Required bankroll for multiple hands

[James989]

Please help me on this :-

Assumed the EV and variance when bet 1 hand in a round is e and v respectively. So the required Bankroll for one hand, BR1= vbk/e, where k is Kelly Ratio and b is bet size. What is required Bankroll(BR5) if I bet 5 hands, each hand same bet size, b in a round ? Should less than 5vbk/e ? Should related to covariance ?

Please explain with an example if possible.

James

[ZenMaster_Flash]

It is doubtful that you know a casino that permits 5 spots being bet.

I assume that this is strictly an academic question.

[James989]

You can easily find a Asia casino that allows you to bet up to 5 or 6 spots !

[DSchles]

Please help me on this :-

Assumed the EV and variance when bet 1 hand in a round is e and v respectively. So the required Bankroll for one hand, BR1= vbk/e, where k is Kelly Ratio and b is bet size. What is required Bankroll(BR5) if I bet 5 hands, each hand same bet size, b in a round ? Should less than 5vbk/e ? Should related to covariance

Please explain with an example if possible.

James

You need to find the total variance for the five simultaneous bets. Variance varies according to rules, but a common value for one hand is 1.32 squared units. And the covariance is generally approximated to be 0.50 between two hands. The general formula would then be: V(n) = 1.32n + .50n(n-1). For five hands, total variance = 1.32(5) + .50(5)(5-1) = 6.60 + 10 = 16.6, assuming one unit bet on each hand.

From here, use your formula above, plugging in bet size, Kelly fraction, and e.v. to get required bankroll.

Don

[DSchles]

It is doubtful that you know a casino that permits 5 spots being bet.

What earthly difference would that possibly make?

I assume that this is strictly an academic question.

Why would you care? Why not just answer the question that was asked? He didn't ask for a comment; he asked for an answer. That's what wrong with 95% of the answers on this forum. Everyone has a compulsion to pontificate, usually with drivel. The mantra for this board ought to be: Answer the goddamn question!

Don

[James989]

You need to find the total variance for the five simultaneous bets. Variance varies according to rules, but a common value for one hand is 1.32 squared units. And the covariance is generally approximated to be 0.50 between two hands. The general formula would then be: V(n) = 1.32n + .50n(n-1). For five hands, total variance = 1.32(5) + .50(5)(5-1) = 6.60 + 10 = 16.6, assuming one unit bet on each hand.

From here, use your formula above, plugging in bet size, Kelly fraction, and e.v. to get required bankroll.

Don

Don, thanks for your reply.

1) What is the covariance value for games other blackjack ?
2) With the combine variance of 16.6( higher than 5 x 1.32), the total required bankroll for five simultaneous bets is MUCH HIGHER than 5 * single bet bankroll ! I thought the total required bankroll for five simultaneous bets should be lower than 5 x the bankroll of single bet(because you can share the bankroll between these 5 hands) ? Am I missing something ?

James

[Gronbog]

I thought the total required bankroll for five simultaneous bets should be lower than 5 x the bankroll of single bet(because you can share the bankroll between these 5 hands)

You can only share bankroll between independent hands, such as when a team of 5 players are all on different tables. This is because the covariance is then zero.

[Gronbog]

With the combine variance of 16.6( higher than 5 x 1.32), the total required bankroll for five simultaneous bets is MUCH HIGHER than 5 * single bet bankroll ! I thought the total required bankroll for five simultaneous bets should be lower than 5 x the bankroll of single bet

Don't forget that variance is measure in squared units. You can not compare them in a linear way. If you take the square roots of 16.6 and 1.32 to get the respective standard deviations, you will see that, for 5 hands, the standard deviation is less than 5 times the standard deviation of a single hand.

[James989]

So if 5 players bet at different table(covariance = 0), then the bankroll of 5 players should be exactly same as 5 * a single player bankroll ?

[Gronbog]

No, in that case the 5 players can all share the single player bankroll with no additional risk. The variance stays the same and there is no covariance.

[James989]

No, in that case the 5 players can all share the single player bankroll with no additional risk. The variance stays the same and there is no covariance.

So if single player bankroll = 5000, then 5 players bet at 5 different tables(one player at each table) can just share the same bankroll 5000 with no additional risk ?

[James989]

Don't forget that variance is measure in squared units. You can not compare them in a linear way. If you take the square roots of 16.6 and 1.32 to get the respective standard deviations, you will see that, for 5 hands, the standard deviation is less than 5 times the standard deviation of a single hand.

Although the 5 hands standard deviation is less than 5 times the standard deviation of a single hand, but the bankroll is measure by VARIANCE, and therefore the 5 hands bankroll(bet 5 hands simultaneously at same table) still much higher than 5 x single hand bankroll, is that correct ? I think I must made a mistake somewhere.

[DSchles]

So if 5 players bet at different table(covariance = 0), then the bankroll of 5 players should be exactly same as 5 * a single player bankroll ?

Yes. Also, if you play five simultaneous hands at the same table, it goes without saying that, to bet optimally, each of the bets will be much smaller than the one-hand wager. For two simultaneous hands, we know that wager to be 73%. For five hands, it's 43%.

Don