[Ganchrow]

I've for some time been a bit embarrassed by my rather egregiousness overcomplication of the mathematical curiosity that is the unconstrained simultaneous independent event Kelly solution, which I detailed over 3 years ago in this thread.

So to rectify in brief:

Given N independent events, x₁, x₂, ..., x_N, with corresponding single bet Kelly stakes of κ₁, κ₂, ..., κ_N, the unconstrained Kelly solution (for any Kelly multiplier > 0) consists of the 2^N-1 parlays such that the wager on a given parlay comprised of all events in set S would be:

[nbtable] [tr] [td] [/td] [td] κ_i[/td] [td]   [/td] [td]   [/td] [td] × [/td] [td]   [/td] [td] [/td] [td] (1-κ_i) [/td] [/tr] [/nbtable]

So given, for example, events A, B, C, D, and E, with corresponding single-bet Kelly stakes of κ_A, κ_B, κ_C, κ_D, and κ_E, then the Kelly stake for the 1-team parlay consisting of only bet A would be:

κ_A * (1-κ_B) * (1-κ_C) * (1-κ_D) * (1-κ_E)

While the Kelly stake for the 3-team parlay consisting of bets A, B, and C would be:

κ_A * κ_B * κ_C * (1-κ_D) * (1-κ_E)

Much simpler, no?

Now if only I had used this logic in my old-school JavaScript Kelly calculator, it would run a hell of a lot faster. Well, c'est la vie.

If anyone's interested in the C-code for this (and or/a DLL linkable from Excel) let me know and I'll post it here. (Although as I've said, this is really more of a curiosity than anything else.)

[trixtrix]

ganch, i'll give 450 more SBR pts if you can deliver that halves/quarter .5 pt calculator you've been working forever and a day on

[Ganchrow]

ganch, i'll give 450 more SBR pts if you can deliver that halves/quarter .5 pt calculator you've been working forever and a day on

That's sweet of you, but as I'm no longer in SBR's employ I'll have to pass for the time being.

[Flying Dutchman]

So, Ganch. Are you back?

...as in the semi-Mod position you were "once" in?

In any regard, welcome back. We really, really missed you.

[Flying Dutchman]

Whoops, I guess we posted about the same time...

...anyway, glad to see you back regardless.

[Ganchrow]

So, Ganch. Are you back?

...as in the semi-Mod position you were "once" in?

In any regard, welcome back. We really, really missed you.

Nope ... just wasting time that would be better spent working ...

[trixtrix]

ganch 4 mod!!11

pls don't desert us again

[JR007]

Dude....you know your stuff !!!!!!!!!!!!

[Art Vandeleigh]

I've for some time been a bit embarrassed by my rather egregiousness overcomplication of the mathematical curiosity that is the unconstrained simultaneous independent event Kelly solution.

No need to be embarrassed sir. People here will let you know when your egregerious overcomplications are worthy of a red face, this is not one of those times.

Hope all is well, glad to see you back at SBR.

[Wrecktangle]

Ganch, we are busy gathering all your posts into a little red book which we will title: Sayings of Chairman Ganchrow. And then when we venture into Players Talk, we will hold it out in front of us to keep JJ's flesh eaters at bay.

[Peeig]

Ganch should start his own forum and charge $$$ to join

[statnerds]

yes!!!!!!!!!!!!!!!!

read many of your posts from the time long ago sir. didn't even read this thread yet. just count me among the extremely excited that you may have possibly potentially returned.

[durito]

yes!!!!!!!!!!!!!!!!

read many of your posts from the time long ago sir. didn't even read this thread yet. just count me among the extremely excited that you may have possibly potentially returned.

lol

[Chris_B]

I signed up to this forum just to say that the work behind this post by Ganchrow is incredible.

I've being trying to find a closed form solution for independent, simultaneous events (but without the parlay option) for sometime now. I got my first class in maths at uni and i'm certainly no slouch, but this problem is so messy and unpleasant to navigate that i'm not suprised it has eluded you for 3 years in a closed form.

I would really like to know how you solved this. You can see that it has most of the desired properties, although I would have thought that if exactly one individual Kelly stake is 1 - for a certain event - then the all the parlay Kelly stakes would be zero except for a single parlay of size 1 for the certain event. But as you've shown this wouldn't then maximise growth.

I've tried multiple times to work through a similar problem (excluding parlays), for a small number of independent, simultaneous events, but there are pages full of terms and I just never had the patience to go through it all. Every time I try it again I seem to forget just how many damn terms there are. It's fairly straight forward computationally and so I put it off and never thought I would see a closed form solution...

It's just so elegant and simple, I can't believe out of all the mess something so simple comes out the other side. Well I just wanted to say that although I didn't get to solve this one myself, it's very gratifying to finally see it.

One last thing, if it's possible to post the method you used to solve this problem then I would really appreciate it, as I havn't been able to see anything other than by brute force only...

