Probability of session outcome
Is it possible to calculate the probability of a session outcome knowing the EV and variance of a game?
For example, if I played a game with a 10% edge, betting $10 a hand with a variance of 12, what is my probability of playing 200 hands and losing $300 (random numbers)?
Thanks
Quote: gunbjHello,
Is it possible to calculate the probability of a session outcome knowing the EV and variance of a game?
For example, if I played a game with a 10% edge, betting $10 a hand with a variance of 12, what is my probability of playing 200 hands and losing $300 (random numbers)?
Thanks
Calculate your z-score then look it up on a table. This question if for a one sided zscore test.
Mean loss = 10% * $10 * 200 = $200
Standard deviation = sqrt(12*200) = $48.99
Z-score = ($300 - $200)/$48.99 = 2.04
Look that up here: https://faculty.washington.edu/heagerty/Books/Biostatistics/TABLES/Normal/index.html
Gets you a probability of 2.02% of being down at least $300.
Quote: unJonCalculate your z-score then look it up on a table. This question if for a one sided zscore test.
Mean loss = 10% * $10 * 200 = $200
Standard deviation = sqrt(12*200) = $48.99
Z-score = ($300 - $200)/$48.99 = 2.04
Look that up here: https://faculty.washington.edu/heagerty/Books/Biostatistics/TABLES/Normal/index.html
Gets you a probability of 2.02% of being down at least $300.
When Gunbj writes that the variance is 12, he might mean that the variance is 12 based on a unit bet.
If so, the standard deviation would be 10 * sqrt(12*200) = $489.90.
Then Z-score = ($300 - $200) / $489.90 = 0.204.
Then the probability of being down at least $300 would be 41.9%.
Quote: ChesterDogWhen Gunbj writes that the variance is 12, he might mean that the variance is 12 based on a unit bet.
If so, the standard deviation would be 10 * sqrt(12*200) = $489.90.
Then Z-score = ($300 - $200) / $489.90 = 0.204.
Then the probability of being down at least $300 would be 41.9%.
That’s actually a much more likely interpretation.
Quote: ChesterDogWhen Gunbj writes that the variance is 12, he might mean that the variance is 12 based on a unit bet.
If so, the standard deviation would be 10 * sqrt(12*200) = $489.90.
Then Z-score = ($300 - $200) / $489.90 = 0.204.
Then the probability of being down at least $300 would be 41.9%.
Thank you both! That's what I meant.
Sorry for the silly question, but how did you get 41.9% from 0.204 as the exact number is not on the table above?
Quote: gunbjThank you both! That's what I meant.
Sorry for the silly question, but how did you get 41.9% from 0.204 as the exact number is not on the table above?
There’s a formula in excel you can use. Or an online calculator like this one (second one down): https://www.calculator.net/z-score-calculator.html?c2z=.204&c2p=&c2pg=&c2p0=&c2pin=&c2pout=&calctype=converter&x=43&y=28#converter
What about the probability of being UP $300 instead of down? How does the calculation change?
Quote: gunbjOk thanks!
What about the probability of being UP $300 instead of down? How does the calculation change?
Instead of being (300-200) away from the mean, you are (-300-200) away from the mean. I’m other words 500 away from mean instead of 100. Will see if you can calculate the new z-score from there.
