Conditional EV and Advanced Kelly Optimization for 5 treasures
December 30th, 2025 at 1:12:50 PM
permalink
Most published analyses of baccarat side bets assume independence, single-bet
wagering, and closed-form Kelly sizing. These assumptions are false for multi-bet systems
such as 5 Treasures. I'd like describe a framework that removes those assumptions
by enumerating the full outcome space and solving the resulting bankroll optimization problem
numerically.
Kelly sizing when multiple side bets are placed simultaneously
where winning one wager forces losses on others Composite bets whose payoff depends on
other outcomes Cases where positive marginal EV implies zero optimal bet.
For a given shoe composition, the system enumerates all valid baccarat hands consistent with the rules
and remaining cards. Each hand is treated as an equally weighted realization. Expected value is
computed as the mean realized payoff across the full enumerated outcome set. No Monte Carlo
approximation or probability weighting is used.
In interacting side-bet systems, applying single-bet Kelly independently to each wager is
mathematically invalid. The correct formulation is a constrained log-utility maximization problem over all
wagers simultaneously. Closed-form solutions do not exist in general. Numerical optimization is
required.
Independent Kelly sizing overstates growth and understates drawdown in multi-bet systems.
Interaction effects can dominate marginal EV. Positive conditional EV does not imply a
positive Kelly allocation. The mathematically optimal strategy can be to place no side bets.
Most public analyses implicitly assume assumptions that do not hold in live play ( IE Elliot Jacobson and Advanced advanced play).
These outcomes are not paradoxes. They are the correct consequence of optimizing log-growth under
realistic interaction constraints. Users who are surprised by zero-bet solutions are applying
inappropriate intuition.
There are five side bets, as follows:
• Fortune 7 — Wins if the Banker has winning 3-card total of 7. Pays 40 to 1.
• Golden 8 — Wins if the Player has winning 3-card total of 8. Pays 25 to 1.
• Heavenly 9 — Wins if the Player or Banker have a three-card total of 9. Pays 75 to 1 if both do
and 10 to 1 if one does.
• Blazing 7s — Wins if the Player or Banker both have a total of 7 composed of the same number
of cards. Pays 200 to 1 if both hands are composed of three cards and 50 to 1 if both hands are
composed of two cards.
• Cover All — This bet pays 6 to 1 if any of the four bets above win.
The win probabilities are also detailed on that website (and generally known I guess), so I’ll take them
as given.
This will most likely need a two-step approach:
1. Find formulas for mutually exclusive bets (ignoring Cover All),
2. Add the Cover All bet
3 Mutually exclusive outcomes
I’m ignoring Cover All in this section. The other 4 bets can be rephrased as 6 mutually exclusive
bets (the individual odds of these are all given at /games/baccarat/
side-bets/5-treasures/):
1. Fortune 7 — Wins if the Banker has winning 3-card total of 7. Pays 40 to 1.
2. Golden 8 — Wins if the Player has winning 3-card total of 8. Pays 25 to 1.
3. Heavenly 9A — Wins if BOTH the Player AND Banker have a three-card total of 9. Pays 75 to
1.
4. Heavenly 9B — Wins if EITHER the Player OR Banker (NOT BOTH) have a three-card total
of 9. Pays 10 to 1.
5. Blazing 7sA — Wins if the Player and Banker both have a total of 7 composed of THREE cards.
Pays 200 to 1.
6. Blazing 7sB — Wins if the Player and Banker both have a total of 7 composed of TWO cards.
Pays 50 to 1.
Googling the problem, I found this formulation
(/article/kelly-criterion-multiple-mutually-exclusive-outcomes):
I’ll use this formulation of E, but I’ll add a term for the case that none of the bets wins. Since the
bets are all mutually exclusive, the probability that none of them wins is (1 − p1 − p2 − ... − p6), and
adding this to E .
Unfortunately the partial derivatives in the source above are wrong. To get things right, I first
derive the optimal Kelly bets for a system with only two mutually exclusive bets, and then derive the
more general formulas for 6 bets later.
The logarithm of the wealth is given by
E2 = (1 − p1 − p2) ln(1 − f1 − f2) + p1 ln (1 + f1b1 − f2) + p2 ln (1 + f2b2 − f1),
2
and the partial derivatives are
dE2
df1
= −
1 − p1 − p2
1 − f1 − f2
+
p1b1
1 + f1b1 − f2
−
p2
1 + f2b2 − f1
,
∂E2
∂f2
= −
1 − p1 − p2
1 − f1 − f2
+
p2b2
1 + f2b2 − f1
−
p1
1 + f1b1 − f2
.
