Multiple Bets (Not BJ, but definitely theory and math)

London Colin

Well-Known Member
#1
Suppose you have the opportunity to place multiple bets, all related to a single event.

As in roulette, for example, some bets represent mutually exclusive outcomes like red/black, specific numbers etc., while others have a degree of overlap (red/odd, even/3rd-dozen, etc.). The bets have a variety of payoffs and crucially, unlike roulette, some of them offer a postive EV.

How do you go about determining the optimal total amount to bet, and how to distribute this among the available bets? (Assuming that, if faced with just a single bet, you would be employing a utility function to assign a fraction of your bankroll to the bet, proportional to the EV.)

The more I've tried to ponder this, the more I've managed to confuse myself. Presumably the key is to identify the variance of various individual and combined bets, in search of the highest overall Certainty Equivalent, but this is very much where I start to venture outside of my mathematical comfort zone.:confused:
 

Kasi

Well-Known Member
#2
London Colin said:
Suppose you have the opportunity to place multiple bets, all related to a single event.

As in roulette, for example, some bets represent mutually exclusive outcomes like red/black, specific numbers etc., while others have a degree of overlap (red/odd, even/3rd-dozen, etc.). The bets have a variety of payoffs and crucially, unlike roulette, some of them offer a postive EV.

How do you go about determining the optimal total amount to bet, and how to distribute this among the available bets? (Assuming that, if faced with just a single bet, you would be employing a utility function to assign a fraction of your bankroll to the bet, proportional to the EV.)

The more I've tried to ponder this, the more I've managed to confuse myself. Presumably the key is to identify the variance of various individual and combined bets, in search of the highest overall Certainty Equivalent, but this is very much where I start to venture outside of my mathematical comfort zone.:confused:
I'm only quite sure I have no idea lol.

I know how'd I'd bet $500 roll with a goal of a $250 win or go bust in trying in at most 3 spins of single-zero roulette wheel, say, to maximize my probability of success compared to betting all $500 on just 1 spin (say $250 on each of 2 dozens) vs another way of betting the $500 given I have at most 3 spins to win the $250.

That's voodoo though but fine with me. I understand it. I know if I have 3 spins of time available I can bet in a way to increase my chances of winning the same $250 with the same starting roll of $500 by exposing a smaller avg amount of money I bet to the same (neg) house edge.

Obviously, with the same roll and goal and having at least that ampount of time avialable, why would I stupidly bet all $500 in the form of $250 on 2 of the 3 dozens and risk all on only 1 spin?

What does "optimal" mean to you - betting with highest EV or lowest risk to reward ratio?

Really clueless on what CE means. Maybe just being able to bet a little more with same risk?

Why are you throwing in a +EV bet? Just bet on that when it happens maybe lol?

Otherwise I suppose it's how much the -EV bets pay, what, the probilities of those -EV happenings are, what the bet range limit is, maybe what total roll is and maybe how long you will be playing (forever vs 5 more minutes) and, maybe like you say, the variance attached to each event.

I mean if you are talking about how much of a roll to bet at TC+7 with some side bet also available while also being able to back bet 3 other people at the table and being able to spread to multiple hands at the other 3 spots that are available, all I can say, if I know I don't know, I just don't and don't ever worry about it lol.
 

London Colin

Well-Known Member
#3
In case there's any confusion, I ought to clarify this isn't about roulette (since it's a negative-expectation game), nor blackjack (despite the forum I posted it in! :))

Kasi said:
What does "optimal" mean to you - betting with highest EV or lowest risk to reward ratio?
Well, not highest EV, or the answer would be very straightforward :- bet your entire bankroll on the highest EV bet available. (Or, slightly more reasonably, put whatever amount you are prepared to risk on that single bet, rather than spread it around.)

The main thrust of my question was the notion that a player who would use the Kelly Criterion (or some other utility function) to work out how much to put on a single bet would ideally wish to create a combined bet that fulfills the same criteria.

In BJ, the amount you bet per hand decreases as you spread to multiple hands, because of the covariance of the hands; there is some tendency for hands to be won/lost together, because of the influence of the dealer's hand in determining all their outcomes.

In other games you may have situations in which some of the available bets have a tendency to either both win or both lose, as in BJ, but others have the opposite tendency, or are completely mutually exclusive - if you win bet A then you lose bet B, and vice versa.

