Optimum Bet Size

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November 5th, 2014 at 12:13:52 AM permalink

Quote: ssho88
A game with three sidebets :-

Sidebet 1 : EV=6.5%, Var=5.5, +ve edge when TC>=5
Sidebet 2 : EV=8.3%, Var=7.8, +ve edge when TC>=3
Sidebet 3 : EV=7.2%, Var=6.1, +ve edge when TC>=7

Player can ONLY win one of the sidebet in a round, regardless of how many sidebets you are going to bet. For example, if you bet for all three sidebets, you may loss all OR only win one of it and loss the other two sidebets.

Bankroll, BR=$300,000, Kelly ratio, a = 0.5. The bet size = EV/Var * BR * a

a) If TC=4, what is the optimum bet size for sidebet 2 ?
b) If TC=6, what is the optimum bet size for sidebet 1 and sidebet2 ?
b) If TC=8, what is the optimum bet size for sidebet 1, sidebet2 and sidebet3 ?

Should we calculate the bet size for each sidebet base on original BR=300k or smaller bankroll, say 300k/3=100k ?

u use the original BR.

You have to know how much EV changes by count, first of all. The you gotta adjust EV down to account for the fact that a win may actually lose, so you have to know how often that is going to happen. You haven't provided enough information for anyone to help.

ssho88

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November 5th, 2014 at 12:30:21 AM permalink

Thanks for your reply.

Additional info :-

Sidebet 1 : averg.EV=6.5%, Var=5.5, +ve edge when TC>=5, bet freq=8.2 hands/shoe
Sidebet 2 : averg.EV=8.3%, Var=7.8, +ve edge when TC>=3, bet freq=11.3 hands/shoe
Sidebet 3 : averg.EV=7.2%, Var=6.1, +ve edge when TC>=7, bet freq=7.2 hands/shoe

I know that edge is higher when the count is higher but I think in this case we just use the average edge as stated above. The question here is what BR should we use to compute the bet size.

Did you mean to say player still use BR=300k to compute the bet size(for each sidebet) although the TC justify to bet all three sidebets at the same(in same round) ? Should we use smaller BR(say 300k/3=100k) since the BR is share for three sidebets ?

Sonuvabish

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November 5th, 2014 at 3:12:15 AM permalink

Quote: ssho88
Thanks for your reply.

Additional info :-

Sidebet 1 : averg.EV=6.5%, Var=5.5, +ve edge when TC>=5, bet freq=8.2 hands/shoe
Sidebet 2 : averg.EV=8.3%, Var=7.8, +ve edge when TC>=3, bet freq=11.3 hands/shoe
Sidebet 3 : averg.EV=7.2%, Var=6.1, +ve edge when TC>=7, bet freq=7.2 hands/shoe

I know that edge is higher when the count is higher but I think in this case we just use the average edge as stated above. The question here is what BR should we use to compute the bet size.

Did you mean to say player still use BR=300k to compute the bet size(for each sidebet) although the TC justify to bet all three sidebets at the same(in same round) ? Should we use smaller BR(say 300k/3=100k) since the BR is share for three sidebets ?

U use the same BR, $300,000. But you cannot use the stated EV because it doesn't reflect your actual edge. For example, say you bet on #1 and #2. Both of them have a 90% chance of losing, and offer only 1 payout. There is a 1% chance that you will lose a bet that actually won. You win 10 of 100. One of these, will actually be a loss. So now it's 9 out of a 100. Divide by two, since the other still wins. That's 5% off the total edge; I believe that's how you would work it. That's going to be about 1/3 of your edge taken away. My best guess is that this scenario follows the same pattern as does betting multiple hands of blackjack. When playing 3 hands of blackjack, you should bet 2/3 your normal on each hand. So, to make it easier, 200K would be the proper bankroll. However, I believe you'd better off going with 150-175, because that is risk-averse when betting 3. When betting 2, should be 200-225. When betting 1, 300. I am not a math guy, this is my best educated guess.

