[Ganchrow]

In case anyone's interested, following is the closed-form solution for simultaneous bet Kelly staking, given the simplest case where the single-bet Kelly stakes for each simultaneous bet are equivalent, all bets are uncorrelated, and the only bound on wagers is the size of the bankroll.

I've yet to work out the closed-form solution for the general case of correlation and differing single-bet Kelly stakes (if it even exists). That would obviously be considerably more difficult, and probably better left calculated by an optimizer.

If anyone's really interested in seeing the proof (not that I expect that), I could probably write it up. I've also created a Kelly calculator as proof of concept.

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Given n uncorrelated binary bets, we define the Kelly-optimal allocation as the set of weightings for each of the 2ⁿ-1 n-or-fewer-team parlays (where a single bet is considered a 1-team parlay) that can be created from the n-single bets, which maximizes the expected logarithm of the bankroll.

Let o_i = decimal odds on the i^th bet,
Let p_i = win probability of the i^th bet,
Let k_i = i^th single-bet Kelly stake = p_i + (1 - p_i)/(1 - o_i),

If k_i = k_j for all i,j on the interval [1,n],

then the Kelly-optimal weighting of each and every m-team parlay (as a percentage of the total bankroll), K_n^m, is given by:

       n
 Knm = ∑ combin(n-m, m-i) * k1+n-m * (-1)m-i
       i=n-m+1

[Dark Horse]

Err....

Interested, but is there a way to say this in plain English?

[Ganchrow]

Originally Posted by Dark Horse

Err....

Interested, but is there a way to say this in plain English?

I'm sure there is ... I just can't think of anything right now.

Basically, I'm saying that if you're looking at a bunch of ucnorrelated bets where all the single-bet Kelly stakes would be the same, then the simultaneous bet Kelly stake will be a collection of single bets and parlays where the weightings are the same for any two parlays of the same size. The summation above represents the simultaneous n-bet Kelly weighting (call it K_n^m) for all parlays of size m.

The linked spreadsheet might shed some light.

[Lucas]

I have turned off my antivirus... I am scared of Ganchvirus that causes that everytime I switch on my compy I will see only strange hieroglyphs

[Dark Horse]

LOL

--------------------------------
Thanks, Ganch. Much appreciated.

[Lucas]

OK, I downloaded it... 2 MB for excel file with 2 columns and 9 rows??

Ganch I hope you enjoy my porn anthology... But my passwords are on a list of paper

[Ganchrow]

This is the more general case where we relax the constraint that all single-bet Kelly stakes need to be equal. For the sake of sanity, the weightings are defined recursively.

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Given n uncorrelated binary bets, the "Kelly-optimal allocation" is the set of weightings for each of the 2ⁿ-1 n-or-fewer-team parlays (where a single bet is considered a 1-team parlay) that can be created from the n-single bets, which maximizes the expected logarithm of the bankroll.

Let o_i = decimal odds on the i^th bet,
Let p_i = win probability of the i^th bet,
Let k_i = i^th single-bet Kelly stake = MAX[(p_i*o_i-1)/(o_i-1), 0],

Define κ(n,m,{B}) as the sum of the Kelly optimal weights for all m-team parlays made up of all bets included the set {B}, then

                       n 
κ(n,m,{B}) = ∏ ki  -  ∑ κ(n,i,{B})
           i Є {B}   i=m+1

Example:

given:
k1 = 1%
k2 = 2%
k3 = 3%
k4 = 4%
k5 = 5%

κ(5,5,{1,2,3,4,5})  = (weighting of the 5-team parlay as % of bankroll)
  = k1*k2*k3*k4*k5 
  = 1%*2%*3%*4%*5%
  = 0.0000012%

κ(5,4,{1,2,3,4})  = (weighting of the 4-team parlay consisting of bets {1,2,3,4} as % of bankroll)
  = k1*k2*k3*k4 - κ(5,5,{1,2,3,4})
  = 1%*2%*3%*4% - κ(5,5,{1,2,3,4,5})
  = 0.0000228%

κ(5,4,{1,2,3,5})  = (weighting of the 4-team parlay consisting of bets {1,2,3,5} as % of bankroll)
  = k1*k2*k3*k5 - κ(5,5,{1,2,3,5})
  = 1%*2%*3%*5% - κ(5,5,{1,2,3,4,5})
  = 0.0000288%

κ(5,3,{1,2,3})  = (weighting of the 3-team parlay consisting of bets {1,2,3} as % of bankroll)
  = k1*k2*k3 - κ(5,4,{1,2,3}) - κ(5,5,{1,2,3})
  = 1%*2%*3% - κ(5,4,{1,2,3,4}) - κ(5,4,{1,2,3,5}) - κ(5,5,{1,2,3,4,5})
  = 0.00054720%

etc.

And now a non-recursive statement of the same. Please forgive the abuse of notation.

---

Given n uncorrelated binary bets, define the "Kelly-optimal allocation" is the set of weightings for each of the 2n-1 n-or-fewer-team parlays (where a single bet is considered a 1-team parlay) that can be created from the n-single bets, which maximizes the expected logarithm of the bankroll.

Let o_i = decimal odds on the i^th bet,
Let p_i = win probability of the i^th bet,
Let {k} = the set of all n single-bet Kelly stakes,
where k_i = i^th single-bet Kelly stake = MAX[(p_i*o_i-1)/(o_i-1), 0],

Define {P(k)} = the power set of {k}

Define {S({B},i)} = the set of all sets, {s} Є {P(k)} such that |{s}| = i, {S} ⊇ k_{B},
where k_{B} is the set of the single-bet Kelly weights associated with the elements of {B}

Define κ(n,m,{B}) as the Kelly optimal weight for the m-team parlay made up of all bets included in the set {B} (where |{B}| = m).

             n
κ(n,m,{B}) = ∑( (-1)i-m * ∏ kj)
             i = m              j Є {S({B},i)}

[Arilou]

So in practical terms, how does someone apply this to, say, a night of NBA action without spending way too much time on math?

[Sam Odom]

this aint fun anymore

[Ganchrow]

Originally Posted by Arilou

So in practical terms, how does someone apply this to, say, a night of NBA action without spending way too much time on math?

Write some software implementing either the above recursion or the methodology I outlined here.

[rjp]

Thanks

[Big Razorback]

What is this in VERY simple terms..

[Ganchrow]

Originally Posted by Big Razorback

What is this in VERY simple terms..

My Kelly calculator does these calculations for you.

[Big Razorback]

thats a hell of a lot easier to understand the main purpose without the derivitives and ECT...


I sent it to a friend that is into math to try to explain to me...lol

[rjp]

Just going over this again, and it'd be really sweet if you could take into account a bet already made, such that say I find a single +EV bet that I wager on at the optimal wager size, and then later I find say three +EV bets I wish to bet on, while the first single +EV bet is still pending. Obviously you can't re-bet the single one at a lower amount, but it'd be nice to take that into account too.

[Ganchrow]

Originally Posted by rjp

Just going over this again, and it'd be really sweet if you could take into account a bet already made, such that say I find a single +EV bet that I wager on at the optimal wager size, and then later I find say three +EV bets I wish to bet on, while the first single +EV bet is still pending. Obviously you can't re-bet the single one at a lower amount, but it'd be nice to take that into account too.

When you get into these more complicated boundary problems, you need to start using a nonlinear optimizer to approximate glabal solutions.

I think if you search around you can find a post where I outlined this procedure as it relates to hedging and line movements. And some point I'll write an article explaining it in the more general case.

[rjp]

Cool, I'll start poking around.

[rjp]