And two such mutually exclusive bets might have the same EV, but one derives from a big payoff and a low probability, while the other derives from a small payoff and a high probability.

Kasi said:
Really clueless on what CE means. Maybe just being able to bet a little more with same risk?
If you read this page, then you will know everything I know. :)
http://www.bjmath.com/bjmath/kelly/kellyfaq.htm (Archive copy)

Kasi said:
Why are you throwing in a +EV bet? Just bet on that when it happens maybe lol?
Multiple, simultaneous +EV bets is the whole reason for this (hypothetical) question.

Kasi said:
Otherwise I suppose it's how much the -EV bets pay, what, the probilities of those -EV happenings are, what the bet range limit is, maybe what total roll is and maybe how long you will be playing (forever vs 5 more minutes) and, maybe like you say, the variance attached to each event.

I mean if you are talking about how much of a roll to bet at TC+7 with some side bet also available while also being able to back bet 3 other people at the table and being able to spread to multiple hands at the other 3 spots that are available, all I can say, if I know I don't know, I just don't and don't ever worry about it lol.
Not sure if there are any blackjack/sidebet scenarios that this question would relate to. As I said, I wasn't thinking of blackjack at all when I started my ramblings.
 

Kasi

Well-Known Member
#4
London Colin said:
If you read this page, then you will know everything I know. :)
http://www.bjmath.com/bjmath/kelly/kellyfaq.htm (Archive copy)
Pretty much sums up all I know too - no wonder neither of us can figure this out :grin:

Schlesinger does give a nice treatment of CE when discussing Risk Averse indexes in case it helps you more than it has helped me lol.

London Colin said:
Multiple, simultaneous +EV bets is the whole reason for this (hypothetical) question.
Well I figured you weren't dealing with any sissy -EV game like roulette lol.

So, say, what if single-zero roulette paid 37-1 on a single number, 3-1 on any dozen and 1.5 to 1 on an "even-money" bet? (Or make the pay-offs on each bet such that each bet yields the same HA if that in any way makes anything easier).

You want to know how much of your roll you'd bet assuming you wanted to make some bet on each of those 3 prop bets each of which is +EV now?

Or, maybe something like, like in the article, you have to make 3 bets every time but the first one vs a coin flip with 10% edge, the second 20% edge and 3rd 30% edge?

Is anything like this what you have in mind when saying multiple simultaneous bets?

Maybe stuff like that wouldn't be much different than an AP guy figuring out his N0 after playing a bunch of different games each for so long each with it's own EV and SD?

Stll Clueless in Seattle and very likely to remain so lol.
 

London Colin

Well-Known Member
#5
Hey Kasi,

Thanks for the responses.

Kasi said:
Pretty much sums up all I know too - no wonder neither of us can figure this out :grin:

Schlesinger does give a nice treatment of CE when discussing Risk Averse indexes in case it helps you more than it has helped me lol.
I'll take a look. Does anyone have any recommendations for good, general books on the mathematics of gambling, pitched at the mathematically mediocre like myself. :)

Kasi said:
Well I figured you weren't dealing with any sissy -EV game like roulette lol.

So, say, what if single-zero roulette paid 37-1 on a single number, 3-1 on any dozen and 1.5 to 1 on an "even-money" bet? (Or make the pay-offs on each bet such that each bet yields the same HA if that in any way makes anything easier).

You want to know how much of your roll you'd bet assuming you wanted to make some bet on each of those 3 prop bets each of which is +EV now?
That's the kind of thing, yeah. However, that example highlights one extreme aspect of the spectrum of possibilities. You can guarantee a profit on each spin by covering all the numbers!

Even if some -EV bets had to be made, you might still be able to construct a hedge, covering all possible outcomes, including the -EV, and still guaranteeing an overall profit on each spin. With your bankroll growing on each spin, you can steadily crank up the overall bet size (ignoring the practical difficulties of table limits, for the purposes of this discussion), and in doing so you may have increased your win rate, despite lowering your EV compared to betting just the +EV bets.

In fact, if you can't lose, you'd really like to be betting your entire bankroll each time!

The more general case I am trying to comprehend (and ideally discover a formula for) is the surrender of some EV for reduced risk, meaning that the overall bet size can be larger, and thus the win rate greater.

