"Probability of Being Ahead by a Certain Amount" Question

[Overkill]

Hello all,

I'd like to calculate the probability of having a +5.00% EV after flat betting 5,000 hands while playing a -.45% player advantage game of double-deck blackjack. Is it less than 1%? Can someone please show the math (z-statistic, etc.)?

I used Norm's Outcome Calculator https://www.qfit.com/calcrisk.htm
to answer this question. But his calculator requires entering monetary parameters (such as bet size and desired financial outcome) and not just the above terms, and I am not sure I entered the correct numbers.

Again, using the above numbers:

When flat betting $10 for 5,000 hands (by the way, can "hands" be said to be exactly equivalent to "rounds?") and ending up with a player EV of +5.00% while playing a -.45% player advantage game of double-deck blackjack, does this mean the player ends up with $2,500 ($10 X 5,000 hands X 5.00%)?

If this is true, again, is the probability of this occuring less than 1%?

[Dog Hand]

Hello all,

I'd like to calculate the probability of having a +5.00% EV after flat betting 5,000 hands while playing a -.45% player advantage game of double-deck blackjack. Is it less than 1%? Can someone please show the math (z-statistic, etc.)?

I used Norm's Outcome Calculator https://www.qfit.com/calcrisk.htm
to answer this question. But his calculator requires entering monetary parameters (such as bet size and desired financial outcome) and not just the above terms, and I am not sure I entered the correct numbers.

Again, using the above numbers:

When flat betting $10 for 5,000 hands (by the way, can "hands" be said to be exactly equivalent to "rounds?") and ending up with a player EV of +5.00% while playing a -.45% player advantage game of double-deck blackjack, does this mean the player ends up with $2,500 ($10 X 5,000 hands X 5.00%)?

If this is true, again, is the probability of this occuring less than 1%?

Overkill,

Ok, I'll give it a shot!

First, if you play 5000 rounds at $10/round on a -0.45% game, what is the expected value, or EV? That's simply the "handle" (total amount of initial bets) times the edge, so

EV = (# of rounds)*(flat bet per round)*(edge) = 5000*$10*(-0.45%) = -$225

So, if you were to repeat this 5000-round experiment many, Many, MANY times, on average you should expect to lose $225 each time. However, the chance of any one 5000-round trial ending at precisely -$225 is very small.

Now the missing piece of information in your problem is the standard deviation (or, alternatively, the variance) per round for flat-betting this hypothetical DD game. Typical values are in the 1.1 to 1.2 range: the actual value depends on the house rules. I'll use 1.14 units in this post.

So, if the average result is -$225, what range of results should you expect? Here's where we need the standard deviation, or SD. The SD is calculated by multiplying the SD per round (in units) times the flat bet per round times the square root of the number of rounds, so:

SD = (1.14 units)*($10)*sqrt(5000) = $806.10

Now if your results follow a typical bell curve, statistics tells us that about two-thirds (actually 68%) of the time, you should expect your actual result to fall within one SD of the EV. Using the numbers above, one SD below EV is -$225 -$806 = -$1031; one SD above EV is -$225 +$806 = +$581.

Furthermore, about 95% of the time you should expect to finish within two SD's of EV, so in the range of -$1837 to +$1387,
and 99.7% of the time within 3 SD's of EV, so in the range of -$2643 to +$2193: see https://www.wikipedia.org/wiki/68–95–99.7_rule

If your results showed a +5% edge over these 5000 rounds at $10/round, then you won $2500: how unlikely is this result for your hypothetical game? To answer this question, we need to calculate the "z-score": how many SD's above EV is the result?

Well, you expected to finish at -$225, so finishing at +$2500 is $2725 above EV. Since the SD is $806, the z-score is

Z = +$2725/$806 = +3.38

Now the Complimentary Cumulative table of the Standard Normal Distribution (see https://www.wikipedia.org/wiki/Stand...ary_cumulative) will tell us the probability of finishing at a given Z or higher. For +3.38, the table gives a probability of 0.00036, or 0.036%.

