Probability of winning in BJ

[Thunder]

Would it be correct to tell a basic strategy player in BJ that they only have a 1/10,000 chance of making $200 or more in 2 hours of playing blackjack assuming that they flat bet $10/hand, play where the house edge is .40%, play 60 hands/hr and have a $1000 bankroll.

I used binomial distribution to come up with this answer but I have a hard time the believing the odds are this unlikely.

 

[Jack_Black]

Would it be correct to tell a basic strategy player in BJ that they only have a 1/10,000 chance of making $200 or more in 2 hours of playing blackjack assuming that they flat bet $10/hand, play where the house edge is .40%, play 60 hands/hr and have a $1000 bankroll.

I used binomial distribution to come up with this answer but I have a hard time the believing the odds are this unlikely.

I will say yes, although my PC just crashed and I haven't put CVdata back on yet. Sounds reasonable given the bankroll size vs. bet unit size. Although I think I would still need pen. % and decks used to determine an exact number. My guess without it, 6D 75% pen., would have a win rate of -$8/hr.

 

[kewljason]

Would it be correct to tell a basic strategy player in BJ that they only have a 1/10,000 chance of making $200 or more in 2 hours of playing blackjack assuming that they flat bet $10/hand, play where the house edge is .40%, play 60 hands/hr and have a $1000 bankroll.

I used binomial distribution to come up with this answer but I have a hard time the believing the odds are this unlikely.

a 1 in 10,000 chance of winning 20 units in 2 hours? Me thinks you be doing something wrong. :laugh: 20 units in 2 hours happens all the time. Just part of normal fluctuation. I'll bet it's a lot closer to 1 in 10 chance than 1 in 10,000.

 

[Thunder]

I'd agree Jason but I don't know of another way to calculate the likelihood of this happening. Perhaps someone with a strong math background could answer.

 

[sagefr0g]

Would it be correct to tell a basic strategy player in BJ that they only have a 1/10,000 chance of making $200 or more in 2 hours of playing blackjack assuming that they flat bet $10/hand, play where the house edge is .40%, play 60 hands/hr and have a $1000 bankroll.

I used binomial distribution to come up with this answer but I have a hard time the believing the odds are this unlikely.

the odds for making $200 in 2 hours at the game conditions above is circa 11% or 12%.
it would be a standard deviation of about 1.63, a probability of about 5.16%
according to my cvcx sim of a similar game

 

[Thunder]

how can the odds be 11 or 12% if the probability is 5.16%???? They're the same thing.

 

[sagefr0g]

how can the odds be 11 or 12% if the probability is 5.16%???? They're the same thing.

the 5.16% probability is the probability if the result was $200.
the 11 or 12% is the odds of making $200 before going broke or hitting the specified number of hours to play (2hrs) under the stated conditions.

 

[Thunder]

Hmm thanks for the explanation. I'm curious though why you wouldn't be able to use binomial distribution to solve this problem.

 

[sagefr0g]

Hmm thanks for the explanation. I'm curious though why you wouldn't be able to use binomial distribution to solve this problem.

i haven't the foggiest, lol. don't even know really what the binomial distribution is, so i looked it up.
noticed one thing, it has to do with yes or no stuff sort of thing, so i guess that is like win or lose.
thing is with blackjack, the pushes, the double downs, splits, perhaps surrenders, maybe insurance bets and the way a snapper pays makes for not an exactly situation where you go by yes, no or win or lose only sort of stuff.
just rambling here, so but what ever, maybe it's a rather complicated thing to try and employ the binomial distribution?
edit: errh also blackjack as it's being played isn't really independent as far as wins, loss's, successful doubles and splits, snappers, successful insurance bets ect. and whatever the binomial distribution thing was talking about independent events sort of stuff, more on the order of black or red roulette, or craps sort of stuff, maybe?

 

[Sonny]

I'm curious though why you wouldn't be able to use binomial distribution to solve this problem.

We can. The mighty curve tells all! It is the seer of all seers...as long as the results are normally distributed.

We are looking for the probability of a $200 win betting $10 after 120 hands.

EV = $10 * 120 * -0.004 = -$4.80
SD = $10 * sqrt(1.33 * 120) = $126.33

Now we can use the standard z-score to find out how many standard deviations a $200 win would be.

z = ($200 + $4.80 + 0.5) / $126.33
z = 1.63

Looking that up in a chart or spreadsheet, we see that 94.84% of the results are to the left of it (winning less than $200) and 5.16% of the results are to the right of it (winning $200 or more). So a BS player has a 5.16% chance of experiencing a swing of 1.63 standard deviations or more, which in this case is enough to produce a $200 win. As sagefrog pointed out, this does not take into consideration the possibility of ruin (which is fairly negligible in this case) or the probability that the player will win $200 or more before the 120th hand.

-Sonny-

 

[sagefr0g]

We can. The mighty curve tells all! It is the seer of all seers...as long as the results are normally distributed.

We are looking for the probability of a $200 win betting $10 after 120 hands.

