Math Question BJ

Don't know if it's variance, volatility, median average, whatever, Say a player plays 3 hours, 70 hands per hour, does it 1,000
times? If I want to know his range for 500 of those sessions, how would I figure if he had 1,000 sessions of the following games.
BJ strip rules $10 , Switch $5, Spanish 21 $10. flat betting. Is it just the house edge, bell curves, or one of them formulas that looks like alphabet soup inbred with symbols. Just looking for a range, like -$12 to -$13

If I understand your question correctly...

A session is 210 hands, and you want to know what is the range of outcomes, centered about the mean, that will contain 50% (500 out of 1000) of the sessions. If this is the case, then I think it really is just house edge and bell curves.

Given a normal distribution, 50% of the distribution lies between -0.675 sigma and +0.675 sigma of the mean.

For $10 BJ, I'll assume a 0.5% house edge = $0.05 per hand, and a standard deviation of $11.4 per hand

For 210 hands, the expected return is 210 x -0.05 -$10.50, and the standard deviation is sqrt(210) x 11.4 = $165.20.

The range that will contain 50% is -10.5 - 0.675 x 165.20 = -$122 on the low end and -10.5 + 0.675 x 165.20 = +$101 on the high end.

I'm not going to bother looking up the means and standard deviations for the other games, but the same methodology would apply.

Quote: buzzpaff

or one of them formulas that looks like alphabet soup inbred with symbols.

LOL.

Quote: PapaChubby

If I understand your question correctly...

A session is 210 hands, and you want to know what is the range of outcomes, centered about the mean, that will contain 50% (500 out of 1000) of the sessions. If this is the case, then I think it really is just house edge and bell curves.

Given a normal distribution, 50% of the distribution lies between -0.675 sigma and +0.675 sigma of the mean.

For $10 BJ, I'll assume a 0.5% house edge = $0.05 per hand, and a standard deviation of $11.4 per hand

For 210 hands, the expected return is 210 x -0.05 -$10.50, and the standard deviation is sqrt(210) x 11.4 = $165.20.

The range that will contain 50% is -10.5 - 0.675 x 165.20 = -$122 on the low end and -10.5 + 0.675 x 165.20 = +$101 on the high end.

I'm not going to bother looking up the means and standard deviations for the other games, but the same methodology would apply.

Damn. I actually think I understand that. Well, some of it Thanks. Seems reasonable to think that half the 3 hours sessions will
end up about $100 each way of even. I am working on a BJ game with liberal rules and jackpots.Naturally BJ has to pay
even money. My concern was not with the $10,000 jackpot but the effect of the $100 jackpot. The trigger is set at 1 in
every 221 hands. That's why I chose 3 hours at 70 hph. Also read that somewhere that's the average time a visitor
spends gambling each day.
Since once the trigger is set, a $100 is only paid out 1 in 25 times, it should not have a dramatic effect on distribution.
Free advice is worth the price, but this time is an exception to the rule.
THANKS !!!!!!!!!!!!!!!!!!!!!!!!!!!