As an aside, I just happened to see a casino promotion in which one specific roulette number is given enhanced odds of 50:1 for a brief period each day. One can speculate about the best way to play that.

But going back to contrived examples, suppose slightly raised payoffs are offered on the middle column (which contains 8 black and 4 red numbers), and on betting black. A bet of 1 unit on both then has 4 possible outcomes -

Both win : 4/37
Just red wins: 14/37
Just column wins: 8/37
Both lose: 11/37

So we can consider the combined bet as a single entity, with its own variance. The amount that we bet, and how we divide it between the two components ought presumably to be different if the raised payoffs applied to red, rather than to black.



Kasi said:
Or, maybe something like, like in the article, you have to make 3 bets every time but the first one vs a coin flip with 10% edge, the second 20% edge and 3rd 30% edge?

Is anything like this what you have in mind when saying multiple simultaneous bets?

Maybe stuff like that wouldn't be much different than an AP guy figuring out his N0 after playing a bunch of different games each for so long each with it's own EV and SD?

Stll Clueless in Seattle and very likely to remain so lol.
I think I'd equate the task of formulating a multi-part bet for a specific circumstance to that of formulating a betting ramp and unit size for a specific BJ game. The goal in both cases is to play with a known level of risk, while maximising the win rate.
 

blackchipjim

Well-Known Member
#6
math books

If you are fairly profecient in math which I am not there is a web site for math that deals with gambling in addition to other formulas. It' wolframs mathworld and is a site that I go to for fun.
 

Kasi

Well-Known Member
#7
London Colin said:
In fact, if you can't lose, you'd really like to be betting your entire bankroll each time!
As an aside, I just happened to see a casino promotion in which one specific roulette number is given enhanced odds of 50:1 for a brief period each day. One can speculate about the best way to play that.
Well, in my example if 0 or 00 came up, one could lose all lol.

In your example, with one number paying 50-1 on a 00 wheel, I'd speculate something like EV=14/38=36.8% EV. SD=8.17 so var=66.76 and therefore maybe a full Kelly bet might be 34.2%/66.76=.0051. So bet 0.51% of roll on the next spin on the theory a full-Kelly bet is adv%/variance as a % of roll to bet.

And, uh, trust me, it'd be a freakin' miracle and then some if any of what I said even makes sense for a Kelly bet for your 50-1 example let alone "Certainty Equivalent" stuff like you asked about lol.
 

London Colin

Well-Known Member
#8
blackchipjim said:
If you are fairly profecient in math which I am not there is a web site for math that deals with gambling in addition to other formulas. It' wolframs mathworld and is a site that I go to for fun.
Thanks for that. A quick look didn't show up anything that directly applies to what I'm currently struggling with, but could prove a useful resource. (Not sure I'd describe it as fun, though:))


Kasi said:
Well, in my example if 0 or 00 came up, one could lose all lol.
Not if one covered those numbers with a hedge bet. Your example allowed for a guaranteed profit, with zero risk. Nice work, if you can get it.:grin:

Kasi said:
In your example, with one number paying 50-1 on a 00 wheel, I'd speculate something like EV=14/38=36.8% EV. SD=8.17 so var=66.76 and therefore maybe a full Kelly bet might be 34.2%/66.76=.0051. So bet 0.51% of roll on the next spin on the theory a full-Kelly bet is adv%/variance as a % of roll to bet.
I was thinking about hedging when I speculated about how to play this bet too. But, thinking about it some more, I don't think it could ever make sense to use negative-expectation bets as part of a hedge, unless the overall game still comes out with a positive expectation. The 50:1 payout is not enough to achieve that; if you tried to cover all the other numbers you'd lock into an overall loss.

But I think it would be a different and more complicated matter if there were a group of +EV bets available. You may not be able to lock into a profit, but you presumably should bet some amount on all the +EV bets; probably a greater proportion of your bankroll, in total, than you could justify if you just placed a single bet with the highest EV available.
 

London Colin

Well-Known Member
#9
It might help to focus, one at a time, on some of the individual, small questions that go to make up the totality of my cluelessness.:)

Suppose you have identified just two +EV bets that can be made at one time. These bets, A and B, are mutually exclusive. Both may lose, but just one may win.

