Expected Loss/Win Range

Can somebody give me the formula to figure this out. I saw it once in a thread about a year ago and I think it was Romes that broke it down. I think I may have found what I was looking for on Wikipedia, but I guess I wasn't understanding the formula. I know that it involves the EV and the Standard deviation and the total amount wagered.

Just using this example, can somebody please give me the formula.

$5 a hand BJ, at a rate of say 100 HPH for 10 hours = 5000 Wagered.
EV is say .5%
Standard Deviation is 1.15

Thanks

What’s the question?

Quote: unJon

What’s the question?

I need the formula to get win loss range within say 3 SD.

I know I have seen it posted here before, I just dont remember where.

Quote: VCUSkyhawk

I need the formula to get win loss range within say 3 SD.

I know I have seen it posted here before, I just dont remember where.

That standard deviation is low. But anyway, be simple and model it as a normally distributed random walk:

Confidence interval equals = bet size * number of bets * EV +/- t-score * standard deviation * square root (number of bets).

There’s a fair chance I am misremembering the formula.

If you want three standard deviations then set your t-score to 3.

(1,000 * -.005 units) +/- (1,000^.5 * 1.15 * 3) =

-5 units +/- 109 =

- $25 +/- $545.

High variance with respect to EV since the edge and trials are low.

Fun exercise is how many hands need to be played before you are “guaranteed” to be ahead. Guaranteed meaning it would be X standard deviations to be down.

Just set the random walk formula edge equal to the downside standard deviation:

number of bets (N) * EV = standard deviation * X * square root (N).

Solve for N at 3 standard deviations:
N = (3* std / EV)^2

For std of 1.15 and EV of 0.005 it gives an N of 476,100. At a super fast 100 hands an hour that’s 4,761 hours at the table.

ETA: because of the squaring, note that doubling your edge has the effect of reducing by 4 times the required number of hands (N).

OriginalSD = AvgBet * 1.15

EV(x hands) = (x * AvgBet)*(HouseEdge)
SD(x hands) = Sqrt(x) * OriginalSD

So for your example above...

AvgBet = $5
OriginalSD = 5*1.15 = 5.75

EV(1000 hands) = (1000*5)*(-.005) = -$25
SD(1000 hands) = Sqrt(1000) * 5.75 = $181.83... 3SD = $545.49

Thus, with 99% certainty, after 10 hours, 100 hands per hour (so 1000 hands total), flat betting $5 at a .5% HE blackjack game... You can expect to lose $25+/- $545.49.

So the BEST you could hope to do is -25 + 545.49 = +$520.49
And the WORST you could do would be -25 - 545.49 = -$570.49

Quote: Romes

OriginalSD = AvgBet * 1.15

EV(x hands) = (x * AvgBet)*(HouseEdge)
SD(x hands) = Sqrt(x) * OriginalSD

So for your example above...

AvgBet = $5
OriginalSD = 5*1.15 = 5.75

EV(1000 hands) = (1000*5)*(-.005) = -$25
SD(1000 hands) = Sqrt(1000) * 5.75 = $181.83... 3SD = $545.49

Thus, with 99% certainty, after 10 hours, 100 hands per hour (so 1000 hands total), flat betting $5 at a .5% HE blackjack game... You can expect to lose $25+/- $545.49.

So the BEST you could hope to do is -25 + 545.49 = +$520.49
And the WORST you could do would be -25 - 545.49 = -$570.49

Thank you Romes. This is exactly what I was looking for.

Quote: Romes

OriginalSD = AvgBet * 1.15

EV(x hands) = (x * AvgBet)*(HouseEdge)
SD(x hands) = Sqrt(x) * OriginalSD

So for your example above...

AvgBet = $5
OriginalSD = 5*1.15 = 5.75

EV(1000 hands) = (1000*5)*(-.005) = -$25
SD(1000 hands) = Sqrt(1000) * 5.75 = $181.83... 3SD = $545.49

Thus, with 99% certainty, after 10 hours, 100 hands per hour (so 1000 hands total), flat betting $5 at a .5% HE blackjack game... You can expect to lose $25+/- $545.49.

So the BEST you could hope to do is -25 + 545.49 = +$520.49
And the WORST you could do would be -25 - 545.49 = -$570.49

I think OP means EV +0.5% not -0.5%.

No, I meant -.5%. I am not a counter so I imagine the best I could do would be around .5% negative EV. That was all just hypothetical anyway for me to understand the formula.

Quote: VCUSkyhawk

No, I meant -.5%. I am not a counter so I imagine the best I could do would be around .5% negative EV. That was all just hypothetical anyway for me to understand the formula.

Got it. Withdrawn.

Quote: Romes

So the BEST you could hope to do is -25 + 545.49 = +$520.49
And the WORST you could do would be -25 - 545.49 = -$570.49

the above IS 100% wrong.
99.0% is NOT = to 100%
no matter who says it does.
why NOT use 8 sigma instead, that is what I say (I am more right than wrong, most times)
*****
so we would expect out of 1 million such sessions and this happens about every weekend
(say 1 million players with one session each, flat-betting-ugh! flat etting. no one flat bets that many bets. but...)
for many
to win MORE or LOSE more than 3 SD.

the math says yes

it can be exactly calculated too instead of an approximation.
computers have feelings too
*****
another problem here is every gambling writer (including the Wizard)
says one WILL lose just $25 over that many bets as long as each bet is made
and basic strategy is followed perfectly every round and...
look at all the writings!
it is right there (Oh, some do add the 'average' word)

problem IS, there IS NO average of just 1
1000 hand BJ session.
average
"A calculated "central" value of a set of numbers."
https://www.mathsisfun.com/definitions/average.html

'there will never be ONE player losing more or less'
this has been written more times than the original Bible
so it must be true
lol
come on
get real!

the truth hurts no one
except those that feel hurt
Sally

OP, imo,
use the 8 sigma range
place yourself above the rest and shine on!!

or calculate this yourself
if I can do that (in R of course using a computer)
anyone can

Quote: VCUSkyhawk

No, I meant -.5%. I am not a counter so I imagine the best I could do would be around .5% negative EV. That was all just hypothetical anyway for me to understand the formula.

Yes, but if you did want to count, you could use the same formulas and modify the HouseEdge with your average advantage (typically about 1% for most counters). So then you could just put in a positive .01 for that -.005, and you'd get a positive expectation.

Oh Sally, I know you know that I know that you've seen me post time and time again about running proper simulations for exact amounts. The 1.15 is subjectively the sd in blackjack and could be calculated out more. You could run worse than 3SD, but for all intensive purposes that would seriously get in to the grounds of cheating, as we all know and accept.

I do also agree you are more right that wrong... but I do believe for the OP and his question the calculations provided will be more than accurate enough.

Actually this is a good question, is a 1% chance of failure acceptable risk?

*if risking your life, I would say obviously no, in almost any situation

*if risking your life savings, likewise

*if risking what you can afford to lose, it's probably OK for most of us, which is why 3 SD is used so often

which reminds me of that old thing about 99.9% not being good enough in many things. "If you can land a plane without crashing 99.9% of the time, there would be several crashes a day at O'Hare " ... or 100 crashes a day worldwide, see link. For many things not only is 99% not good enough, but not the often used 99.9% either.

https://www.quora.com/How-many-commercial-aircraft-take-off-per-day-worldwide