average lowest point of bankroll

[ThodorisK]

Suppose a player plays 5000 hands, and places 1$ on each hand.
What is the average lowest point a player's bankroll will reach, during the course of these 5000 hands?
Or, in other words, what is the average greatest loss a player will meet during a course of 5000 hands?
(this could be, e.g. -100$)
Suppose that the bankroll is infinite.
(I am not talking about the average loss at the end of the 5000 hands)
I want the whole procedure which gives the result.
Do you know any math-blackjack forums I can ask this?

 

[Sonny]

Suppose a player plays 5000 hands, and places 1$ on each hand.
What is the average lowest point a player's bankroll will reach, during the course of these 5000 hands?

If he is playing at a disadvantage then the average lowest point will be at the end of those 5,000 hands. The longer you play, the longer you will lose on average.

I think your question is more about variance than average results. Instead of average results you're asking about the probability within a range of outcomes. Here is a good article to get you started with variance:

http://www.blackjackforumonline.com/content/Blackjack_Basic_Strategy_Betting_And_Risk.htm

You can use the standard deviation to find the probability that you will lose X bets over the course of N hands. You can either use the binomial distribution or a Trip Risk of Ruin formula. There is a great discussion of the formula in Don Schlesinger’s Blackjack Attack. There are also various risk calculators here:

http://www.qfit.com/blackjack-online-calculators.htm

You can also find some heavily math-related articles here:

http://www.bjmath.com/bjmath/ror/ror.htm (Archive copy)

-Sonny-

 

[ThodorisK]

I meant that after the end of the 5000 hands, one might end up having lost -40$, but the greatest loss he met, was at the hand No 3524, where he was losing -150$. That's the lowest point of his bankroll during the play of the 5000 hands. So what is the average value for the lowest point? This does not refer to a loss AFTER the 5000 hands are completed. This loss corresponds to the -40$.

Of course the standard deviation is a variable included in the formula or the procedure that gives the answer.

 

[Sonny]

So what is the average value for the lowest point? This does not refer to a loss AFTER the 5000 hands are completed.

I still don’t understand the question. Again, you’re not looking for the average lowest point. The average point will be equal to your theoretical expected value and will max out after the last hand has been played.

So maybe you’re asking about actual results instead of average results. In that case the answer is the same. The worst possible loss would be $5,000 after the 5,000th hand (assuming the player always bets exactly $1). Obviously that would be unlikely, but it is possible.

Any other results will end up somewhere within the range of standard deviations with varying probabilities. For example, you will be within one standard deviation of your expectation about 68% of the time. That means a loss of 1 SD will occur about 34% of the time. You will be within two SDs of your EV about 95% of the time and within three SDs over 99% of the time. The probability of any other results can be calculated as a percent of your SD.

I don't know if that answered your question, but hopefully you can give me a better idea of what you're looking for.

-Sonny-

 

[ThodorisK]

Always flatbetting 1$, basic strategy, reshuffle after every hand.

Player 1, ends up with a balance of -40$ after 5000 hands.
But the lowest point his balance reached, was -150$ at hand No 3524

Player 2, ends up with a balance of -15$ after 5000 hands.
But the lowest point his balance reached, was -70$ at hand No 2026

Player 3, ends up with a balance of +17$ after 5000 hands.
But the lowest point his balance reached, was -65$ at hand No 3730

... and so on. So when the number of players-samples tends to infinity, the values of the ending balances tend to an average value, and the values of these lowest points tend to another average value.

So how can we find the average value for this lowest point?

I think the analysis you gave refers to how probable it is for the ending balance (after the completion of the 5000 hands) to deviate that much or that much from the average value of the ending balance.

I think these are two different things.

My problem has the condition that 5000 hands must be completed.

 

[Kasi]

So how can we find the average value for this lowest point?

Can't say I really understand ur question either but Sonny's answer sounds pretty good to me.

Otherwise buy a sim and run a million 5000 hands sessions. (can sims do this?!)

Or run it 100 times. Would it matter to u if the lowest point was -1 99 times out of the hundred and that always occurred within the first 100 hands and once it was -4901 after 4000 hands? So many different combinations that would give the same average low point.