Norm's calculator gives 0.03%, so we see that our calculations are pretty close to Norm's. The slight difference is due to two factors: some rounding on my part, and no doubt a slightly different value for the per-round SD (recall I simply assumed a value of 1.14).

At any rate, we see you have only a tiny chance of being ahead $2500 after flat-betting $10/round for 5000 rounds. How can we improve your chances while sticking to flat-betting? What if you flat bet $500 per round for 100 rounds? That would still be a handle of $50,000, so your EV would still be -$225. However, your SD would be a lot higher:

New SD = (1.14 units)*($500)*sqrt(100) = $5700

So your Z would be much smaller:

New Z = +$2725/$5700 = +0.48

From the Complimentary Cumulative table, the probability of a Z of +0.48 or higher is about 31.5%.

Hope this helps!

Dog Hand

[James989]

Hi Dog Hand,

Are you assuming he have more than 5000 units bankroll ? I think bankroll will affect the probability to reach his goal before he loss his entire bankroll.

James

[James989]

1) Is your game Variance = 1.31 ?
2) what is your bankroll in units ? If you have very small bankroll, you will probably bankrupt FIRST before you reach your goal of 250 units.
3) I think you should ask : What is the probability to win MORE THAN 250 units after 5000 rounds( flat bet 1 unit each round) ?

[21forme]

Great explanation, Doghand!

[Dog Hand]

Hi Dog Hand,

Are you assuming he have more than 5000 units bankroll ? I think bankroll will affect the probability to reach his goal before he loss his entire bankroll.

James

James989,

My calculations are based on an infinite BR.

If his BR is not infinite, then you are correct. Don S. covers this topic in his seminal Blackjack Attack 3 in his discussion of the "double barrier problem".

Hope this helps!

Dog Hand

[Overkill]

Thanks so much, Dog Hand. I appreciate the time you put into your response. And thank you for showing your work. The answer and interpretation you arrived at were what I thought.

I used an online sim (I think it was bjsim) a few years back to investigate one of my strategies. (By the way, when I last checked, bjsim had changed some.). Anyway, as I recall it would only let me do sims with small numbers of rounds. When checking one of my strategies. the sim said my E.V. was 5.00% (actually 4.98%) for 5,000 rounds.

Regarding interpretation: I believe you will all say that the 0.036% probability (please see Dog Hand's post) is meaningless as only 5,000 hands were run in the sim. I no longer have Norm's nice sim that I purchased, so maybe I will ask one of you to run a billion or so rounds if you don't mind.

But in the meantime, what is the value of all that calculating that Dog Hand did if it is "meaningless." I was under the impression that the z score calculations ALREADY TOOK INTO ACCOUNT the (small) sample size.

Is it perhaps true that using z scores, etc. for the game of blackjack is an inappropriate use of z scores. etc., unless one is working with, say, 100 million rounds or so or more? But again, though we are only using 5,000 rounds in the above example, standard deviation, etc. is indeed BUILT IN to the calculations, no?, thereby obviating the need for millions of rounds?

If you believe the 5,000 round probability result (0.036%) is meaningless, does that also mean that you would predict that a sim of one billion rounds with my flat betting, non-progression strategy is no more likely to result in a positive player E.V. than just 'regular' (i.e., no 'strategy' or system) flat-betting, basic strategy Blackjack?

[DSchles]

The validity of the results from a SIM of 5,000 hands is meaningless. But that has nothing to do with what Dog Hand calculated. His work had nothing to do with a sim, and the results are accurate, no matter what the number of hands.

Don

[Dog Hand]

The validity of the results from a SIM of 5,000 hands is meaningless. But that has nothing to do with what Dog Hand calculated. His work had nothing to do with a sim, and the results are accurate, no matter what the number of hands.

Don

Just to elaborate on Don's comment concerning 5000-hand sims:

I have an Excel file containing 300,000 hands of BJ played by a heads-up B.S. player flat-betting $5/round for a 6D, H17, DAS game with 75% penetration. For this game, billion-round simulation results give the player's IBA as -0.5580% and his SD as 1.16.

