10,000 hand of bad beats, whats the odds?

[winky51]

Curious if you are playing correctly vs fish what the chances are that in 10,000 hands you have a -EV of lets say 2. Assume low limit like 3/6 or 2/4.

[TomCollins]

What is a -EV of 2?

[meow_meow]

[ QUOTE ]
Curious if you are playing correctly vs fish what the chances are that in 10,000 hands you have a -EV of lets say 2. Assume low limit like 3/6 or 2/4.

[/ QUOTE ]

Major problems with your question:
1) EV = 'expected' value, not results.
2) 'playing correctly' is ill defined.
3) specifying the limit is not important.

Let me reword and and try to answer:
Q. What is the probability that a player playing against a specific set of opponents with a playing style that yields a true winrate of 2BB/100hands (a good winrate) and a standard deviation of 13BB/100 will be a 2BB/100 loser over a 10k hand stretch?
A. Less than one in five hundred

[BruceZ]

[ QUOTE ]
Let me reword and and try to answer:
Q. What is the probability that a player playing against a specific set of opponents with a playing style that yields a true winrate of 2BB/100hands (a good winrate) and a standard deviation of 13BB/100 will be a 2BB/100 loser over a 10k hand stretch?
A. Less than one in five hundred

[/ QUOTE ]

Actually 1 in 956 that he will be a -2 bb/100 loser or worse in 10K hands. I believe that you found the probabilty for being +/- (4/1.3) standard deviations from the mean, rather than at least this many standard deviations strictly below the mean, so your answer is off by a factor of 2.

The standard deviation of the win rate, or standard error, is 13/sqrt(10,000/100) = 1.3. A win rate of -2 bb/100 would be 4/1.3 standard errors below the average win rate of +2 bb/100. From Excel =NORMSDIST(-4/1.3) =~ 1 in 956.

[meow_meow]

[ QUOTE ]
[ QUOTE ]
Let me reword and and try to answer:
Q. What is the probability that a player playing against a specific set of opponents with a playing style that yields a true winrate of 2BB/100hands (a good winrate) and a standard deviation of 13BB/100 will be a 2BB/100 loser over a 10k hand stretch?
A. Less than one in five hundred

[/ QUOTE ]

Actually 1 in 956 that he will be a -2 bb/100 loser or worse in 10K hands. I believe that you found the probabilty for being +/- (4/1.3) standard deviations from the mean, rather than at least this many standard deviations strictly below the mean, so your answer is off by a factor of 2.

The standard deviation of the win rate, or standard error, is 13/sqrt(10,000/100) = 1.3. A win rate of -2 bb/100 would be 4/1.3 standard errors below the average win rate of +2 bb/100. From Excel =NORMSDIST(-4/1.3) =~ 1 in 956.

[/ QUOTE ]

Actually, I did a much rougher calc, which is why I went conservative and added "less than"
/images/graemlins/blush.gif

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Let me reword and and try to answer:
Q. What is the probability that a player playing against a specific set of opponents with a playing style that yields a true winrate of 2BB/100hands (a good winrate) and a standard deviation of 13BB/100 will be a 2BB/100 loser over a 10k hand stretch?
A. Less than one in five hundred

[/ QUOTE ]

Actually 1 in 956 that he will be a -2 bb/100 loser or worse in 10K hands. I believe that you found the probability for being +/- (4/1.3) standard deviations from the mean, rather than at least this many standard deviations strictly below the mean, so your answer is off by a factor of 2.

The standard deviation of the win rate, or standard error, is 13/sqrt(10,000/100) = 1.3. A win rate of -2 bb/100 would be 4/1.3 standard errors below the average win rate of +2 bb/100. From Excel =NORMSDIST(-4/1.3) =~ 1 in 956.

[/ QUOTE ]

Actually, I did a much rougher calc, which is why I went conservative and added "less than"
/images/graemlins/blush.gif

[/ QUOTE ]

With the nice numbers that you chose, you should be able to do this quickly in your head and see that the answer is very close to 1 in 1000. That's how I noticed the problem.

In case anyone's interested in this thought process, it went like this. The SD is given in units of bb/100 hands, so to get the standard error, we need to divide the SD of 13 by the square root of the number of 100 hand units in 10K hands. 10K/100 is 100, and the square root of 100 is 10, so we divide the SD of 13 by 10 to get a standard error of 1.3. Now -2 bb/100 is 4 bb/100 below the average of 2 bb/100, and 4 bb would correspond to 4/1.3 standard deviations below the mean, which is a little more than 3 standard deviations. I remember that 3 standard deviations one sided is about 99.9%, so the probability of being more than this many standard deviations below average is about 0.1% or 1 in 1000.

[winky51]

Nice reword, thats exactly what I meant, the math is almost over my head, but I estimated the answer to be about that.

Interesting.

Now I am curious what are the chances he makes 2BB/100 or better in hands in 10,000 hands? Around 60%?

[elitegimp]

[ QUOTE ]
Now I am curious what are the chances he makes 2BB/100 or better in hands in 10,000 hands? Around 60%?

[/ QUOTE ]

It's exactly 50% (because we assume his "true" win-rate is 2 BB/100, so the distribution of his sample win-rate after 10K hands is centered at 2 BB/100)

[mdplayah]

i always remembered 3 SD covering 99.7%. This would make one sided 0.15% and the answer likely in the middle of the two replies (~1/700). Anyway cool refresher with the two posters. I had forgotten a lot of the logic behind that. also, let me know if i'm off on this. that number just sticks out in my head...

Q: If you are playing correctly against lots of fish what is the probability you can play 10K hands and lose 2BB/100?

A: you are the fish.

[BruceZ]

[ QUOTE ]
i always remembered 3 SD covering 99.7%. This would make one sided 0.15% and the answer likely in the middle of the two replies (~1/700). Anyway cool refresher with the two posters. I had forgotten a lot of the logic behind that. also, let me know if i'm off on this. that number just sticks out in my head...

[/ QUOTE ]

It isn't exactly 3 SD, it is 4/1.3 SD. I told you that the exact answer was computed from NORMSDIST(-4/1.3) = 1 in 956.

I only used 3 SD as an approximation for doing it in your head, and I rounded the 1-sided 99.87% to 99.9% or 1 in 1000, since 4/1.3 is a little larger than 3.