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

[London Colin]

Originally Posted by Kasi

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]

Originally Posted by London Colin

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]

Originally Posted by London Colin

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]

Originally Posted by Kasi

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.

Originally Posted by Kasi

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]

Originally Posted by London Colin

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...

Originally Posted by London Colin

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.

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]

Originally Posted by Sonny

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.

Originally Posted by Sonny

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.

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.)

Originally Posted by Sonny

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?

Originally Posted by Sonny

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, ....

Originally Posted by Sonny

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.


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: http://www.blackjackinfo.com/bb/showthread.php?p=155695. 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]

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.

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]

Originally Posted by London Colin

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.

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.

[London Colin]

Originally Posted by sagefr0g

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?

Originally Posted by sagefr0g

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.

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

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]

Originally Posted by London Colin

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.

i appreciate your consideration of the paradox thing.