I used Excel to look at every 5000-hand set in these data: thus hands 1-5000, 2-5001, 3-5002, etc. The best set had the player winning $1,285 on total initial bets of $24,425 (4,885 rounds, EV = -$136.29), for an edge of a whopping +5.2610% and a z-score of 3.5061. The worst set had the poor guy losing $1,557.50 on total initial bets of $24,375 (4,875 rounds, EV = -$136.01), for an edge of -6.3897% and a z-score of -3.5102.

The wide range of results for 5000-hand sims indicates clearly why such small sample sizes are not useful when the edge is so close to zero.

In fact, for these 300,000 hands the player was "lucky": instead of losing the expected $8,133.35, he lost "only" $7,070.00, for a z-score of 0.3396. Thus, even 300,000-hand sim are not sufficient to generate useful results.

Hope this helps!

Dog Hand

[Wave]

The wide range of results for 5000-hand sims indicates clearly why such small sample sizes are not useful when the edge is so close to zero.

What would a sufficient number of trials be (not blackjack related, but in general) to establish reliable Means and the SD for those Means? Is it possible that the type of statstic one is attempting to identify has an impact on the number of trials needed to be considered statistically reliable? 5000 trials seems to me to be sufficient to establish Mean and SD, and your examples clearly show that 5000 trials is not a large enough sample to establish EV, but I am not a statistician and am wondering if different types of statistics require different sample sizes in order to be considered reliable.

[DSchles]

What would a sufficient number of trials be (not blackjack related, but in general) to establish reliable Means and the SD for those Means? Is it possible that the type of statistic one is attempting to identify has an impact on the number of trials needed to be considered statistically reliable? 5000 trials seems to me to be sufficient to establish Mean and SD, and your examples clearly show that 5000 trials is not a large enough sample to establish EV, but I am not a statistician and am wondering if different types of statistics require different sample sizes in order to be considered reliable.

The term(s) that you're looking for is how many trials it takes for the values to "converge." And yes, they are quite different, depending on the statistic you're trying to determine. Norm will tell you that he wants to run hundreds of millions of hands before "declaring" any number to be accurate or reliable, but I will allow him to weigh in further on this.

Suffice it to say that 5,000 hands is, obviously, good for absolutely nothing whatsoever.

Don

[Wave]

The term(s) that you're looking for is how many trials it takes for the values to "converge." And yes, they are quite different, depending on the statistic you're trying to determine. Norm will tell you that he wants to run hundreds of millions of hands before "declaring" any number to be accurate or reliable, but I will allow him to weigh in further on this.

Suffice it to say that 5,000 hands is, obviously, good for absolutely nothing whatsoever.

Don

Thanks Don.

I am in no way trying to make this political, but we are constantly bombarded with public opinion polls in the media where they act so sure of the results of polls based on "1000 registered voters", or "1000 likely voters", +/- 3% margin of error, blah, blah, blah...so if I'm interpreting your response correctly, these types of polls are basically worthless?

[DSchles]

Thanks Don.

I am in no way trying to make this political, but we are constantly bombarded with public opinion polls in the media where they act so sure of the results of polls based on "1000 registered voters", or "1000 likely voters", +/- 3% margin of error, blah, blah, blah...so if I'm interpreting your response correctly, these types of polls are basically worthless?

No, didn't say that! With 1,000 people polled, the typical margin of error is, indeed, a bit higher than 3%. So, it might say that candidate A leads B by 55 to 45, give or take 3 percentage points. That means the vote could actually be as close as 52-48 or as far apart as 58-42 ... or more (see below). And that's a big difference (as Hillary might tell you!)!! So, the values simply aren't very reliable.

For 5,000 observations, standard error drops to 1.4%, and so, if I tell you that your SCORE in a game is, say 50, you now can feel reasonably confident that it probably is somewhere between 50.71 and 49.29. For most, that might be acceptable, but for researchers, and those of us who obsess over every decimal point, it isn't nearly accurate enough. And, the true result will lie outside of these upper and lower limits almost one-third of the time, so, again, the results can leave much to be desired.

Don