If you know the EV (and the probability and payoff from which it derives) and the variance for both bets, what proportion of your bankroll should you bet in total, and how should you divide it between the two bets?
 

Kasi

Well-Known Member
#10
London Colin said:
It might help to focus, one at a time, on some of the individual, small questions that go to make up the totality of my cluelessness.:)
Pretty much what I think too lmao.

Good point on the 0,00 stuff with hedging - hadn;t hought of that lol.

Like you, I wonder what the details are that make up for the fact I have no idea what to do or how to proceed lol.

IEven more micro-cosmically than you propose, for the life of me, I have never been able to think theoretically (abstractly I'd guess you'd say).

So "Suppose you have identified just two +EV bets that can be made at one time. These bets, A and B, are mutually exclusive. Both may lose, but just one may win." just makes me wish you could define it even more in terms of a concrete example lol.

Like "2 coins flipped., you must bet on both for some reason. One coin will always lose. The other coin will sometimes win."

Btw, not that even providing a concrete example would help anyway as I am just not thta much of a "math" guy in the first place lol.

Wherever it is now that I may be, and it ain't that far, I have only gotten to wherever it is I now may be in a series of the smallest of the most incremental steps one can imagine.

I was hoping someone could even comment on what I said to that 50-1 roulette thing you suggested as far as it went about what % to bet of roll for a Kelly bet lol.

No big deal - I'm not expecting to ever answer your question in any meaningful way anyway.

My math skills essentially stopped in the first grade when I was so scared to death by a neighbor who said if I did not know how to add, subtract, multiply and divide on the first day of first grade really horrible things would happen to me. My parents later told me I threw up alot when I was 4 or 5 but they didn't know why. I don't remember throwing up alot but I do remember exactly who it was that scared the living Be-Jesus out of me. So I asked my father how to add/s.m/d when I was 4 or so and could do all that the first day of first grade because I was scared to death. Never learned another damn thing since then. But I can still a/s/m/d today as good as I could when I was 6 - haven't lost a step on adding subtracting since then lmao.

But I like the question!

Hey it's "Theory" forum so whatever I can or can't learn from people who actually seem to know what they are doing so much the better for me.

Long way of saying, you can't ask a question small enough I can't focus on to make me aware of my cluelessnes. :laugh:
 

London Colin

Well-Known Member
#11
Kasi said:
So "Suppose you have identified just two +EV bets that can be made at one time. These bets, A and B, are mutually exclusive. Both may lose, but just one may win." just makes me wish you could define it even more in terms of a concrete example lol.
Certainly ...

Suppose you are betting on the outcome of the roll of a die. Fair odds (i.e. 0 EV) would be 5:1. From what I understand, the Kelly formula for a single-payoff wager is -

f = (bp -q) / b

where,
f is the fraction of your bankroll to bet
b is the payoff (5 in the above example)
p is the probability of winning
q is the probability of losing (i.e. 1-p)

With the 5:1 odds, that gives (5*1/6 - 5/6) / 5 = 0. [i.e. don't bet anything.]

Suppose you receive 6:1 if you bet on rolling a six, and win. It then becomes (6*1/6 - 5/6) / 6 = 0.0278. i.e. bet $278 on a $10K bankroll.

Suppose instead you receive 7:1 if you bet on rolling a one, and win. It then becomes (7*1/6 - 5/6) / 7 = 0.0476. i.e. bet $476 on a $10K bankroll.

Now suppose both bets are available at the same time. Only one can win; both might lose. How much of your $10K do you place on each?


For what it's worth, my guess would be that, since the EV of betting on a one (7/6 - 5/6 = +0.333) is twice that of betting on a six (6/6 - 5/6 = +0.167), twice as much should be bet on that outcome. That would then just leave the question of calculating the total amount to be bet, before dividing it up in that ratio.
 

London Colin

Well-Known Member
#12
London Colin said:
For what it's worth, my guess would be that, since the EV of betting on a one (7/6 - 5/6 = +0.333) is twice that of betting on a six (6/6 - 5/6 = +0.167), twice as much should be bet on that outcome. That would then just leave the question of calculating the total amount to be bet, before dividing it up in that ratio.
I just realised that that was, at best, an oversimplification. There's more than the relative EVs of the two bets to consider.