Setting these to zero to find local optima, we get the solutions
∂
∂f1
= 0
⇐⇒ − (1 − p1 − p2)(1 + f1b1 − f2)(1 + f2b2 − f1) + p1b1(1 − f1 − f2)(1 + f2b2 − f1)
− p2(1 − f1 − f2)(1 + f1b1 − f2) = 0
and (in the same way)
−(1−p1−p2)(1+f1b1−f2)(1+f2b2−f1)−p1(1−f1−f2)(1+f2b2−f1)+p2b2(1−f1−f2)(1+f1b1−f2) = 0.
Solving this gets complex quickly, so chances of getting an analytic solution for the more complex
overall problem are extremely slim. So we changed our approach and solved the problem numerically.
The function to be maximized is the expectation of the logarithm of the bankroll after the bets
(according to Kelly’s problem formulation), so in our case this is
E = (1 − pF − pG − pHa − pHb − pBa − pBb) ln (1 − fF − fG − fH − fB − fC )
+ pF ln (1 + 40fF + 6fC − fG − fH − fB)
+ pG ln (1 + 25fG + 6fC − fF − fH − fB)
+ pHa ln (1 + 75fH + 6fC − fF − fG − fB)
+ pHb ln (1 + 10fH + 6fC − fF − fG − fB)
+ pBa ln (1 + 200fB + 6fC − fF − fG − fH)
+ pBb ln (1 + 50fB + 6fC − fF − fG − fH)
The first term corresponds to none of the bets winning, and the other terms correspond to one of the
bets winning and the other ones losing. We know that 1 − pF − pG − pHa − pHb − pBa − pBb > 0,
and ln() only exists for positive numbers, with lim(x) −→ −∞ as x −→ 0. This means that the first
term (1 − pF − pG − pHa − pHb − pBa − pBb) ln (1 − fF − fG − fH − fB − fC ) will approach −∞ as
fF + fG + fH + fB + fC approaches 1. Since we have a maximization problem, −∞ will never be the
optimal solution. This means we don’t have to worry about the constraint fF +fG +fH +fB +fC ≤ 1
– the optimum of our function E will automatically respect that constraint. We should still make sure
the individual bet amounts stay within the [0, 1] range.
For an analytical solution, we’d now take the partial derivatives of E w.r.t. the bet amounts, set
them to zero and hope that this boils down to formulas for the optimal bet amounts. The partial
derivate of E w.r.t. fF is
∂E
∂fF
= −
1 − pF − pG − pHa − pHb − pBa − pBb
1 − fF − fG − fH − fB − fC
+
40pF
1 + 40fF + 6fC − fG − fH − fB
−
pG
1 + 25fG + 6fC − fF − fH − fB
−
pHa
1 + 75fH + 6fC − fF − fG − fB
−
pHb
1 + 10fH + 6fC − fF − fG − fB
−
pBa
1 + 200fB + 6fC − fF − fG − fH
−
pBb
1 + 50fB + 6fC − fF − fG − fH
.
The other derivatives are very similar. For example,
∂E
∂fG
= −
1 − pF − pG − pHa − pHb − pBa − pBb
1 − fF − fG − fH − fB − fC
−
pF
1 + 40fF + 6fC − fG − fH − fB
+
25pG
1 + 25fG + 6fC − fF − fH − fB
−
pHa
1 + 75fH + 6fC − fF − fG − fB
−
pHb
1 + 10fH + 6fC − fF − fG − fB
−
pBa
1 + 200fB + 6fC − fF − fG − fH
−
pBb
1 + 50fB + 6fC − fF − fG − fH
Setting those derivatives to zero, we end up with infeasibly large equations. However, since most terms
are the same in ∂E
∂fF
and ∂E
∂fG
, setting
∂E
∂fF
=
∂E
∂fG
reduces to
∂E
∂fF
=
∂E
∂fG
⇐⇒
40pF
1 + 40fF + 6fC − fG − fH − fB
−
pG
1 + 25fG + 6fC − fF − fH − fB
= −
pF
1 + 40fF + 6fC − fG − fH − fB
+
25pG
1 + 25fG + 6fC − fF − fH − fB
⇐⇒
41pF
1 + 40fF + 6fC − fG − fH − fB
=
26pG
1 + 25fG + 6fC − fF − fH − fB
⇐⇒ 41pF (1 + 25fG + 6fC − fF − fH − fB) = 26pG(1 + 40fF + 6fC − fG − fH − fB)
⇐⇒ (−41pF − 1040pG)fF + (1025pF + 26pG)fG + (246pF − 156pG)fC + (−41pF + 26pG)fH
+ (−41pF + 26pG)fB + (41pF − 26pG) = 0,
which is a linear equation involving all bet amounts. Unfortunately that trick won’t work with the
other partial derivatives, since they differ in more than just two terms, and the equations quickly
become too large to handle.