ChesterDog

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November 5th, 2014 at 7:02:03 AM permalink

Quote: ssho88
...Should we calculate the bet size for each sidebet base on original BR=300k or smaller bankroll, say 300k/3=100k ?

Winning side bets' being mutually exclusive allows you to bet more than you would than if you consider the bets separately because of the effect of the negative covariance between the bets on your overall variance. Here's an example using your side bets 1 and 2 with numbers changed slightly (I had to start with the odds instead of variance.):
EV1 = 0.065
odds1 = 5
w1 = prob.(win) = (1+EV1)/(1+odds1) = 0.178
var1 = w1*(odds1 - EV1)^2 + (1-w1)*(-1 - EV1)^2 = 5.26

EV2 = 0.083
odds2 = 7
w2 = 0.135
var2 = 7.49

Covariance: cov1,2 = w1*(odds1 - EV1)*(-1 - EV2) + w2*(-1 - EV1)*(odds2 - EV2) + (1 - w1 - w2)*(-1 - EV1)*(-1 - EV2) = -1.15

Let 1 unit be the total money bet on the side bets. Of that 1 unit, suppose you bet, say, 0.533 on bet 1 and 0.467 on bet 2. Your total EV would be 0.533 * EV1 + 0.467 * EV2 = 0.0734. And your overall variance would be:
0.533^2 * var1 + 2*0.533*0.467*cov1,2 + 0.467^2 * var2 = 2.55

Then the total money bet on the side bets would be 0.5 * $300,000 * 0.0734 / 2.55 = $4314, with $2299 on bet 1 and $2014 on bet 2. Compare these with your bets if you had considered them separately; you would have bet 0.5*$300,000*0.065/5.26 = $1855 on bet 1 and $1662 on bet 2.

I used 0.533 at the fraction of money on side bet 1 but you can vary that fraction to see its effect on your profit.

I didn't include the main bet in the analysis but you would use a similar approach.

ssho88

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November 5th, 2014 at 8:07:28 AM permalink

Hi ChesterDog, thanks for your explanation.

Player can ONLY win one of the sidebet in a round, regardless of how many sidebets you are going to bet. For example, if you bet for all three sidebets, you may loss all OR only win one of it and loss the other two sidebets.

Have you take above into consideration?

ChesterDog

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November 5th, 2014 at 8:45:04 AM permalink

Quote: ssho88
Thanks for your explanation.

Player can ONLY win one of the sidebet in a round, regardless of how many sidebets you are going to bet. For example, if you bet for all three sidebets, you may loss all OR only win one of it and loss the other two sidebets.

Have you take above into consideration?

Yes. In my example, there are only two bets, and you can either win the first, win the second, or lose both. See the formula for covariance below:

cov1,2 = w1*(odds1 - EV1)*(-1 - EV2) + w2*(-1 - EV1)*(odds2 - EV2) + (1 - w1 - w2)*(-1 - EV1)*(-1 - EV2)

The first term is for winning odds1 on bet 1 and losing 1 on bet 2. The second term is for losing 1 on bet 1 and winning odds2 on bet two. And the third term is for losing 1 on bet 1 and losing 1 on bet 2.

ssho88

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November 5th, 2014 at 9:22:15 AM permalink

The formula will be very complex when there are 3 or more side bets !
Any simple method to calculate covariance ?

ChesterDog

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November 5th, 2014 at 11:24:42 AM permalink

Quote: ssho88
The formula will be very complex when there are 3 or more side bets !
Any simple method to calculate covariance ?