If the 6:1 bet on a six were replaced with a 5:2 bet on either one of a pair of numbers being rolled (e.g., win if you roll a five or a six), then the EV would remain the same. The Kelly fraction for that bet in isolation, however, goes up considerably: ((5/2 * 2/6) - 4/6) / (5/2) = 0.0667. That's more than would be bet on the 7:1 single-number.
 

Kasi

Well-Known Member
#13
London Colin said:
I just realised that that was, at best, an oversimplification. There's more than the relative EVs of the two bets to consider.

If the 6:1 bet on a six were replaced with a 5:2 bet on either one of a pair of numbers being rolled (e.g., win if you roll a five or a six), then the EV would remain the same. The Kelly fraction for that bet in isolation, however, goes up considerably: ((5/2 * 2/6) - 4/6) / (5/2) = 0.0667. That's more than would be bet on the 7:1 single-number.
Well, too late for me tonite lol but I like what you are saying lol.

Don't forget, I'd venture in general after reading this, maybe a "Kelly" bet also must take into account variance. Somehow lmao.

If no one besides you or me chooses to express their thoughts here, trust me, we are both doomed here lol.
 

London Colin

Well-Known Member
#14
Kasi said:
Don't forget, I'd venture in general after reading this, maybe a "Kelly" bet also must take into account variance. Somehow lmao.
Absolutely. That's implicit. I was trying to hint that maybe the total bet (however one calculates that) should be divided up in the ratio of the individual Kelly fractions. Alternatively, maybe the correct approach is not to start with a grand total and divide it up, but to calculate the individual Kelly bets and scale them down by some factor.

Kasi said:
If no one besides you or me chooses to express their thoughts here, trust me, we are both doomed here lol.
Thanks for your efforts to help me kickstart a discussion. I must admit I'm disappointed by the lack of response. I had all manner of follow-up questions ready, but I guess they will have to go unasked.
 

Sonny

Well-Known Member
#15
London Colin said:
I must admit I'm disappointed by the lack of response.
Me too! I guess that means you’re stuck listening to my sloppy math. :( Here goes...

London Colin said:
Suppose you receive 6:1 if you bet on rolling a six, and win. It then becomes (6*1/6 - 5/6) / 6 = 0.0278. i.e. bet $278 on a $10K bankroll.

Suppose instead you receive 7:1 if you bet on rolling a one, and win. It then becomes (7*1/6 - 5/6) / 7 = 0.0476. i.e. bet $476 on a $10K bankroll.

Now suppose both bets are available at the same time. Only one can win; both might lose. How much of your $10K do you place on each?
If I had to bet both numbers and I was aiming for a full Kelly bet then I would bet $350 on the 6:1 shot and $430 on the 7:1 shot. Now get ready for the long and twisted road of how I got those numbers. :eek:

If we flat bet 1 unit on each bet then I get an EV of 25%. We no longer have a fixed odds bet so we cannot use the odds as a divisor to find the Kelly bet. Instead we can use the variance of 3.15 units, although for bet sizing it’s more correct (and easier) to use the average squared result of 3.21 units. That gives us a full Kelly bet of about $780. Let me explain how I got those numbers.

We will win the #6 bet 1/6 = 17% of the time. When we win, we get 6 units but we also lose 1 unit on the #1 bet. This gives a result of 0.17*(6-1) = 83.33% and an average squared result (ASR) of 0.17*(6-1)^2 = 4.17 units. For the #1 bet we get 0.17*(7-1) = 100% with an ASR of 0.17*(7-1)^2 = 6 units. We will lose 2 units 67% of the time for a result of 0.67*-2 = -133.33% and an ASR of 0.67*(-2)^2 = 2.67 units. Summing those will give us an EV of 50% per bet (which agrees with the sum of the two individual bets 0.33+0.17 = 0.5), or 50%/2 = 25% per round if we consider the combined bets to be 1 unit (which agrees with the average advantage for the two bets), and an ASW of 12.83 per bet, or 12.83/2^2 = 3.21 units per round. That gives us a full Kelly bet of around $10000*0.25/3.21 = $779.22 per round. Let’s call it $780. That’s how much we’re planning to bet per round, but how will we divide that between the two numbers?

