 2 units. 1 unit is \$25.

I ran a very quick simplified sim of 10M hands with "rand() < 0.5" as the factor to increasing a unit or going back to 2 units, and I arrived at 2.875 or so units (\$71.88) as the average bet.

50: 4,999,300
75: 2,498,470
100: 1,250,711
125: 1,251,519
(yes I see what they converge to)

I get in the actual world, there are splits and doubling in BJ and the game is not 50-50 but something along the lines of 42-48-10. I assume all this overcomplicates the equation when all I want is a rough estimate (within 5% is fine) on one's ROR after 320 hands.

So my question is approx what are the odds of tapping out (within 5% accuracy) if he brought
a) \$1500
b) \$2000
c) \$2500

My assumption is there exists a ROR formula where I can just plug in
- sample size
- expected value,
- average bet,
- stdev, and
- session bankrol
to arrive at an approx answer.

My gut tells me \$1500 and even \$2500 is not enough (impractical) as a session bankroll, because he is likely to go bust a high frequency of the time, but I am unable to convince him that without the actual math. He seems fixated on the idea that the tail end of 1stdev is a good estimate as to what a session bankroll should be.

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Originally Posted by FishBear
Thanks Don.

I'm asking for a recreational player who is not card counting. I understand this is theoretically a losing game. The question is how do I calculate the ROR given a 320 hand constraint. I'm assuming near perfect basic strategy and the house edge to be around -0.67%. If I had to guess, the game is probably a standard shoe game, so: h17, das, da2, rsa, no surr.

He is using a betting system where I gauged \$75 to be a rough estimate as the average bet. Based on your tone, it seems like I cannot just use 1.14 unit as stdev.

The betting system to be precise is to start with 2 units and increase the bet for every previous win. The maximum to be risked is 5 units. If he loses a bet then he'll return to 2 units. 1 unit is \$25.

I ran a very quick simplified sim of 10M hands with "rand() < 0.5" as the factor to increasing a unit or going back to 2 units, and I arrived at 2.875 or so units (\$71.88) as the average bet.

50: 4,999,300
75: 2,498,470
100: 1,250,711
125: 1,251,519
(yes I see what they converge to)

I get in the actual world, there are splits and doubling in BJ and the game is not 50-50 but something along the lines of 42-48-10. I assume all this overcomplicates the equation when all I want is a rough estimate (within 5% is fine) on one's ROR after 320 hands.

So my question is approx what are the odds of tapping out (within 5% accuracy) if he brought
a) \$1500
b) \$2000
c) \$2500

My assumption is there exists a ROR formula where I can just plug in
- sample size
- expected value,
- average bet,
- stdev, and
- session bankrol
to arrive at an approx answer.

My gut tells me \$1500 and even \$2500 is not enough (impractical) as a session bankroll, because he is likely to go bust a high frequency of the time, but I am unable to convince him that without the actual math. He seems fixated on the idea that the tail end of 1stdev is a good estimate as to what a session bankroll should be.
The correct answer is if you are not counting, or shuffle tracking, or hole carding, or ace steering, do not play. Otherwise, in the long run, your friend is throwing their money away.

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Originally Posted by Wave
The correct answer is if you are not counting, or shuffle tracking, or hole carding, or ace steering, do not play.
No that's not the correct answer. He has a right to lose money if he wants to and wants to be better educated about it.
I guess you'll now post another stupid picture.

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Originally Posted by 21forme
No that's not the correct answer. He has a right to lose money if he wants to and wants to be better educated about it.
I guess you'll now post another stupid picture.
Originally Posted by 21forme
What you call a part-time AP, I call a recreational AP. The part I quoted I call a gambler who is making a half-assed attempt trying to limit his losses from gambling.

P.S. OSN has a lot of misinformation in it. Don't wear your entry as a merit badge.

So, let me get this straight. OP has a right to LOSE money without your scorn...but I don't have a right to WIN less money when I occasionally play for lower EV without your scorn?

WTF.jpg

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Originally Posted by Wave
Can you please tell me what the point is of posting this photo over and over again, ad nauseam? Do you think that maybe its usefulness has run its course?

Don

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Originally Posted by DSchles
Can you please tell me what the point is of posting this photo over and over again, ad nauseam? Do you think that maybe its usefulness has run its course?

Don
I'm laughing at whatever I posted it in response to.

No, there will be more things I will laugh at.

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Originally Posted by FishBear
Thanks Don.

I'm asking for a recreational player who is not card counting. I understand this is theoretically a losing game. The question is how do I calculate the ROR given a 320 hand constraint. I'm assuming near perfect basic strategy and the house edge to be around -0.67%. If I had to guess, the game is probably a standard shoe game, so: h17, das, da2, rsa, no surr.

He is using a betting system where I gauged \$75 to be a rough estimate as the average bet. Based on your tone, it seems like I cannot just use 1.14 unit as stdev.

The betting system to be precise is to start with 2 units and increase the bet for every previous win. The maximum to be risked is 5 units. If he loses a bet then he'll return to 2 units. 1 unit is \$25.

I ran a very quick simplified sim of 10M hands with "rand() < 0.5" as the factor to increasing a unit or going back to 2 units, and I arrived at 2.875 or so units (\$71.88) as the average bet.

50: 4,999,300
75: 2,498,470
100: 1,250,711
125: 1,251,519
(yes I see what they converge to)

I get in the actual world, there are splits and doubling in BJ and the game is not 50-50 but something along the lines of 42-48-10. I assume all this overcomplicates the equation when all I want is a rough estimate (within 5% is fine) on one's ROR after 320 hands.

So my question is approx what are the odds of tapping out (within 5% accuracy) if he brought
a) \$1500
b) \$2000
c) \$2500

My assumption is there exists a ROR formula where I can just plug in
- sample size
- expected value,
- average bet,
- stdev, and
- session bankrol
to arrive at an approx answer.

My gut tells me \$1500 and even \$2500 is not enough (impractical) as a session bankroll, because he is likely to go bust a high frequency of the time, but I am unable to convince him that without the actual math. He seems fixated on the idea that the tail end of 1stdev is a good estimate as to what a session bankroll should be.

OK. Clearer. And your gut is absolutely correct. I'm assuming you don't have BJA3 and are not familiar with the "premature bumping into the barrier syndrome" that I describe beginning on page 122. It's a very important idea, and it's precisely applicable to your question.

In any event, I recalculated the s.d. based on your description of the betting style and frequencies of the various bet sizes. Instead of 1.14, I get about 1.76, so you can see how that's rather important. We now have that e.v. for 320 hands is -\$160, while s.d. is more along the lines of \$2,361 (which is: 1.76 x \$75 x 320^0.5). So, you can see that a one-s.d. loss AT THE END of the four hours is -\$2,521, and that occurs roughly 16% of the time. But, what you don't know is that the probability of losing that amount is roughly DOUBLE the 16%, because of the barrier phenomenon that I mentioned above.

Bottom line: I'd bring easily \$4,000 or more for such a trip. Bringing \$2,500, your friend will tap out almost once every three trips he takes. Give me a little more time, and I'll get more numbers for you. The problem with using all the calculators is that they aren't designed for negative e.v.; rather, they're designed for counting, where the e.v. is always positive. But I know how to trick them to be able to get answers. So, I'll get back to you.

Don

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