Good question! You're right. This formula is complicated:
cov1,2 = w1*(odds1 - EV1)*(-1 - EV2) + w2*(-1 - EV1)*(odds2 - EV2) + (1 - w1 - w2)*(-1 - EV1)*(-1 - EV2)

But I just now found that if we substitute into the above w1 = (1+EV1)/(odds1 +1 ) and w2 = (1+EV2)/(odds2 +1) and simplify we get:
cov1,2 = - (1 + EV1) * (1 + EV2)

So, the covariances between your mutually exclusive side bets 1 and 2, 1 and 3, and 2 and 3 are -1.15, -1.14, and -1.16, respectively.

The odds cancel out, so we don't need them. We can use your original values of the variances and EV's together with these covariances to find the optimal side bets if we disregard the main bet.

To find the optimal main bet and side bets, we need the covariances between the main bet and each side bet, as well as the EV and variance of the main bet. A computer simulation would give those.

ssho88

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November 5th, 2014 at 2:17:34 PM permalink

Player can just bet the side bets, NOT necessary to bet the main bet.

The above example only for two sidebets. If player bet for three side bets, bet 1, bet 2 and bet 3 at the same time :-

1) How to find covariance ? Please check the formula for cov1,2,3.

cov1,2,3=w1*(odds1 - EV1)*(-1 - EV2)*(-1 - EV3) + w2*(-1 - EV1)*(-1 - EV3)*(odds2 - EV2) + w3*(-1 - EV1)*(-1 - EV2)*(odds3 - EV3)+ (1 - w1 - w2 - w3)*(-1 - EV1)*(-1 - EV2)*(-1 - EV3)

2) How to find the overall variance ? What is the formula ?

3 What fraction should we use ? base on ratio EV1/Var1 : EV2/Var2 : EV3 /Var3 ?

Your reply would be much appreciated.

ChesterDog

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November 5th, 2014 at 3:04:18 PM permalink

Quote: ssho88
Player can just bet the side bets, NOT necessary to bet the main bet.

The above example only for two sidebets. If player bet for three side bets, bet 1, bet 2 and bet 3 at the same time :-

1) how to find covariance ? What is the formula ?
2) how to find the overall variance ? What is the formula ?
3 what fraction should we use ? base on ratio EV1/Var1 : EV2/Var2 : EV3 /Var3 ?

Your reply would be much appreciated.

That's good that the players can bet only the side bets if they wish! That makes the calculation much easier.

We use the same formula for each of the covariances. cov1,2 = -(1 + EV1)*(1 + EV2); cov1,3 = -(1 + EV1)*(1 + EV3); and cov2,3 = -(1 + EV2)*(1 + EV3)

The overall variance is based on a total of the sidebets of 1 unit. Let a be the fraction of the total bet on sidebet 1. Let b be sidebet 2, and let c be sidebet 3. (These fractions will be varied by you in a spreadsheet or even by trial-and-error.) Then the overall variance is:
Var = a^2 * var1 + b^2 * var2 + c^2 * var3 + 2*a*b*cov1,2 + 2*a*c*cov1,3 + 2*b*c*cov2,3

And the EV for a total of 1 unit of sidebets is:
EV = a*EV1 + b*EV2 + c*EV3

Then your total bet on the sidebets would be:
Bet = 0.5 *300000 * EV / Var

The profit, which must be maximized, would be:
Profit = Bet * EV = 0.5 * 300000 * ( a*EV1 + b*EV2 + c*EV3 )^2 / ( a^2 * var1 + b^2 * var2 + c^2 * var3 + 2*a*b*cov1,2 + 2*a*c*cov1,3 + 2*b*c*cov2,3 )

In a spreadsheet you will vary the values of a and b to maximize your profit. I used Excel and put the values of a starting at 0 and incremented by 0.1 down the left side. I put the values of b at the top likewise starting at 0 and incremented by 0.1. The value of c is simply 1 - a - b.

Using this procedure I arrived at these answers for your 3 sidebet question:
a = 0.3547
b = 0.3017
c = 0.3436

EV = 7.284%; Var = 1.356

Bet = $8055

Sidebet1 = a * Bet = $2857
Sidebet2 = b * Bet = $2430
Sidebet3 = c * Bet = $2768

Profit = $587

ssho88

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November 5th, 2014 at 5:27:38 PM permalink

Wow ! Thanks for your answer !