I did a quick numerical check setting all partial derivatives to zero and having Python find the
solution – some of the optimal bet amounts were −∞. This means we also have to take into account
the bounds on the bet amounts, and just solving these equations would not give us the solution we
want. So I finally give up on solving (even parts of) this problem analytically, and use Python to solve it numerically instead.
Sorry for the bad math formatting.
Credit to Pia Kempker for help on the heavy math.
I'm happy to share the full formulation and numerical results. I won't be publishing the underlying python code implementation as the goal here is to discuss the correctnedd of the model and assumptions rather than the mechanics of the specific implementation.
wagering, and closed-form Kelly sizing. These assumptions are false for multi-bet systems
such as 5 Treasures. I'd like describe a framework that removes those assumptions
by enumerating the full outcome space and solving the resulting bankroll optimization problem
numerically.
Kelly sizing when multiple side bets are placed simultaneously
where winning one wager forces losses on others Composite bets whose payoff depends on
other outcomes Cases where positive marginal EV implies zero optimal bet.
For a given shoe composition, the system enumerates all valid baccarat hands consistent with the rules
and remaining cards. Each hand is treated as an equally weighted realization. Expected value is
computed as the mean realized payoff across the full enumerated outcome set. No Monte Carlo
approximation or probability weighting is used.
In interacting side-bet systems, applying single-bet Kelly independently to each wager is
mathematically invalid. The correct formulation is a constrained log-utility maximization problem over all
wagers simultaneously. Closed-form solutions do not exist in general. Numerical optimization is
required.
Independent Kelly sizing overstates growth and understates drawdown in multi-bet systems.
Interaction effects can dominate marginal EV. Positive conditional EV does not imply a
positive Kelly allocation. The mathematically optimal strategy can be to place no side bets.
Most public analyses implicitly assume assumptions that do not hold in live play ( IE Elliot Jacobson and Advanced advanced play).
These outcomes are not paradoxes. They are the correct consequence of optimizing log-growth under
realistic interaction constraints. Users who are surprised by zero-bet solutions are applying
inappropriate intuition.
There are five side bets, as follows:
• Fortune 7 — Wins if the Banker has winning 3-card total of 7. Pays 40 to 1.
• Golden 8 — Wins if the Player has winning 3-card total of 8. Pays 25 to 1.
• Heavenly 9 — Wins if the Player or Banker have a three-card total of 9. Pays 75 to 1 if both do
and 10 to 1 if one does.
• Blazing 7s — Wins if the Player or Banker both have a total of 7 composed of the same number
of cards. Pays 200 to 1 if both hands are composed of three cards and 50 to 1 if both hands are
composed of two cards.
• Cover All — This bet pays 6 to 1 if any of the four bets above win.
The win probabilities are also detailed on that website (and generally known I guess), so I’ll take them
as given.
This will most likely need a two-step approach:
1. Find formulas for mutually exclusive bets (ignoring Cover All),
2. Add the Cover All bet
3 Mutually exclusive outcomes
I’m ignoring Cover All in this section. The other 4 bets can be rephrased as 6 mutually exclusive
bets (the individual odds of these are all given at /games/baccarat/
side-bets/5-treasures/):
1. Fortune 7 — Wins if the Banker has winning 3-card total of 7. Pays 40 to 1.
2. Golden 8 — Wins if the Player has winning 3-card total of 8. Pays 25 to 1.
3. Heavenly 9A — Wins if BOTH the Player AND Banker have a three-card total of 9. Pays 75 to
1.
4. Heavenly 9B — Wins if EITHER the Player OR Banker (NOT BOTH) have a three-card total
of 9. Pays 10 to 1.