EV = $10 * 120 * -0.004 = -$4.80
SD = $10 * sqrt(1.33 * 120) = $126.33

Now we can use the standard z-score to find out how many standard deviations a $200 win would be.

z = ($200 + $4.80 + 0.5) / $126.33
z = 1.63

Looking that up in a chart or spreadsheet, we see that 94.84% of the results are to the left of it (winning less than $200) and 5.16% of the results are to the right of it (winning $200 or more). So a BS player has a 5.16% chance of experiencing a swing of 1.63 standard deviations or more, which in this case is enough to produce a $200 win. As sagefrog pointed out, this does not take into consideration the possibility of ruin (which is fairly negligible in this case) or the probability that the player will win $200 or more before the 120th hand.

-Sonny-

like from my sim, would the way to get the (i think) generic standard deviation per hand figure you use of 1.33 be $11.51/$10=1.151 then square it = 1.325 pretty much same as 1.33? if so why square it?
whatever i guess the 1.33 figure kind of generically takes care of the split, double, insurance bet, snapper pay & surrender sort of complications?
errhh also whats the 0.5 from your z-score formula? some sort of component of the mean?

 

[Sonny]

like from my sim, would the way to get the (i think) generic standard deviation per hand figure you use of 1.33 be $11.51/$10=1.151 then square it = 1.325 pretty much same as 1.33? if so why square it?

Right, the SD is $11.51/$10 = 1.151 units. We can square that to get the variance, which is close to 1.33 units (I used 1.33 as a generic number, not taken from your sim). I used variance in my calculation, although I could have written it using SD since they are equivalent:

SD = $10 * sqrt(1.33 * 120) = $126.33

SD = $10 * sqrt(1.33) * sqrt(120) = $126.33

SD = $10 * 1.15 * sqrt(120) = $126.33

errhh also whats the 0.5 from your z-score formula? some sort of component of the mean?

I have no idea. I read a book that told me to do it, then I saw The Wizard doing it so I thought I must be on to something. If I had to take a guess, I would say that it has something to do with "straddling the line" between two numbers, but that explanation probably only makes sense to me (and barely even then!). Because we are looking at the results on one side of a line, we have to remember that our results do not include the area that the line is actually on. For example, if we want to know the probability of 1 SD or less, we have to step a little bit farther than 1 SD (1 SD + 0.5) so that our results will include 1 SD and everything under it. Otherwise, we will only be looking at everything under but not including 1 SD. The difference is probably very small, but like I said, everybody told me to do it. I envoke Nuremburg defense (better known in AP circles as the Imperial Palace defense).

[EDIT: Hmm, now that I had to explain it I think I actually understand it a little better. Thanks wise one!]

-Sonny-

 

[Nynefingers]

I have no idea. I read a book that told me to do it, then I saw The Wizard doing it so I thought I must be on to something. If I had to take a guess, I would say that it has something to do with "straddling the line" between two numbers, but that explanation probably only makes sense to me (and barely even then!). Because we are looking at the results on one side of a line, we have to remember that our results do not include the area that the line is actually on. For example, if we want to know the probability of 1 SD or less, we have to step a little bit farther than 1 SD (1 SD + 0.5) so that our results will include 1 SD and everything under it. Otherwise, we will only be looking at everything under but not including 1 SD. The difference is probably very small, but like I said, everybody told me to do it. I envoke Nuremburg defense (better known in AP circles as the Imperial Palace defense).

[EDIT: Hmm, now that I had to explain it I think I actually understand it a little better. Thanks wise one!]

-Sonny-

I'm a little clueless about this as well, but from some quick googling, it looks like it may have to do with discrete events, such as coin flips, etc. In other words, to figure out the probability of x number of heads outcomes after n number of flips, you would basically integrate the probability density function from x - .5 to x + .5, or if you want to know the probability of getting at least x heads, you would integrate from x + .5 to infinity. That would probably also apply here since with a flat bet we are working with discrete outcomes. In this case, I would think you would use

z = ($200 + $4.80 - $2.50) / $126.33

since $2.50 represents half of the smallest difference in outcomes we could possibly see ($10 unit, BJ or surrender would result in our outcomes being multiples of $5). I subtracted because we want results equal to or greater than $200, so we want to start integrating halfway to the next possible outcome to the left. Perhaps instead of -$2.50, it makes more sense to use -$2.30 since that puts us exactly halfway between our discrete outcomes of +$195 and +$200.

Just guessing here. Makes sense to me, but I'm really not sure one way or the other.

Edit: I think $2.30 would be wrong. Since the possible outcomes are ...$195, $200, $205..., and we want to know the probability of being at $200 or better, I think we would pick the point halfway between $195 and $200 and integrate from there to infinity. This has nothing to do with the house edge of the game, which is it's own term. I think the correct formula is as I posted above, z = ($200 - $2.50 + $4.80) / $126.33. Of course, since we are assuming that our results are represented by a normal distribution, instead of integrating the probability density function from z to infinity, we simply use a table to look up the value or invoke the normsdist function. But still...after putting the numbers in Excel, I see that Sonny's formula gives the correct z score to match the risk numbers given in Sagefr0g's sim, and my formula does not. So maybe I'm just way off in left field somewhere. Sonny, let me know if you ever get it figured out...