Really great work!

[20Four7]

Ganch should start his own forum and charge $$$ to join

I would be there in a minute. WB Ganch

Hopefully this gets posted as I am on Post Review......

[Chris_B]

I managed to find a method used to solve this problem in the paper http://www.afaanz.org/research/AFAANZ%200676.pdf

They introduce a lot of notation, which is very natural and helps to break down the otherwise enormous mess of the situation into a compact, succinct structure that makes the theory much more approachable.

If your method is better than this Ganchrow I would really like to see it, but their approach seems to be quite powerful and comprehensive and definately worth a look.

[boxcar]

THANK YOU! This is awesome for a stats-deficient person like myself.

[bztips]

ok, a really dumb question. In Ganchrow's post, when he talks about a "parlay", does he literally mean it in the usual sense where you must win all events in the parlay in order to cash? Or is he simply referring to n simultaneous independent wagers which are won or lost individually?

[Data]

κ_N, the unconstrained Kelly solution (for any Kelly multiplier > 0) consists of the 2^N-1 parlays such that the wager on a given parlay comprised of all events in set S would be:

[nbtable] [tr] [td] [/td] [td] κ_i[/td] [td]   [/td] [td]   [/td] [td] × [/td] [td]   [/td] [td] [/td] [td] (1-κ_i) [/td] [/tr] [/nbtable]

OMFG

7of9 called. She said that was the second most beautiful thing she has ever seen.

[RickySteve]

I managed to find a method used to solve this problem in the paper http://www.afaanz.org/research/AFAANZ%200676.pdf

They introduce a lot of notation, which is very natural and helps to break down the otherwise enormous mess of the situation into a compact, succinct structure that makes the theory much more approachable.

If your method is better than this Ganchrow I would really like to see it, but their approach seems to be quite powerful and comprehensive and definately worth a look.

It appears to be the same solution.

I'm not sure why they're so concerned about fixed % vig in that paper since that doesn't actually exist.

[saintjames]

i dont get it...whats the purpose for the kelly calculator & can it give me the best value when i bet on a team with multiple pointspreads to find out which spread gives me the highest value on my wager

[Dash2in1]

Am I missing something here?

Suppose you have a series of simultaneous independent bets where you have odds of 2 and the real probability of winning is 80% (unrealistic, i know).
In the case of a single bet the Kelly criterion states that you should bet 60% of your bankroll.

My expectation would be, that as the amount of possible bets approaches infinity, the total amount of your bankroll waged should go up and ultimatly reach all of your bankroll. However, I find the opposite to be the case if you follow the solution given above.

1 bet - 1 bet of 60% of BR
2 bets - 2 bets of 24% = 48% of BR
3 bets - 3 bets of 9.6% = 28.8% of BR

My guess would be that my expectation of what should be happening is wrong, but what's wrong with it?

[Dash2in1]

...

[bookiebust]

I attempted to quote the |-| symbol in your formulas. I couldn't capture it. This is not kappa I'm referring to. What is the mathematical terminology for the |-| symbol used in your formulas? The parallel vertical bars connected by a single horizontal line at the top. Thanks. --

[Dash2in1]

@bookiebust: It's the product of a sequence: See http://en.wikipedia.org/wiki/Multipl...al_Pi_notation

[bztips]

Am I missing something here?

Suppose you have a series of simultaneous independent bets where you have odds of 2 and the real probability of winning is 80% (unrealistic, i know).
In the case of a single bet the Kelly criterion states that you should bet 60% of your bankroll.

My expectation would be, that as the amount of possible bets approaches infinity, the total amount of your bankroll waged should go up and ultimatly reach all of your bankroll. However, I find the opposite to be the case if you follow the solution given above.

1 bet - 1 bet of 60% of BR
2 bets - 2 bets of 24% = 48% of BR
3 bets - 3 bets of 9.6% = 28.8% of BR

My guess would be that my expectation of what should be happening is wrong, but what's wrong with it?

You need to go back and read Ganch's original post/comments that he linked to above. I (and probably most others) have long had a fundamental misunderstanding, I think, of the "simplification" provided here.

Long and short: there's nothing wrong with your calculations per se, but the rub is that Ganch's formulas are optimal assuming you bet not only the individual events ("1-team parlay" in his terminology) but ALSO all of the multiple-team parlays!! So in your example of 2 simultaneous independent events, you would need to bet .6*(1-.6)=.24 on each single, but also .6*.6=.36 on the two-team parlay, giving a total bet of 84% of payroll.

Sort of a depressing realization, really, since in the most common situations people are not going to be betting all the different parlay combinations if they have a handful of simultaneous individual plays. Looking on the bright side, you're really going conservative (from a Kelly standpoint) by not betting the multi-team parlays, which is probably a good thing. Of course, you're probably also giving up a good amount of your expected bankroll growth.