5. Blazing 7sA — Wins if the Player and Banker both have a total of 7 composed of THREE cards.
Pays 200 to 1.
6. Blazing 7sB — Wins if the Player and Banker both have a total of 7 composed of TWO cards.
Pays 50 to 1.
Googling the problem, I found this formulation
(/article/kelly-criterion-multiple-mutually-exclusive-outcomes):
I’ll use this formulation of E, but I’ll add a term for the case that none of the bets wins. Since the
bets are all mutually exclusive, the probability that none of them wins is (1 − p1 − p2 − ... − p6), and
adding this to E .
Unfortunately the partial derivatives in the source above are wrong. To get things right, I first
derive the optimal Kelly bets for a system with only two mutually exclusive bets, and then derive the
more general formulas for 6 bets later.
The logarithm of the wealth is given by
E2 = (1 − p1 − p2) ln(1 − f1 − f2) + p1 ln (1 + f1b1 − f2) + p2 ln (1 + f2b2 − f1),
2
and the partial derivatives are
dE2
df1
= −
1 − p1 − p2
1 − f1 − f2
+
p1b1
1 + f1b1 − f2
−
p2
1 + f2b2 − f1
,
∂E2
∂f2
= −
1 − p1 − p2
1 − f1 − f2
+
p2b2
1 + f2b2 − f1
−
p1
1 + f1b1 − f2
.
Setting these to zero to find local optima, we get the solutions
∂
∂f1
= 0
⇐⇒ − (1 − p1 − p2)(1 + f1b1 − f2)(1 + f2b2 − f1) + p1b1(1 − f1 − f2)(1 + f2b2 − f1)
− p2(1 − f1 − f2)(1 + f1b1 − f2) = 0
and (in the same way)
−(1−p1−p2)(1+f1b1−f2)(1+f2b2−f1)−p1(1−f1−f2)(1+f2b2−f1)+p2b2(1−f1−f2)(1+f1b1−f2) = 0.
Solving this gets complex quickly, so chances of getting an analytic solution for the more complex
overall problem are extremely slim. So we changed our approach and solved the problem numerically.
The function to be maximized is the expectation of the logarithm of the bankroll after the bets
(according to Kelly’s problem formulation), so in our case this is
E = (1 − pF − pG − pHa − pHb − pBa − pBb) ln (1 − fF − fG − fH − fB − fC )
+ pF ln (1 + 40fF + 6fC − fG − fH − fB)
+ pG ln (1 + 25fG + 6fC − fF − fH − fB)
+ pHa ln (1 + 75fH + 6fC − fF − fG − fB)
+ pHb ln (1 + 10fH + 6fC − fF − fG − fB)
+ pBa ln (1 + 200fB + 6fC − fF − fG − fH)
+ pBb ln (1 + 50fB + 6fC − fF − fG − fH)
The first term corresponds to none of the bets winning, and the other terms correspond to one of the
bets winning and the other ones losing. We know that 1 − pF − pG − pHa − pHb − pBa − pBb > 0,
and ln() only exists for positive numbers, with lim(x) −→ −∞ as x −→ 0. This means that the first
term (1 − pF − pG − pHa − pHb − pBa − pBb) ln (1 − fF − fG − fH − fB − fC ) will approach −∞ as
fF + fG + fH + fB + fC approaches 1. Since we have a maximization problem, −∞ will never be the
optimal solution. This means we don’t have to worry about the constraint fF +fG +fH +fB +fC ≤ 1
– the optimum of our function E will automatically respect that constraint. We should still make sure
the individual bet amounts stay within the [0, 1] range.
For an analytical solution, we’d now take the partial derivatives of E w.r.t. the bet amounts, set
them to zero and hope that this boils down to formulas for the optimal bet amounts. The partial
derivate of E w.r.t. fF is
∂E
∂fF
= −
1 − pF − pG − pHa − pHb − pBa − pBb
1 − fF − fG − fH − fB − fC
+
40pF
1 + 40fF + 6fC − fG − fH − fB
−
pG
1 + 25fG + 6fC − fF − fH − fB
−
pHa
1 + 75fH + 6fC − fF − fG − fB
−
pHb
1 + 10fH + 6fC − fF − fG − fB
−
pBa
1 + 200fB + 6fC − fF − fG − fH
−
pBb
1 + 50fB + 6fC − fF − fG − fH
.
