First of all, I'm down 150 buy-ins liftetime at Pokerstars in EV with 1.3 millions hands played. I'm trying to better understand whether I have been unlucky or not.

Anyway, Let's say that I flip a coin 10 000 times and there are 2 possible outcomes: - heads +10$
- tails -10$

Heads came out 4950 times and tails appeared 5050 times.
So, let's say that I won $10 4950 times,and I lost $10 5050 times.
The result is (10*4950)-(10*5050)=-$1000

So, After 10 000 coin flips I should lose $1000 and be down 100 buy-ins in EV.

Running below EV irritates me; however, getting 4950 heads after 10000attempts is not that unlucky if I'm not mistaken.

Am I correct?

Probability of getting 4950 or less heads is 16.1%.

Your example is a binomial experiment. 2 possible outcomes.
The math is quite easy.

mean = P*N
variance = N*P*(1-P)
standard deviation is just the square root of the variance.

you should get 5000 for the mean and 50 as the SD
1SD range of possible outcomes = 4950 to 5050
Less than the 4950, or more than 1SD worse is 16.1% as noted in the above post.
2SD range of possible outcomes = 4900 to 5100

There is quite a lot of math to this part.
a simple online calculator can be faster
try this one
Binomial Calculator

From your example, if 1 million players did your experiment, we would expect about 161,000 of them to be worse than 4950 after 10,000 trials.

we would also expect about 161,000 of them to be better than 5050 after 10,000 trials.

No matter what they tried to do.

Thank you for your posts. I also wanted to post your same link I had found earlier today, but I didn't know what to say. And I agree with your 16.1% result.

Example n°2: Let's suppose that I go all-in with pocket rockets versus my opponent's cowboys (AAvsKK;82%18% sharp).
We do it 10 000 times, and I only win 8150 times, and I lose 1850 times.
So, I win 8150*$10 and I lose 1850 times $10 = +63000
However, I shoulda won 64000; therefore, I ran 100 buy-ins below ev again.

I used the same binominial calculator and entered 0.82, 10000 and 8150.
The result is P(X<=8150)= 0.099

So I ran both 100 buy-ins below EV in my example 1 and 2; however, I got a probability of 16.1% for my example 1 and 9.9% in my example 2.

That's because your variance is greater in example 1, so it is more likely that you will run 100 buy-ins below EV. Your variance will be greatest for a coin flip. The variance is 10000*1/2*1/2 = n/4 for the coin flip, and the standard deviation is the square root of that. For example 2, the variance is 10000*0.82*0.18 which is smaller.

An easy way to see this is by comparing the EV/SD of both results.

Example #1
ev = 5000
sd = 50
results 4950
formula (simple one)
(results - mean) / SD

4950-5000 = -50
-50/50 = -1
1SD to the left of the mean

The probability of being -1sd or less (between 0 and 4950.5)
(we will skip the standard normal correction for the moment)

Example #2
ev = 8200
sd = 38.41874542
results 8150
4950-5000 = -50
-50/38.41874542 = -1.301448016

The farther away any value is from 0, the lower the probability.
In this case we are negative and to the left of the mean.
Look at the 2 values.
-1
-1.301448016
Of the two, -1 is closer to 0, so it has a higher probability of happening.


I'm sure Bruce can correct me or make this concept even clearer if needed.

If you want to know more about using the normal distribution to approximate the binomial, especially for large values of N, ask Bruce.
I am sure he has a link to a very easily understandable post.

Excel has a BINOMDIST function
and a
NORMSDIST function (uses the ev/sd value)
that can give probabilities also

Sally

I get it now; thank you for all of your help.

Sorry for bringing back an old post, but this was one that came up when I was searching through google. Furthermore people in this thread have background in Stats and offer helpful insight rather than just trolling. I'm in a very similar position with OP. I've played about 1.15 million hands on Party through various stakes, and am down about 120 buy-ins below EV, and am extremely frustrated and questioning the normality of things.