I have a question :-

cov1,2= TERM 1 + TERM2 + (1 - w1 - w2)*(-1 - EV1)*(-1 - EV2) = -1.15

For TERM3, should we use (1-w1)*(1-w2)*(-1 - EV1)*(-1 - EV2) ?

ChesterDog

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November 5th, 2014 at 6:18:23 PM permalink

Quote: ssho88
Wow ! Thanks for your answer !

I have a question :-

cov1,2= TERM 1 + TERM2 + (1 - w1 - w2)*(-1 - EV1)*(-1 - EV2) = -1.15

For TERM3, should we use (1-w1)*(1-w2)*(-1 - EV1)*(-1 - EV2) ?

That's a good question, too. I had thought about that. Term 1 refers to sidebet 1 winning (with sidebet2 losing), term 2 refers to sidebet 2 winning (with sidebet1 losing), and that last term refers to both sidebet 1 and 2 losing. The probability that both sidebets lose is 1 - w1 - w2. As a check, we know the sum of the probabilities of those three events has to be 1, which mine is. That's w1 + w2 + (1 - w1 - w2) = 1. But try this: w1 + w2 + (1-w1)*(1-w2) = w1 + w2 + 1 - w2 -w1 +w1*w2 = 1 + w1*w2 which is greater than 1.

Sonuvabish

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November 5th, 2014 at 6:44:20 PM permalink

You seem to be neglecting the fact that if both side bets win, 1 side bet actually loses. This negatively impacts EV. Even tho covariance allows you to bet more than the half or 33% of your bet on each spot, the lowered EV is a counterweight. Covariance allows you to bet more in total. Your conclusion says that you can bet more on each bet, which would result in a much larger total bet. This is patently wrong. Your EV is lower per bet due to the fact a win may result in a loss; betting 3 bets is detrimental to one particular bet, but beneficial as a whole. In other words, it is impossible you would bet more on one bet than if you considered it separately.

ssho88

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November 5th, 2014 at 6:54:18 PM permalink

I think Term 1(and term 2) have problem.

For Term 1, the probability = w1* (1-w2) (win on bet 1, lose on bet 2).

OR

For Term 1, the probability = w1* (1) (win on bet 1, SURE lose on bet 2).

The tricky part is win you win bet 1 , you will SURE lose bet 2 ! dependent or independent case ?

For this game, ONLY one bet can be won in a round and SURE you will lose all other(if you bet more than one sidebet) bets.

Which one is correct ?

Sonuvabish

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November 5th, 2014 at 6:56:32 PM permalink

Quote: ssho88
I think Term 1(and term 2) have problem.

For Term 1, the probability = w1* (1-w2) (win on bet 1, lose on bet 2).

OR

For Term 1, the probability = w1* (1) (win on bet 1, SURE lose on bet 2).

The tricky part is win you win bet 1 , you will SURE lose bet 2 ! dependent or independent case ?

Which one is correct ?

I have no idea. His math is above me. I'm absolutely sure he made a mistake somewhere. His conclusion makes no sense. And I'm reasonably sure he could figure out the correct answer, he is a bright individual. My guesstimate will not be far off for you to ad hoc it.

ssho88

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November 5th, 2014 at 7:02:18 PM permalink

The calculations is not done by you ?

For this game, ONLY one bet can be won in a round and SURE you will lose all other(if you bet more than one sidebet) bets.

Just an example(NOT my actual game), In blackjack, dealer's final total point, 17,18,19,20,21 or BJ.

Quote: Sonuvabish
Quote: ssho88
I think Term 1(and term 2) have problem.

For Term 1, the probability = w1* (1-w2) (win on bet 1, lose on bet 2).