The other derivatives are very similar. For example,
∂E
∂fG
= −
1 − pF − pG − pHa − pHb − pBa − pBb
1 − fF − fG − fH − fB − fC
−
pF
1 + 40fF + 6fC − fG − fH − fB
+
25pG
1 + 25fG + 6fC − fF − fH − fB
−
pHa
1 + 75fH + 6fC − fF − fG − fB
−
pHb
1 + 10fH + 6fC − fF − fG − fB
−
pBa
1 + 200fB + 6fC − fF − fG − fH
−
pBb
1 + 50fB + 6fC − fF − fG − fH
Setting those derivatives to zero, we end up with infeasibly large equations. However, since most terms
are the same in ∂E
∂fF
and ∂E
∂fG
, setting
∂E
∂fF
=
∂E
∂fG
reduces to
∂E
∂fF
=
∂E
∂fG
⇐⇒
40pF
1 + 40fF + 6fC − fG − fH − fB
−
pG
1 + 25fG + 6fC − fF − fH − fB
= −
pF
1 + 40fF + 6fC − fG − fH − fB
+
25pG
1 + 25fG + 6fC − fF − fH − fB
⇐⇒
41pF
1 + 40fF + 6fC − fG − fH − fB
=
26pG
1 + 25fG + 6fC − fF − fH − fB
⇐⇒ 41pF (1 + 25fG + 6fC − fF − fH − fB) = 26pG(1 + 40fF + 6fC − fG − fH − fB)
⇐⇒ (−41pF − 1040pG)fF + (1025pF + 26pG)fG + (246pF − 156pG)fC + (−41pF + 26pG)fH
+ (−41pF + 26pG)fB + (41pF − 26pG) = 0,
which is a linear equation involving all bet amounts. Unfortunately that trick won’t work with the
other partial derivatives, since they differ in more than just two terms, and the equations quickly
become too large to handle.
I did a quick numerical check setting all partial derivatives to zero and having Python find the
solution – some of the optimal bet amounts were −∞. This means we also have to take into account
the bounds on the bet amounts, and just solving these equations would not give us the solution we
want. So I finally give up on solving (even parts of) this problem analytically, and use Python to solve it numerically instead.
Sorry for the bad math formatting.
Credit to Pia Kempker for help on the heavy math.
I'm happy to share the full formulation and numerical results. I won't be publishing the underlying python code implementation as the goal here is to discuss the correctnedd of the model and assumptions rather than the mechanics of the specific implementation.
December 30th, 2025 at 2:31:30 PM
permalink
Too many words, too much theory.
{Edit: I did that too. Self-censored.}
Finally, when I get a counting system I use it with a simulator to see how it actually performs in a simulated game, and then most importantly, when I get my results, I don't blabber them all over the internet! This is advanced counting and as a rule, a person who is not experienced and ambitious enough to discover these things with their own work is also not experienced and ambitious enough to apply them successfully at the table. These are high variance bets, most require exotic counts you won't use anywhere else, there are limited tables and stores in which they can be applied, publicity risks the bets or games being shut down and jeopardizes opportunities for those actively exploiting them, and rookies are usually going to be in over their heads and get discouraged or exercise poor judgment, and get themselves in trouble if they try this. Please consider all of this before publicizing this sort of play- who benefits from your publicity, and who might be harmed?
Get your own preparation done, get your own numbers and get out there and play!
{Edit: I did that too. Self-censored.}
Finally, when I get a counting system I use it with a simulator to see how it actually performs in a simulated game, and then most importantly, when I get my results, I don't blabber them all over the internet! This is advanced counting and as a rule, a person who is not experienced and ambitious enough to discover these things with their own work is also not experienced and ambitious enough to apply them successfully at the table. These are high variance bets, most require exotic counts you won't use anywhere else, there are limited tables and stores in which they can be applied, publicity risks the bets or games being shut down and jeopardizes opportunities for those actively exploiting them, and rookies are usually going to be in over their heads and get discouraged or exercise poor judgment, and get themselves in trouble if they try this. Please consider all of this before publicizing this sort of play- who benefits from your publicity, and who might be harmed?
Get your own preparation done, get your own numbers and get out there and play!
Last edited by: AutomaticMonkey on Dec 31, 2025
December 30th, 2025 at 3:08:20 PM
permalink
Thanks for your reply but you missed the point of the thread. The counting system for these side bets are well known see advanced advantage play. What im talking about is Kelly bet sizing to determine optimal outcome.
