ITH Poker Forum

[PauliF]

Note: This thread serves as the footnotes to PauliF's article about bankroll management on the home page.

Calculating your expected earn rate from10 tosses of lucky coins

First lets look at the extremes.

Probability that you win all 10 = probability that you lose all 10= 0.5^10 = 0.1%

We can similarly work out the others.

For example the probability of winning exactly 1 toss out of 10=

The probability of winning 1 then loosing 9
+ The probability losing one and then winning one and then loosing 8
.
.
.
+ The probability loosing 9 then winning 1

= 10×0.5^10



(This is ten times more likely than winning or losing all 10. If you think about this, it seems correct since there is only one way you can win all 10 tosses, but you can win exactly 1 toss 10 different ways.)


We can summarise these in a simple equation:

Probability of winning n out 10 tosses = {10!/n!×(10-n)!}× 0.510

(The exclamation mark! After a number just means that you multiply that number by every number less than it down to 1.
eg 10!=10×9×8×7×6×5×4×4×3×2×1

Let me know if you want me to explain the first term which is called the combination function further)

Then I just plugged the above formula into Excel and got it to produce the table for me.

Number of wins :::::: ::: probability
0 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.1%
1 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 1.0%
2 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 4.4%
3 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 11.7%
4 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 20.5%
5 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 24.6%
6 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 20.5%
7 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 11.7%
8 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 4.4%
9 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 1.0%
10 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.1%

[PauliF]

More on the binomial distribution function

The general formula, or the probability density function, looks like this:

Probability of n successes out N trials = {N!/n!×(N-n)!}× (p)^ n × (1-p) ^( N-p)

p is the probability of success in a single trial so in lucky coins it was ½.

You could also use it, for example, to look at the distribution of how many sixes you would expect to see if you rolled a dice ten times.

Probability of n sixes out 10 rolls = {10!/n!×(10-n)!}× (1/6)^n×(5/6)^(N-n)

Then produce a table like we did for lucky coins:

Number of wins ::: ::: ::: ::: probability
0 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 16.2%
1 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 32.3%
2 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 29.1%
3 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 15.5%
4 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 5.4%
5 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 1.3%
6 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.2%
7 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.024807%
8 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.001861%
9 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.000083%
10 ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: 0.000002%

And then graph it:


Note how because the probability of success is no longer equal to the probability of failure we lose the symmetry in the graph.

[PauliF]

Now that we have a mean and standard deviation how can we use these to calculate expected ranges like the ones I quoted in the article?

Well I basically just ask Excel to answer the question, that I want answered, for me. You can use any software that calculates statistical functions.

Lets have a quick look at what question it is that we are asking.

In lucky coins we used a discrete distribution. That simply means that the graph is a bar chart rather than a continuous line.

So for example we can look up the answers directly from the graph if we wanted.
What’s the probability that get 3 heads?
Or what’s the probability that I get 7 or more heads?





So with a discrete distribution we are just reading off the individual bar or adding the bars that we require to answer our questions.


With continuous distributions its not that simple. Simply reading (or calculating) corresponding probabilities from numerical points on the x axis tells us very little. We need to measure, or calculate, the area under the graph between points to get the probabilities, or likelyhoods we are after.

The probability density function of the normal distribution looks like this:



This equation is just the instruction on how to draw the graph;
μ is your average earn rate
and σ is your standard deviation.


(The area under the whole graph of a probability density function always = 1 or 100%)


So in the graph for the normal distribution with μ = 2 and σ = 19, the probability that in any 100 hands you finish in profit is the area that I have shaded in blue below.




I obviously didn’t calculate the numbers I used in the article by working out how big shaded areas under graphs were.

We use what are called cumulative distribution functions. These are the integral of the probability density function.

Don’t panic. These formula just ask the question what is the probability that of getting at x or less given the distribution and the given parameters.

Now in the old days when there were no computers to calculate the numbers we wanted the process was a bit tedious. First you would need to “standardise” your question which just meant adjusting the numbers so that you could look the answers up on a standard table of answers (the standard normal distribution has a mean of 1 and a standard deviation of 0).

Now days you can just plug the numbers you want into Excel or whatever software you are using.

So lets look at how I got the numbers I used in the article.

1) The easiest one is the probability of losing 30 or more bb after 100 hands

So I asked excel what is the probability that with a mean of 2 and a standard deviation of 19 I get -30 or less.
The way to ask excel this is =NORMDIST(-30,2,19,TRUE)
The answer it gave is 4.6% ie one time in 22.