Our calculations so far have assumed that we split our money evenly on both bets. If we continue with this assumption then we would place $390 on each bet and earn $195 per round. That’s pretty good, but we can squeeze out a little more profit by betting more on the #1 bet. The problem is that we don’t want to bet too much on the #1 bet because that will negate our variance reduction. The whole reason we are making the #6 bet is to smooth out the fluctuations. If we bet too much on the #1 bet then winning the #6 bet won’t help us very much. For example, if we bet $70 on the #6 bet and $420 on the #1 bet then a roll of 6 will win $70*6 = $420 but the #1 bet will lose $420. We have only broken even. A roll of 1 would produce a win of $420*7-70 = $2870, which is still a big swing. We are betting more on the #1 bet so our EV jumps way up to almost 31%, but the lopsided bets keep the variance pretty high so we need to use a smaller unit size, which drops the full Kelly win rate down to about $152.

So we want to split up the bets in proportion to the advantages. Since the #1 bet represents 0.33/(0.33+0.17) = 55% of our profit I would consider betting 55% ($430) on #1 and 45% ($350) on #6 for an EV of just over $200 for similar levels of risk. That’s better than splitting the bets equally ($195), using a large spread ($152), or even playing the #1 bet only ($159). After fooling around with some other betting options, the 55/45 split seems like a pretty good choice. Maybe she ain’t the prettiest there is, but she’s as good as a guy like me is gonna find. Hopefully it's helpful to you.

-Sonny-
 

London Colin

Well-Known Member
#16
Thanks

Sonny said:
Me too! I guess that means you’re stuck listening to my sloppy math. :( Here goes...
That's perfectly fine, so long as you can tolerate my doubly-sloppy feedback. :)


Sonny said:
If I had to bet both numbers and I was aiming for a full Kelly bet then I would bet $350 on the 6:1 shot and $430 on the 7:1 shot. Now get ready for the long and twisted road of how I got those numbers. :eek:
Just to be clear, there was not meant to be any suggestion of compulsion. My working assumption has been that a bet on both numbers will be preferable, in that it will allow more money to be won than a single bet of equal risk. (The percentage EV may fall, but you are able to put more of your bankroll to work.)

Sonny said:
If we flat bet 1 unit on each bet then I get an EV of 25%. We no longer have a fixed odds bet so we cannot use the odds as a divisor to find the Kelly bet. Instead we can use the variance of 3.15 units, although for bet sizing it’s more correct (and easier) to use the average squared result of 3.21 units. That gives us a full Kelly bet of about $780.
That corresponds with the sort of approach I had started to consider, but I was not very clear about how to calculate the overall Kelly bet. I know it's defined as EV/Var for a hand of blackjack, but I got the impression from various sources that this is an approximation of some fiendishly-complicated, more general formula, and I wasn't sure if it was applicable here.

Could you clarify why it's better to use the average squared result than the variance?

Sonny said:
Let me explain how I got those numbers.

We will win the #6 bet 1/6 = 17% of the time. When we win, we get 6 units but we also lose 1 unit on the #1 bet. This gives a result of 0.17*(6-1) = 83.33% and an average squared result (ASR) of 0.17*(6-1)^2 = 4.17 units. For the #1 bet we get 0.17*(7-1) = 100% with an ASR of 0.17*(7-1)^2 = 6 units. We will lose 2 units 67% of the time for a result of 0.67*-2 = -133.33% and an ASR of 0.67*(-2)^2 = 2.67 units. Summing those will give us an EV of 50% per bet (which agrees with the sum of the two individual bets 0.33+0.17 = 0.5), or 50%/2 = 25% per round if we consider the combined bets to be 1 unit (which agrees with the average advantage for the two bets), and an ASW of 12.83 per bet, or 12.83/2^2 = 3.21 units per round. That gives us a full Kelly bet of around $10000*0.25/3.21 = $779.22 per round. Let’s call it $780.
That all makes sense (and thanks for taking the time to go through it).

However, ....

Sonny said:
That’s how much we’re planning to bet per round, but how will we divide that between the two numbers?