The play is well known and popularized by Elliot Jacobson. What im getting at is optimization of the betting systems for anyone who is currently attacking the 5 treasure side bet.
Take for example a team that is currently attacking the 5 treasure and counting all 5 side bets let's say you currently have the following true counts:
Fortune 7 +6
Golden 8 +12
Heavenly 9 +2
Blazing 7s +12
Cover all +6
Traditional Kelly sizes would suggest you bet the following, given a $50k roll:
F7 $128
G8 $44
Blazing 7s $115
Cover all $162
This exposes you to too much risk. With an advanced Kelly formula I suggest then your Kelly sizes would looks more like this:
F7 $25
G8 $10
Blazing 7s $50
Coverall $600
It moves your money to the more likely percentage chance of winning and off the high variance side bets. Keep in mind this is still at full Kelly 13.5% ror, so of course this isn't a for anyone without at least a $200k bankroll.
The play is well known and popularized by Elliot Jacobson. What im getting at is optimization of the betting systems for anyone who is currently attacking the 5 treasure side bet.
Take for example a team that is currently attacking the 5 treasure and counting all 5 side bets let's say you currently have the following true counts:
Fortune 7 +6
Golden 8 +12
Heavenly 9 +2
Blazing 7s +12
Cover all +6
Traditional Kelly sizes would suggest you bet the following, given a $50k roll:
F7 $128
G8 $44
Blazing 7s $115
Cover all $162
This exposes you to too much risk. With an advanced Kelly formula I suggest then your Kelly sizes would looks more like this:
F7 $25
G8 $10
Blazing 7s $50
Coverall $600
It moves your money to the more likely percentage chance of winning and off the high variance side bets. Keep in mind this is still at full Kelly 13.5% ror, so of course this isn't a for anyone without at least a $200k bankroll.
December 30th, 2025 at 3:28:01 PM
permalink
Following the lead of a failed AP who has become an enemy of APs is not a sound approach either.
But neither is it in my best interest to correct anyone's sloppy spitework, so I will say no more.
But neither is it in my best interest to correct anyone's sloppy spitework, so I will say no more.
December 30th, 2025 at 5:02:48 PM
permalink
Quote: LolpoplobsterThanks for your reply but you missed the point of the thread. The counting system for these side bets are well known see advanced advantage play. What im talking about is Kelly bet sizing to determine optimal outcome.
The play is well known and popularized by Elliot Jacobson. What im getting at is optimization of the betting systems for anyone who is currently attacking the 5 treasure side bet.
Take for example a team that is currently attacking the 5 treasure and counting all 5 side bets let's say you currently have the following true counts:
Fortune 7 +6
Golden 8 +12
Heavenly 9 +2
Blazing 7s +12
Cover all +6
Traditional Kelly sizes would suggest you bet the following, given a $50k roll:
F7 $128
G8 $44
Blazing 7s $115
Cover all $162
This exposes you to too much risk. With an advanced Kelly formula I suggest then your Kelly sizes would looks more like this:
F7 $25
G8 $10
Blazing 7s $50
Coverall $600
It moves your money to the more likely percentage chance of winning and off the high variance side bets. Keep in mind this is still at full Kelly 13.5% ror, so of course this isn't a for anyone without at least a $200k bankroll.
link to original post
This is interesting. Yes given the multiplicity of side bets at baccarat it is worthwhile examining the relative risk involved when making multiple side-bets, that's a worthy project.
I'll comment later when I have time to look at your formulas properly but I wanted you to know someone understood what you are trying to do here.
Edit:
So what I can tell offhand
1 Correct identification of the core problem: Standard analyses do treat side bets as independent when they're structurally linked. Applying single-bet Kelly to each wager separately ignores the interaction effects and can produce nonsensical aggregate positions.
2. Mathematical framework is sound: The log-utility maximization formulation is correct for Kelly optimization. The approach of decomposing the composite bets (Heavenly 9, Blazing 7s) into mutually exclusive outcomes is the right way to handle the problem structure.
3. The observation that positive marginal EV can yield zero optimal Kelly allocation is counterintuitive but mathematically correct. When bets are negatively correlated (one winning forces others to lose), the variance drag can dominate the edge.
How much this matters in practice I'm not sure.