May I ask why u care or what ur ultimately trying to get to? Perhaps some betting system?

 

[sagefr0g]

So how can we find the average value for this lowest point?

i sure don't know but i'm guessing it would be close to your expectation minus one standard deviation.

 

[Sonny]

So when the number of players-samples tends to infinity, the values of the ending balances tend to an average value, and the values of these lowest points tend to another average value.

So how can we find the average value for this lowest point?

Let me try to explain this another way. Start with a game that has a 0.5% house edge. Think of the house edge as the average loss for a bet. If you bet $1 you would expect to lose $0.005 on average. If you bet $10 (or 10 hands of $1 each) you would expect to lose $0.05 on average. Betting $100 would average a $0.50 loss.

As you can see, the average loss (or EV) increases as you play more hands. Therefore the highest average loss will always be after the most number of hands, in this case 5,000. As you said, there may be times when you lose (or win) more than that, but the average results will be equal to the house edge.

I think the analysis you gave refers to how probable it is for the ending balance (after the completion of the 5000 hands) to deviate that much or that much from the average value of the ending balance.

Yes, although you can also use it to find the range of results for any point along the way. You can check the SD at 100 hands, 500 hands, 1000 hands, etc. and see what sort of fluctuations you will experience along the way. You can also use the Trip ROR formula to see how likely it is to lose $X at any point during those 5,000 hands.

-Sonny-

 

[Sonny]

Euroka!

Oh, now I get it!

I think I finally understand your question now. You are asking about the value of the average negative variance. I don’t know that there’s an exact number for that. Variance is usually calculated as a probability. For example, there is a 34% chance that you will experience a loss of 1 SD and a 16% change that you will lose more than that. About 48% of the time you will be within 2 SDs, so that’s pretty close to 50% of the time. I guess you could say that, on average, your results negative variance will be about 2 SDs or less.

You could also approach the problem from another way. Using the Trip ROR formula you can see that there is a 90% chance that you will lose $100 before 5,000 hands. There is a 50% chance that you will lose $365, a 30% chance that you will lose $500, and about a 1% chance that you will lose $1,000. Any other loss can be similarly calculated using the formula. This should give you an idea about what sort of average losses to expect.

-Sonny-

 

[Canceler]

Brute force method

I wrote a program in Excel to simulate playing a game with a .5% house advantage. Each game of 5000 hands was considered to be one trial. After 100,000 trials the average maximum (most low!) low point turned out to be -193.152. If anyone wants more details on this, I can post them later.

 

[ThodorisK]

I think it's much less than 193, but anyway, I am looking for the theoretical formulas.
Someone told me its thisDead link: http://www.iop.org/EJ/abstract/1742-5468/2005/06/P06013)
I havent cheched it yet, I'm busy.

 

[Canceler]

I think it's much less than 193

Actually, my crude sim did not allow for pushes, everything was either a win or a loss. With pushes involved, I think it would be somewhat less negative than -193.

If you ever do find the theoretical formula, please let us know.

 

[ThodorisK]

Well, I know 193 is wrong, because according to the calculator http://www.card-counting.com/blackjack-calculator-c3.htm,
when one has a starting bankroll of 193 bets and plays 5000 hands, the risk of ruin is only 3.2%

 

[Kasi]

Well, I know 193 is wrong, because according to the calculator http://www.card-counting.com/blackjack-calculator-c3.htm,
when one has a starting bankroll of 193 bets and plays 5000 hands, the risk of ruin is only 3.2%

I'm still clueless what ur getting at or what the connection of ROR is to average low point.

Originally, u assumed an infinite bankroll. Now u r assuming a 193 unit bankroll.

What answer would make u happy or seem reasonable to u?

And why are u even asking?

No big deal lol.

 

[ThodorisK]

No, I mentioned the ROR to proove wrong his 193 result

 

[Kasi]

No, I mentioned the ROR to proove wrong his 193 result

So, with a million or billion or trillion unit bankroll that would have essentially a ROR of 0%, the average low point might still be 193.

Don't see how that proves anything. Would it matter if the average low point was 10 or 2000?

What ROR would prove it to be right?