Our calculations so far have assumed that we split our money evenly on both bets. If we continue with this assumption then we would place $390 on each bet and earn $195 per round. That’s pretty good, but we can squeeze out a little more profit by betting more on the #1 bet. The problem is that we don’t want to bet too much on the #1 bet because that will negate our variance reduction. The whole reason we are making the #6 bet is to smooth out the fluctuations. If we bet too much on the #1 bet then winning the #6 bet won’t help us very much. For example, if we bet $70 on the #6 bet and $420 on the #1 bet then a roll of 6 will win $70*6 = $420 but the #1 bet will lose $420. We have only broken even. A roll of 1 would produce a win of $420*7-70 = $2870, which is still a big swing. We are betting more on the #1 bet so our EV jumps way up to almost 31%, but the lopsided bets keep the variance pretty high so we need to use a smaller unit size, which drops the full Kelly win rate down to about $152.

So we want to split up the bets in proportion to the advantages. Since the #1 bet represents 0.33/(0.33+0.17) = 55% of our profit I would consider betting 55% ($430) on #1 and 45% ($350) on #6 for an EV of just over $200 for similar levels of risk. That’s better than splitting the bets equally ($195), using a large spread ($152), or even playing the #1 bet only ($159). After fooling around with some other betting options, the 55/45 split seems like a pretty good choice. Maybe she ain’t the prettiest there is, but she’s as good as a guy like me is gonna find. Hopefully it's helpful to you.

-Sonny-
I have a couple of worries about this approach.

Firstly, there seems to be a chicken-and-egg issue. Which should come first, the total bet to then be subdivided, or the individual bets to then be scaled in some way? It seems to me that if we calculate an overall total Kelly bet, based on the assumption of betting equal amounts on each sub-bet, then that total bet is invalidated as soon as we start resizing the sub-bets.

When I started thinking about ways to analyse this, it occurred to me that any combined bet can be thought of as two bets in one: a bet of 1 unit on each, plus a smaller bet on one of the two.

i.e. if you bet the amounts N and M (where N > M), that's 2*M on the combined bet, and (N-M) on the single bet.

But then I had absolutely no idea what, if anything, I could do with that information. :grin:


Secondly, if we do start from the total bet, I'm not sure EV should be the sole factor to determine how to sub-divide it. That was the point I was making in this post: [post]155695[/post]. As a more extreme example, imagine if the two bets we were dealing with had massively different payoffs and probabilities, but similar EVs; say one paid 100:1 and the other even money. Surely you would want most of your money to go on the even-money shot, even if the 100:1 bet had a slightly higher EV?


I don't know if it is possible, but to re-iterate what I said at the start of the thread, what I am trying to work towards is some kind of formula to handle the general case of -
  • Two or more +EV bets available.
  • Varying degrees of correlation between the possible outcomes of those bets.
So here I have concocted the simplest case I can think of; just two bets, and they are mutually exclusive. Yet already it is too much for my poor brain to cope with! :(
 

London Colin

Well-Known Member
#17
Does the following make sense?

How about using the ratio of the individual Kelly fractions (278:476 in the example) to divide up what we consider to be a single unit. That is, rather than start out with the assumption that we bet 0.5 units on each bet (which we must later violate), assert that a unit bet will be split into 278/(278+476) = 0.37 and 476/(278+476) = 0.63.

We can then use these figures to calculate the overall EV and average squared result, yielding the total bet as an overall Kelly fraction. Then split that total in the predetermined ratio.

Code:
Prob.   Result              EV       ASR
0.17    (6*0.37 - 0.63)     0.26     0.42
0.17    (7*0.63 - 0.37)     0.68     2.74
0.67    -1                  0.67     0.67
                            ----     ----
                            0.27     3.82
So the total Kelly bet would be 10,000 * 0.27/3.82 = $712. And this would be divided as follows:

6:1 => 712 * 0.37 = $262
7:1 => 712 * 0.63 = $450
Dollar EV = $44 + $150 = $194

$712 is not too far from your total of $779.

There was a slight error in your calculations -
Since the #1 bet represents 0.33/(0.33+0.17) = 55% of our profit I would consider betting 55% ($430) on #1 and 45% ($350) on #6 for an EV of just over $200 for similar levels of risk.
That should be 66.67%, which gives bets of $519 and $260, and a dollar EV of $173 + $43 = $216, also not too different from my figures.

So for this example, my method would give a slightly higher percentage EV, but a smaller total bet and thus a smaller dollar EV. (Due to the bigger increase in ASR.)


However, it's a different story for the other example I cited ...

If the 6:1 bet is replaced by a 5:2 bet that either of two numbers is rolled, you get a bet with the same EV (2/6 * 5/2 - 4/6 = 0.17).

