hi everyone, following on from the probability modelling work i did last week i wanted to do a quick analysis of how my actual results to date compared to a theoretical individual, let's call him Jack, that was flat-betting and playing perfect basic strategy for 17,000 hands.
For my game the house advantage is 0.55%
I have worked out the following -
p(win) for 1 hand = 0.49725 - P
p(loss) for 1 hand = 0.50275 - Q
# of hands played - 17,000 - N
Pushes are ignored since they are incorporated in the house advantage.
P(loss) - p(win) = 0.50275 - 0.49725 = 0.0055 = house advantage
Therefore, if flat betting and playing perfect BS (without index plays) Jack should expect to win 0.49725 x 17,000 = 8,453.25 hands
The standard deviation is calculated as follows (assuming results are normally distributed):
SD = (N x P x Q) ^ 0.5, thus
= (17000 x 0.49725 x 0.50275) ^ 0.5 = 65.1 hands
Thus confidence intervals for the number of winning hands would be as follows:
68% (8,388 - 8,518)
95% (8,322 - 8,583)
99% (8,258 - 8,649)
Interpreting the above, does this mean if Jack played numerous rounds of 17,000 hands, that he would be winning more than losing in approx 16% of the rounds (the 68 confidence interval is just outside the break-even point of 7,500 hands)?
Does it also mean that it would be highly unlikely (i.e. 0.5% probability) for Jack to be ahead by more than 149 units (8,649-8,500)
I would be keen for someone to comment on whether the above approach is correct or not.
For my game the house advantage is 0.55%
I have worked out the following -
p(win) for 1 hand = 0.49725 - P
p(loss) for 1 hand = 0.50275 - Q
# of hands played - 17,000 - N
Pushes are ignored since they are incorporated in the house advantage.
P(loss) - p(win) = 0.50275 - 0.49725 = 0.0055 = house advantage
Therefore, if flat betting and playing perfect BS (without index plays) Jack should expect to win 0.49725 x 17,000 = 8,453.25 hands
The standard deviation is calculated as follows (assuming results are normally distributed):
SD = (N x P x Q) ^ 0.5, thus
= (17000 x 0.49725 x 0.50275) ^ 0.5 = 65.1 hands
Thus confidence intervals for the number of winning hands would be as follows:
68% (8,388 - 8,518)
95% (8,322 - 8,583)
99% (8,258 - 8,649)
Interpreting the above, does this mean if Jack played numerous rounds of 17,000 hands, that he would be winning more than losing in approx 16% of the rounds (the 68 confidence interval is just outside the break-even point of 7,500 hands)?
Does it also mean that it would be highly unlikely (i.e. 0.5% probability) for Jack to be ahead by more than 149 units (8,649-8,500)
I would be keen for someone to comment on whether the above approach is correct or not.
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