I understand that all-in EV is a normal distribution for the population. But isn't each individual player's all-in EV a normal distribution as well, with mean = 0?? What I mean is the distribution for the population is normal, but this is also because sample sizes for each player is very different. Some people only play a small set of hands and run either terribly below EV or greatly above EV, and quit. They never go through the correction process of EV. Purely hypothetically speaking, for each individual players' distribution, as sample size approaches infinity, actual winnings and all-in EV should converge at mean = 0. Doesn't this just absolutely have to be the case? But at what sample size is such assumption plausible to be made?

If you have a bigger sample size you can expect to come closer to the ev if you look at what the worst or best results could have been and then check how many % off you are.

If you start at EV then after a while you're down 10 BI's ev, at that point if you want to predict the future you can't expect to be back at EV after x hans, instead you should expect to still be 10 BI's down.

The bigger the sample, the closer to 50% (break even) you would expect to be in coin flips, but even 0.1% off in one direction over a sample of 5 million hands would be 250 flips.

I think this is what is misleading about the fact that EV converges over a large enough sample. The percentage it is off will be damn close to exactly EV, but that can still be hundreds of "buy ins."

At least that's how I understand it.

The average deviation from your EV increases as the square root of the number of hands. It doesn't decrease. The longer you play, the more buy-ins you will be from your EV on average, above or below.


Your total winnings do not converge to your EV as the sample becomes large. That's a common misconception. The difference between your actual winnings and your EV becomes infinitely large, and it actually makes larger and larger fluctuations above and below your EV. But if you divide that difference by your actual EV, that goes to zero. If you divide your total winnings by the number of hands to get winnings per hand, that will converge to your EV per hand.

This is how I perceived the situation to be, and makes sense to me. But what Bruce is saying is not implying the same thing. I need a little bit more clarification. I don't understand how it's any different from say flipping a coin 10 million times. Obviously it won't be exactly 5 million heads or tails, but wouldn't the variance be smaller, and actual would be closer to expected (in proportion to sample size) than say 5 million times, 1 million times or 5000 times? Why does the difference between actual and expected get infinitely large? Thanks for the post guys.

To illustrate what he means, consider flipping a coin 10 times. Say it came up 6 heads and 4 tails. That's 60% heads, 10% off of expected. It's also 1 more head than the expected value of 5. Neither of these seem out of line for a 10-flip run.

Now consider flipping the same coin 1,000,000 times. Say this time it comes up heads 500,500 times. That's a rate of 50.05%, or .05% more than expected - much closer than in our 10 flip case. However, there are now 500 more heads than we would expect, much more than would even be possible in the 10 flip case. As before, this result is not far out of line.

To put it more briefly, the difference in rate converges to zero but the difference in number will diverge from zero and is expected be larger as the sample size increases.

He's saying the same thing as me, but when he says that the difference gets closer to 50%, means that the percentage gets closer to 50%, not that the number of heads gets closer to half the total coin flips.

What you say about the variance is actually true of the standard deviation which is the square root of the variance. How far you are away from the EV depends on the standard deviation which is what you are interested in. The standard deviation gets smaller in proportion to the sample size, meaning that the standard deviation divided by the number of samples goes to zero. But the standard deviation of the total number of heads (without dividing by the number of samples) gets bigger, which means that on average your deviation from 5 million when you flip 10 million times will be a bigger number than your deviation from 500,000 when you flip a million times, which in turn is bigger than your average deviation from 2500 when you flip 5000 times, etc. That difference grows in proportion to the square root of the number of samples, but when we divide that by the number of samples, we get something that decreases as the square root of the number of samples, meaning it goes to zero as the number of samples gets big.

So when you say you are 120 buy-ins away from your EV after some huge number of hands, that isn't necessarily surprising because the longer you play, the bigger that number is expected to be. But if you divide your winnings by your total hands, that should get closer and closer to your true EV per hand the more hands you play.

Thanks guys for very helpful posts. The concept is very much clarified . As long as the % gets closer and closer to 50%, it's all good .