ssho88
ssho88
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September 3rd, 2014 at 11:01:14 PM permalink
A card game with five different sidebets, namely A, B, C, D and E. If you bet on ALL five sidebets, there are only two outcomes :-

a) Loss all
b) Win one of it and loss other four sidebets

I have developed five different counting systems, namely S1, S2, S3, S4 and S5, for each sidebet.

However, it's not practical for a player to count five different systems at the same time. Therefore, I have developed a single counting system to cover five sidebets, namely S6. Of course, the S6 is not as good as S1,S2..S5, only able to capture average 73% of the each ev of S1, S2. . .S5.

Base on system S6, trigger running count for sidebets A,B,C,D and E are RC >= 21, RC >= 22, RC >= 24, RC >= 26 and RC >= 29 respectively.

Question:-

1) Should I bet for all five sidebets if RC=32 ?
2) Should I just bet on sidebet A, B and C if RC=25 ?
3) Each sidebet is independent from each other ?

Please be reminded that if player bet on ALL five sidebets, he may only win one of the sidebet and loss other four sidebets, OR loss all five sidebets.
Buzzard
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September 3rd, 2014 at 11:04:16 PM permalink
John Patrick says you should push and pull, then lift and lay!
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet
ssho88
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September 3rd, 2014 at 11:07:17 PM permalink
What do you mean by that ?
andysif
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September 4th, 2014 at 2:51:18 AM permalink
Quote: ssho88

A card game with five different sidebets, namely A, B, C, D and E. If you bet on ALL five sidebets, there are only two outcomes :-

a) Loss all
b) Win one of it and loss other four sidebets

I have developed five different counting systems, namely S1, S2, S3, S4 and S5, for each sidebet.

However, it's not practical for a player to count five different systems at the same time. Therefore, I have developed a single counting system to cover five sidebets, namely S6. Of course, the S6 is not as good as S1,S2..S5, only able to capture average 73% of the each ev of S1, S2. . .S5.

Base on system S6, trigger running count for sidebets A,B,C,D and E are RC >= 21, RC >= 22, RC >= 24, RC >= 26 and RC >= 29 respectively.

Question:-

1) Should I bet for all five sidebets if RC=32 ?
2) Should I just bet on sidebet A, B and C if RC=25 ?
3) Each sidebet is independent from each other ?

Please be reminded that if player bet on ALL five sidebets, he may only win one of the sidebet and loss other four sidebets, OR loss all five sidebets.



how could they be "Each sidebet is independent from each other ?" and "may only win one of the sidebet and loss other four sidebets, OR loss all five sidebets"
ssho88
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September 4th, 2014 at 4:19:51 AM permalink
The main question here is : Should I bet for all five sidebets if RC=32 ?
Dieter
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September 4th, 2014 at 6:36:18 AM permalink
Quote: ssho88

The main question here is : Should I bet for all five sidebets if RC=32 ?



I don't think you've provided enough information for a specifically relevant answer.

You shouldn't bet more than you can win, ever. You shouldn't bet more than you're likely to win, generally.
May the cards fall in your favor.
ssho88
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September 4th, 2014 at 7:04:00 AM permalink
Quote: Dieter

I don't think you've provided enough information for a specifically relevant answer.

You shouldn't bet more than you can win, ever. You shouldn't bet more than you're likely to win, generally.




Additional info :-

Sidebet A : 1 pay 40
Sidebet B : 1 pay 50
Sidebet C : 1 pay 25
Sidebet D : 1 pay 100
Sidebet E : 1 pay 65
ajemeister
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September 4th, 2014 at 7:20:04 AM permalink
Quote: Dieter


You shouldn't bet more than you can win, ever.



isn't this what laying odds in craps is?
ssho88
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September 4th, 2014 at 7:22:03 AM permalink
IT DOES NOT MATTER WHAT GAME IT IS !
Dieter
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September 4th, 2014 at 7:35:32 AM permalink
Quote: ajemeister

isn't this what laying odds in craps is?



Sorry, I specifically meant placing more contrary bets than one can win.

An ex-girlfriend who didn't understand roulette liked to place 12 or 13 non-overlapping corner bets. She thought it was fun. She didn't understand why I got annoyed.

Laying odds at craps is different.
May the cards fall in your favor.
Dieter
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September 4th, 2014 at 7:42:42 AM permalink
Quote: ssho88

Additional info :-



Paytables help, but the necessary information is the win/loss probability for a given count combined with the paytables.

It sounds like your sidebets have only one winning condition each, and one payout each.

You can say that it doesn't matter what the game is, but it's much harder for anyone to give meaningful advice without a more thorough knowledge of the situation.

Push and pull, then lift and lay is probably reasonable advice, given what you've explained.
May the cards fall in your favor.
beachbumbabs
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September 4th, 2014 at 8:01:55 AM permalink
Quote: ssho88

Additional info :-

Sidebet A : 1 pay 40
Sidebet B : 1 pay 50
Sidebet C : 1 pay 25
Sidebet D : 1 pay 100
Sidebet E : 1 pay 65



This almost sounds like a big six wheel more than a card game, except that there's no deck to burn through to change the count. Guess it's irrelevant. Interesting that the bets are mutually exclusive.
If the House lost every hand, they wouldn't deal the game.
MoeHoward
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September 4th, 2014 at 8:06:49 AM permalink
Quote: beachbumbabs

This almost sounds like a big six wheel more than a card game, except that there's no deck to burn through to change the count. Guess it's irrelevant. Interesting that the bets are mutually exclusive.

Super Pay Egalite?

AceTwo
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September 4th, 2014 at 11:29:00 AM permalink
Quote: ssho88

Base on system S6, trigger running count for sidebets A,B,C,D and E are RC >= 21, RC >= 22, RC >= 24, RC >= 26 and RC >= 29 respectively.