Using your method in this case, you get -
EV: 0.25 (naturally)
ASR: 2.19
Total Kelly Bet: $1,143

Since the EVs of the individual bets are unchanged, the split would be in the same ratio:
5:2 => 0.33 * 1,143 = $381
7:1 => 0.67 * 1,143 = $762
Dollar EV = $317

In contrast, my method gives -
Individual Kelly Bets: 5:2 => $667, 7:1 => $476. (we bet more on the lower-EV choice)
Therefore bet 667/(667+476) = 0.58 units on 5:2 and 0.42 units on 7:1.

Which gives -
EV: 0.24
ASR: 1.77
Total Kelly Bet: $1,335
5:2 => 1,335 * 0.58 = $779
7:1 => 1,335 * 0.42 = $556
Dollar EV = $315

[I notice that I have arrived at two higher amounts than the individual Kelly bets that would be made if only one bet were chosen. Can this be right, or does it imply that I've made a mistake somewhere? I've skipped a lot of the working in this post, but I used a spreadsheet to generate the figures; I could upload it if you are at all interested.]
 

sagefr0g

Well-Known Member
#18
London Colin said:
Suppose you have the opportunity to place multiple bets, all related to a single event.

As in roulette, for example, some bets represent mutually exclusive outcomes like red/black, specific numbers etc., while others have a degree of overlap (red/odd, even/3rd-dozen, etc.). The bets have a variety of payoffs and crucially, unlike roulette, some of them offer a postive EV.

How do you go about determining the optimal total amount to bet, and how to distribute this among the available bets? (Assuming that, if faced with just a single bet, you would be employing a utility function to assign a fraction of your bankroll to the bet, proportional to the EV.)

The more I've tried to ponder this, the more I've managed to confuse myself. Presumably the key is to identify the variance of various individual and combined bets, in search of the highest overall Certainty Equivalent, but this is very much where I start to venture outside of my mathematical comfort zone.:confused:
just curious, i don't know what the heck your talking about with this stuff, nothing on you, just confuses the heck out of me, thing is it sort of reminds me of something else i can't understand, Parando's Pardox :
http://www.eleceng.adelaide.edu.au/Groups/parrondo/
i'm sure i'm way off base but just through this in there for the fun of it.:confused::rolleyes:
 

London Colin

Well-Known Member
#19
sagefr0g said:
just curious, i don't know what the heck your talking about with this stuff, nothing on you, just confuses the heck out of me,
Is it the question that you don't understand, or just my faltering attempts to arrive at an answer myself?


sagefr0g said:
thing is it sort of reminds me of something else i can't understand, Parando's Pardox :
http://www.eleceng.adelaide.edu.au/Groups/parrondo/
i'm sure i'm way off base but just through this in there for the fun of it.:confused::rolleyes:
I don't really see much of a connection. Following the various links from that page, the clearest explanation I found of what is going on is this one ...
http://www.maa.org/mathland/mathtrek_3_6_00.html (Archive copy)

It seems to be a straightforward ratchet effect. You just need to imagine yourself playing the game to visualize the process -

Suppose you start out just playing Game 2, where you have to choose between flipping a coin that is biased in your favour, and one that is heavily biased against you, based on whether or not your current bankroll is a multiple of 3. Say you start out with a bankroll of 100; you are therefore flipping the advantageous coin, and your bankroll tends to rise. However, when you reach 102 (or 99 if you are unlucky), you are forced to switch to the disadvantageous coin, with a heavy probability of losing. So your profits fall back, and the net effect is a losing game overall.

But when you start switching backwards and forwards between Game 1 and Game 2, you effectively get to reset Game 2 to a random starting condition, each time you return to it, locking in some profit with each turn of the ratchet.

Or, to put it another way, you get to swap some of the very disadvantageous flips you would be forced to make in Game 2 whenever your bankroll is a multiple of 3, with less disadvantageous flips in Game1.
 

sagefr0g

Well-Known Member
#20
London Colin said:
Is it the question that you don't understand, or just my faltering attempts to arrive at an answer myself?



...
not to worry Colin, i'm thinking it's likely the question sir, way beyond the complexity i'm able or willing to grapple with. i should have just kept out of it.:eek:

i appreciate your consideration of the paradox thing.:)
 
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