[strixee]

Does anyone have source code of this calculator or the paper mentione by Chris_B?

[buby74]

I have worked out a pessimistic sub-optimal closed form no parlay simultaneous Kelly formula which is easy to calculate even for 20 or so bets.

1. For each game work out the individual Kelly fraction k
2. Then work out the bankroll ratio for each bet which is k/(1-k)
3. Sum the bankroll ratios for all the simultaneous bets call this S.
4. The proportion of your bankroll to bet on all the games combined is S/(1+S)
5. The proportion you bet is then allocated to the games in proportion to their bankroll ratios

Example

1. you want to make bets of 10% 20% 30% and 40% on four games
2. the bank roll ratios are 0.11, 0.25, 0.43, 0.66
3. the total is 1.456
4. so the fraction of the bankroll to bet is 1.45/(1+1.45) which is 59.3%.
5. Which is then divided up in the proportion to the bank roll ratios giving 4.5% 10.2% 17.4% and 27.1%

I notice that the resulting bet sizes are always a fixed percentage higher than the single bets in Ganchrow’s formula. Although I haven’t figured out why.
The bet sizes will be less than the true optimum bet sizes I get using solver (at least for three bets) also these true optimal bet sizes are not proportional to the single bets in ganchrows formula.
For very small initial Kelly bet sizes and/or with a few teams this approach is too pessimistic and expected growth is often less than if you used the individual Kelly with no adjustments. But for betting a lot of games and/or high initial Kelly stakes it is close to the optimum no parlay bet sizes Growth rate (as calculated using solver) although the Ganchrow method is superior to both as it includes parlays.
I devised this formula by working out the stake for a given bet if all the other simultaneous bets lost (hence it is a pessimistic approach) which means you would stake less than the original Kelly amount as your bankroll is smaller but then each other bet gets reduced in turn which means you can then start increasing the bet size again. This can be solved with simultaneous equations but then I stumbled across the relationship with bankroll ratio which gives the same answer. I haven’t done the math to prove the link between the simultaneous equations and the bankroll ratio approach.
So this is not the perfect answer to ChrisB’s question but it starts to address it.

[brebbles]

You need to go back and read Ganch's original post/comments that he linked to above. I (and probably most others) have long had a fundamental misunderstanding, I think, of the "simplification" provided here.

Long and short: there's nothing wrong with your calculations per se, but the rub is that Ganch's formulas are optimal assuming you bet not only the individual events ("1-team parlay" in his terminology) but ALSO all of the multiple-team parlays!! So in your example of 2 simultaneous independent events, you would need to bet .6*(1-.6)=.24 on each single, but also .6*.6=.36 on the two-team parlay, giving a total bet of 84% of payroll.

Sort of a depressing realization, really, since in the most common situations people are not going to be betting all the different parlay combinations if they have a handful of simultaneous individual plays. Looking on the bright side, you're really going conservative (from a Kelly standpoint) by not betting the multi-team parlays, which is probably a good thing. Of course, you're probably also giving up a good amount of your expected bankroll growth.

Apologies for replying to an old topic (and with my first post I think), but I only just came over this beautifully simple solution to what could otherwise be a very complex and computationally expensive problem once the number of independent events starts growing. I'm keen to see if this solutions extends to parlaying independent events with >1 exclusive outcome bets with each other (but obviously only including one outcome per event in each parlay) - I'll do some work on this and report back any findings.

Regarding the query of Dash2in1 above, it's easy enough to find a solution for the total bank staked across all parlays, for the special case where the single-bet Kelly stake in each independent event is the same. The formula is:

Total Bank staked across all parlays = 1 - (1 - K_i)^N where:

K_i is the (identical) single-bet Kelly stake for each event (ie, K₁ = K₂ = ..... = K_N)
N is the number of independent events.

So with the example previously mentioned, the total bank staked for each number of events is:

1 event: 60%
2 events: 84%
3 events: 93.6%
4 events: 97.44%
5 events: 98.98%

In this example the total bank gets close to 100% reasonably quickly.

Hope this solutions proves useful for some people and perhaps re-ignites some discussion on this thread. I've only recently graduated from simple single-kelly bets to mutually exclusive and dependent outcomes (eg, multiple alternate lines in the same match), so finding threads like this is after scouring numerous sites is amazing.

[the_fredrik]

Nice brebbles, I would also like to see this beautiful thread come alive again!

I am trying to figure out the solution for the "Simultaneous bet Kelly staking" like the one Ganchrow solved, but not unconstrained - I have constraints as to how much I can bet on the singles, so I guess I should bet more on the doubles instead?

Any ideas on how to approach this problem? Numerically is fine, I don't need a beautiful closed form...

Does anyone know whether it is dealt with in the lost legendary "Generic Kelly Criterion Spreadsheet for Excel"?

Any help is greatly appreciated!