Question:-

1) Should I bet for all five sidebets if RC=32 ?
2) Should I just bet on sidebet A, B and C if RC=25 ?
3) Each sidebet is independent from each other ?



This looks like 1 Sidebet with 5 (or more) possible results and you correctly treat it as 5 different sidebets.
For example BJ, 2 the fiirst 2 cards having a total of 12,13,14,15,16.
And I assume that the counting system you designed provides approximately a linear estimation of EV per count (like counting in BJ)

1) Yes mathematically you should bet on all 5 sidebets and mathematically you should bet different amounts according to the EV of each. ie Bet A will have a bigger EV and more should be bet on that. But practically it might look very strange if you do that if this is not standard practice by civilians who I assume would mostly put a bet only on 1 possible result (or maybe 2).
Thinking about it again, depending on Variance of each, mathematically it might be better just to bet on the one with the higher EV, ie Bet A. You increase you EV but also your variance so it depends on variance. Without further info, I would say Bet A and B with the highest EV and also C which has lower variance and forget D and E.
2) As in 1, Bet on A and B with the higher EV and C with the lower variance. Or maybe C the EV is very small (just above RC) and forger C.
3) If the sidebts are as I explain above (ie 1 sidebet with 5 or more possible results) then the bets are dependent. And most probably even if there completely different sidebets, they are still dependent if they rely on the same cards.

Apart from the Cut off pointy of RC, you need to find out the EV increment per RC.
Bet A is EV=0 at RC=21. How much EV is increased by each subsequent increase in RC.

And finally, with this payouts 25 to 100, the Variance is huge and you need a Big EV to make the bets worthhile.
EV of 2%-3% is useless to make any money.
Say with a 2% EV and $100k bankroll your bet under Full Kelly should be:
For the 25:1 Bet 2%/25 x $100k = $80 ($40 under 1/2 Kelly)
For the 100:1 Bet 2%/100 x $100k = $20 ($10 under 1/2 kelly)
So you need EVs of say 10% and more to make this worthwhile.

And maybe you should concetrate only on the bet with lower payout, ie bet C and use only the specific counting system for bet C.
Assuming fair odds the frequencies of winning the bets are (obviously odds are not fair so frequency is worse than this)
A 40:1 2,4%
B 50:1 2,0%
C 25:1 3,8%
D 100:1 1,0%
E 65:1 1,5%
So the design of a single count to capture all of them (but lose accuracy) might not be the optimum.
The optimum might be to concetrate on the most frequent ones (with more accuracy). Especially D,E are not worth it.
1 system for C only.
Or 1 System for A,B,C
Or 3 systems for A,B,C if you can handle 3 at the same time.
ssho88
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September 4th, 2014 at 8:26:41 PM permalink
Quote: AceTwo

This looks like 1 Sidebet with 5 (or more) possible results and you correctly treat it as 5 different sidebets.
For example BJ, 2 the fiirst 2 cards having a total of 12,13,14,15,16.
And I assume that the counting system you designed provides approximately a linear estimation of EV per count (like counting in BJ)

1) Yes mathematically you should bet on all 5 sidebets and mathematically you should bet different amounts according to the EV of each. ie Bet A will have a bigger EV and more should be bet on that. But practically it might look very strange if you do that if this is not standard practice by civilians who I assume would mostly put a bet only on 1 possible result (or maybe 2).
Thinking about it again, depending on Variance of each, mathematically it might be better just to bet on the one with the higher EV, ie Bet A. You increase you EV but also your variance so it depends on variance. Without further info, I would say Bet A and B with the highest EV and also C which has lower variance and forget D and E.
2) As in 1, Bet on A and B with the higher EV and C with the lower variance. Or maybe C the EV is very small (just above RC) and forger C.
3) If the sidebts are as I explain above (ie 1 sidebet with 5 or more possible results) then the bets are dependent. And most probably even if there completely different sidebets, they are still dependent if they rely on the same cards.

Apart from the Cut off pointy of RC, you need to find out the EV increment per RC.
Bet A is EV=0 at RC=21. How much EV is increased by each subsequent increase in RC.

And finally, with this payouts 25 to 100, the Variance is huge and you need a Big EV to make the bets worthhile.
EV of 2%-3% is useless to make any money.
Say with a 2% EV and $100k bankroll your bet under Full Kelly should be:
For the 25:1 Bet 2%/25 x $100k = $80 ($40 under 1/2 Kelly)
For the 100:1 Bet 2%/100 x $100k = $20 ($10 under 1/2 kelly)
So you need EVs of say 10% and more to make this worthwhile.

And maybe you should concetrate only on the bet with lower payout, ie bet C and use only the specific counting system for bet C.
Assuming fair odds the frequencies of winning the bets are (obviously odds are not fair so frequency is worse than this)
A 40:1 2,4%
B 50:1 2,0%
C 25:1 3,8%
D 100:1 1,0%
E 65:1 1,5%
So the design of a single count to capture all of them (but lose accuracy) might not be the optimum.
The optimum might be to concetrate on the most frequent ones (with more accuracy). Especially D,E are not worth it.
1 system for C only.
Or 1 System for A,B,C
Or 3 systems for A,B,C if you can handle 3 at the same time.




Thanks AceTwo

Yes it's similar to BJ, fiirst 2 cards having a total of 12,13,14,15,16

The most important to me : Yes mathematically you should bet on all 5 sidebets and mathematically you should bet different amounts according to the EV of each. . . .

The betting limits is very small compare to team bankroll, I am not worry about the variance. I will just bet max amount when higher then trigger RC !

BTW, max bet = $50 and bankroll > $500k !
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