2) Win 30 or more bb:
I asked excel what is the probability, with a mean of 2 and a standard deviation of 19, that I get 30 or more.
The way to ask excel this is =1- NORMDIST(30,2,19,TRUE)
The answer it gave is 7% ie one time in 14

3) The probability of finishing in profit after 100 hands
I asked excel what is the probability that with a mean of 2 and a standard deviation of 19 I get zero or less.
The way to ask excel this is =NORMDIST(0,2,19,TRUE)
The answer it gave is 45.8%
So the answer I need ie the probability of being in profit is 100%-45.8% =54.2%

[robrobrob]

Hey Pauli,

What do you think about the assumptions of normally distributed earn rates. It seems that since most hands are +0bb, as they are folded pre-flop, if nothng else, they are quite peaked at 0.

This distributional assumption has always bothered me, and I would love to hear more on the subject.

rob

[PauliF]

sorry for the delay Rob... i only just seen this...

i spend a lot of time thinking about this exact question.
(maybe is shouldnt admit that in public)

i agree that when looking at individual hands we are basically looking at a Binomial distribution with one outcome of value zero with probability 0.75 or so
and the other value with some other value
I allude to this kind of thinking in my new article (Zoom in Zoom out)

this is of course for a winning player... a loosing player may well have a very different distribution when looking at single hands.

Now earn rate bb/100 hands (or over an hour) could well be normally distributed.
there usually has to be a strong reason for a random variable not to behave normally when observed over the long term...

i struggle to believe intuitively that poker earn rate is indeed so neatly distributed...
all the the prevailing literature and wisdom definitely has that believe...and i am not sure not starting off with that assumption is a terrible error when used to understand the earn rate behaviour for the purpose of understanding swing ... and indeed for developing bankroll strategies...

Still i struggle to conceptualise the factors affecting the shape of the distribution and would love to see more discussion of it...

Even if the assumption that winning players... playing sound solid poker... have normally distributed earn rates is correct... or at least correct enough for our purposes.... i definitely do not think that all players have a normal distribution....

when choosing the earn rate distributions for my characters i gave Larry the Loser a normal distribution because i was trying to explain concepts that would have been well muddied had he been a maniac...
i settled for a small loser ... the kind of guy who would be a small winner if there were no rake!!
so he is actually trying to win or at least strategize his way to winning...
unlike the maniac who just bet and raising like a nutter
i think that a maniac's earn rate is probably skewed... although i have no proof this is purely an educated guess...

i think there is a lot still to be discovered in this area and look forward to other peeps comments....

also my opinions and observations are based purely on my experience playing limit hold em
i am sure there many different factors to consider for other pokers....

and i did pose this question in a slightly different form before

http://www.internettexasholdem.com/phpb ... 14280.html

[checkmate]

i spend a lot of time thinking about this exact question.
(maybe is shouldnt admit that in public)

Nothing you say will convince us that your not the man!

[robrobrob]

Pauli, thanks for the followup. I posed this question a long time ago (over a year ago), and was sort of shut down on the topic. As a result, I have been thinking about it myself in the mean time, and I think that it may be reasonable for earn rate for a winning player playing tight agressive poker to be normally distributed over some number of hands. Perhaps 100 is enough. Clearly 1 isn't. I wanted to compute my own distribution a number of times, but I don't have enough hands in my DB to be meaningful. I have lost my PT db twice, and have been playing at sites that don't support PT a lot since the last loss. Add to it the fact that I am starting to specialize in SH play (2-4 players, not 6max) and my own data becomes useless for answering this question. I'd love to see the actual distribution of a long term winning player like pilchard with 100,000+ hands at a consitant earn rate, and see if it needs to be measured in bb/100, bb/500, or even bb/50 to approach a normal distibution. It would be a fun analysis, and I'd love to get the data, yet I suspect anyone with that data is not going to give it to me.

rob

[AlamedaMike]

Nothing you say will convince us that your not the man!

I second that - nice to hear from a smart person willing to share! I love statistics and always have - just not all that smart since University days.

Reading your articles was a pleasure - my stats are 1.36bb/100 and stdev 19.3bb for 18,049 hands at 0.25/0.50. I should not go broke playing at that level. I would like my stdev to be a little lower.

Nothing you say will convince us that your not the man!

Thanks

[PauliF]

wow thanks man
that brings back some memories
i dont know if anyones told you ... but this is actually now an excercise forum

[AlamedaMike]

wow thanks man
that brings back some memories
i dont know if anyones told you ... but this is actually now an excercise forum