OR

For Term 1, the probability = w1* (1) (win on bet 1, SURE lose on bet 2).

The tricky part is win you win bet 1 , you will SURE lose bet 2 ! dependent or independent case ?

Which one is correct ?

I have no idea. His math is above me. I'm absolutely sure he made a mistake somewhere. His conclusion makes no sense. And I'm reasonably sure he could figure out the correct answer, he is a bright individual. My guesstimate will not be far off for you to ad hoc it.

Sonuvabish

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Posts: 1342

Joined: Feb 5, 2014

November 5th, 2014 at 7:19:33 PM permalink

Quote: ssho88
The calculations is not done by you ?

For this game, ONLY one bet can be won in a round and SURE you will lose all other(if you bet more than one sidebet) bets.

Just an example(NOT my actual game), In blackjack, dealer's final total point, 17,18,19,20,21 or BJ.

Quote: Sonuvabish
Quote: ssho88
I think Term 1(and term 2) have problem.

For Term 1, the probability = w1* (1-w2) (win on bet 1, lose on bet 2).

OR

For Term 1, the probability = w1* (1) (win on bet 1, SURE lose on bet 2).

The tricky part is win you win bet 1 , you will SURE lose bet 2 ! dependent or independent case ?

Which one is correct ?

I have no idea. His math is above me. I'm absolutely sure he made a mistake somewhere. His conclusion makes no sense. And I'm reasonably sure he could figure out the correct answer, he is a bright individual. My guesstimate will not be far off for you to ad hoc it.

I didn't realize you were expressing what I said. I really am not sure what it says. He's assuming they are all independent. He's not figuring in the dependent cases when the bet 2 is a sure loss. If there is never a sure loss like that, and EV just shoots up when you reach the threshold of playing another bet, I can see how it would be possible that you'd want to bet more on one bet than you would if you considered it separately. So, he has to be ignoring the fact that EV lowers on each bet when you play more than one. Until you have an answer, you should use a 300K BR to bet 1...and maybe a lower BR to bet multiple. And bet based on individual advantage/variance. You may not need a lower BR, because of covariance. Do not bet MORE.

ssho88

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Posts: 668

Joined: Oct 16, 2011

November 5th, 2014 at 7:31:36 PM permalink

Quote: Sonuvabish
Quote: ssho88
The calculations is not done by you ?

For this game, ONLY one bet can be won in a round and SURE you will lose all other(if you bet more than one sidebet) bets.

Just an example(NOT my actual game), In blackjack, dealer's final total point, 17,18,19,20,21 or BJ.

Quote: Sonuvabish
Quote: ssho88
I think Term 1(and term 2) have problem.

For Term 1, the probability = w1* (1-w2) (win on bet 1, lose on bet 2).

OR

For Term 1, the probability = w1* (1) (win on bet 1, SURE lose on bet 2).

The tricky part is win you win bet 1 , you will SURE lose bet 2 ! dependent or independent case ?

Which one is correct ?

I have no idea. His math is above me. I'm absolutely sure he made a mistake somewhere. His conclusion makes no sense. And I'm reasonably sure he could figure out the correct answer, he is a bright individual. My guesstimate will not be far off for you to ad hoc it.

I didn't realize you were expressing what I said. I really am not sure what it says. He's assuming they are all independent. He's not figuring in the dependent cases when the bet 2 is a sure loss. If there is never a sure loss like that, and EV just shoots up when you reach the threshold of playing another bet, I can see how it would be possible that you'd want to bet more on one bet than you would if you considered it separately. So, he has to be ignoring the fact that EV lowers on each bet when you play more than one. Until you have an answer, you should use a 300K BR to bet 1...and maybe a lower BR to bet multiple. And bet based on individual advantage/variance. You may not need a lower BR, because of covariance. Do not bet MORE.

Thanks for your advice.

Will he re-do the maths again base on dependent cases ?

Sonuvabish