Disclaimer: This is a plausibility check not a rigorous verification.
.
Last edited by: DougGander on Dec 30, 2025
December 30th, 2025 at 6:14:11 PM
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Quote: AutomaticMonkeyFollowing the lead of a failed AP who has become an enemy of APs is not a sound approach either.
You have it backwards, he's challenging Jacobson's flawed assumptions.
December 30th, 2025 at 6:59:03 PM
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Quote: DougGanderQuote: LolpoplobsterThanks for your reply but you missed the point of the thread. The counting system for these side bets are well known see advanced advantage play. What im talking about is Kelly bet sizing to determine optimal outcome.
The play is well known and popularized by Elliot Jacobson. What im getting at is optimization of the betting systems for anyone who is currently attacking the 5 treasure side bet.
Take for example a team that is currently attacking the 5 treasure and counting all 5 side bets let's say you currently have the following true counts:
Fortune 7 +6
Golden 8 +12
Heavenly 9 +2
Blazing 7s +12
Cover all +6
Traditional Kelly sizes would suggest you bet the following, given a $50k roll:
F7 $128
G8 $44
Blazing 7s $115
Cover all $162
This exposes you to too much risk. With an advanced Kelly formula I suggest then your Kelly sizes would looks more like this:
F7 $25
G8 $10
Blazing 7s $50
Coverall $600
It moves your money to the more likely percentage chance of winning and off the high variance side bets. Keep in mind this is still at full Kelly 13.5% ror, so of course this isn't a for anyone without at least a $200k bankroll.
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This is interesting. Yes given the multiplicity of side bets at baccarat it is worthwhile examining the relative risk involved when making multiple side-bets, that's a worthy project.
I'll comment later when I have time to look at your formulas properly but I wanted you to know someone understood what you are trying to do here.
Edit:
So what I can tell offhand
1 Correct identification of the core problem: Standard analyses do treat side bets as independent when they're structurally linked. Applying single-bet Kelly to each wager separately ignores the interaction effects and can produce nonsensical aggregate positions.
2. Mathematical framework is sound: The log-utility maximization formulation is correct for Kelly optimization. The approach of decomposing the composite bets (Heavenly 9, Blazing 7s) into mutually exclusive outcomes is the right way to handle the problem structure.
3. The observation that positive marginal EV can yield zero optimal Kelly allocation is counterintuitive but mathematically correct. When bets are negatively correlated (one winning forces others to lose), the variance drag can dominate the edge.
How much this matters in practice I'm not sure.
Disclaimer: This is a plausibility check not a rigorous verification.
.
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imgur /a/K79NZKz
This is the formatted formulas so your eyes don't bleed from formatting on this forum.
Quote: DougGanderQuote: AutomaticMonkeyFollowing the lead of a failed AP who has become an enemy of APs is not a sound approach either.
You have it backwards, he's challenging Jacobson's flawed assumptions.
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Partially true, when looking at his work, some would think that Blazing 7s is the money maker of the side bets and to allocate their bankroll to blazing 7s. The math for Kelly Bankroll allocation for these mutually exclusive side bets says a completely different story for optimal bankroll growth.
December 31st, 2025 at 1:04:41 AM
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Quote: Lolpoplobster
Traditional Kelly sizes would suggest you bet the following, given a $50k roll:
F7 $128
G8 $44
Blazing 7s $115
Cover all $162
This exposes you to too much risk. With an advanced Kelly formula I suggest then your Kelly sizes would looks more like this:
F7 $25
G8 $10
Blazing 7s $50
Coverall $600
It moves your money to the more likely percentage chance of winning and off the high variance side bets. Keep in mind this is still at full Kelly 13.5% ror, so of course this isn't a for anyone without at least a $200k bankroll.
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These multiple sidebets problem has been bothering me for a long time, I've always used the traditional Kelly betting amount, and I know there must be an optimal betting amount that can lead to better bankroll growth, but I haven't been able to find the answer. Today, finally, someone might be able to help.
1) You mentioned that Traditional Kelly sizes exposes you to too much risk, what is the ROR ?
2) How you calculate the betsize F7 $25, G8 $10, Blazing 7s $50 and Coverall $600 ? With just an advanced Kelly formula or simulations ?
3) Have you used simulation to prove that your advanced Kelly sizes is still at full Kelly 13.5% ror ?
Thanks.
Last edited by: ssho88 on Dec 31, 2025
