(NOTE: Relates to the SNE launches a petition against Spin & Go format thread - see this post & this post)

Hey guy, is this the correct way to calculate the Kelly bet-fraction when there are multiple outcomes:

3-handed "Winner Take All" format


P(1st) = 0.36, then optimal bet fraction is 0.04 = 4% of bankroll.
P(1st) = 0.35, then optimal bet fraction is 0.025 = 2.5% of bankroll.
P(1st) = 0.37, then optimal bet fraction is 0.055 = 5.5% of bankroll.

* This agrees exactly with the standard Kelly Criterion formula.

So, using the same idea (ie: find the bet fraction that maximizes exponential bankroll growth):

Simplified "Spin and Go" SNG (ie: assumes winner take all for top 3 prize-levels)


P(1st) = 0.36, then optimal bet fraction is ~0.0045 = 0.45% of bankroll.
P(1st) = 0.35, then optimal bet fraction is ~0.0035 = 0.035% of bankroll.
P(1st) = 0.37, then optimal bet fraction is ~0.013 = 1.3% of bankroll.

I was wondering about this the other day in relation to the UK lottery (where you get smaller prizes for matching 3/4/5 numbers as well as the 6-number jackpot), so if this is the correct method then I should be able to answer that question too...

Juk

EDIT: I should add that I know the "3-handed WTA" version assumes no rake and the real "Spin and Go" SNG format has a slightly different prize structure for the top 3 prize-levels - just want to keep it simple and know this is the correct way to do calculate it before doing a proper comparison over a more realistic set of 1st percentages...

For your first calculations that's just a 2-outcome wager and your answers are right. For 2 outcomes, f = p - (1-p)/odds

There's no general formula for N outcomes, so you have to do it the way it looks like you're doing. If the probabilities are p's and the payouts are y's and the fractional stake is f:

ln(growth) = p₁*ln(1+y₁f) + p₂*ln(1+y₂f) + ...

Then set the derivative to 0 (or use a numerical solver) to get the maximum.

I'll proofread your 2nd image another time but it looks good at a quick glance.

As for a hypothetical +EV lotto with similar odds as the UK lotto, the kelly bet would be a ridiculously small percentage of your bankroll (a fraction of a penny).

The 2-outcome formula can be expressed: f = edge / odds. So if to get a certain edge you need 200 million to 1 odds, right away you know f will be tiny.

Sorry, was in a rush and my post above maybe didn't make much sense. Here is what I'm asking, using the similar terminology as on the Kelly Criterion wiki page:

We have a set of outcomes, each with an associated probability and bankroll change:



The expected bankroll growth exponent is:



So, to solve numerically:



Or, to solve analytically:



_______________________________________

So for the "3-handed winner take all with no rake" example:

p = {1-0.36,0.36} = {0.64,0.36}
b = {-1,3-1} = {-1,2}

f(x) = 0.64*ln(1+(-1*x)) + 0.36*ln(1+2*x) = 0.64*ln(1-x) + 0.36*ln(1+2*x)

f'(x) = d/dx(0.64*ln(1-x)+0.36*ln(1+2*x)) = 0.72/(2 x+1)+0.64/(x-1)

and solving:

f'(x) = 0

gives:

x = f* = 1/25 = 0.04

and this agrees with the Kelly formula from the wiki page:



f*= (2*0.36-0.64)/2 = 1/25 = 0.04

_______________________________________

So, is this correct for the multiple outcomes case?

Juk

Thanks, yeah that was what I was doing (ignore the post above as I was typing that when you posted this reply).

Yeah, I've worked this out for just the 6-number jackpot before and found the tiny value, but was interested to see how things would change when you factored in the £25 for 3-numbers, £100-ish for 4-numbers, etc.

Interestingly, in the UK lotto it is actually possible to get an EV of about £1:20 per £1 (assuming you don't buy loads of tickets and effect the pool though...) - by playing uncommonly chosen numbers only on Wednesday "rollovers" - it will be interesting to see how much all the smaller prizes increase the bet fraction.

Juk

When you do that, I'll be interested to hear the result.

Betting >= 2x kelly ensures that you go broke, so if it costs 1 euro to play and the kelly bet is <= half a euro (for you with your bankroll), the game might still not be worth playing! Probably don't need Kelly to tell you that, though, because you can just use multinomial distribution to see how far you'll usually fall before eventually hitting a big prize. If you're usually going broke before hitting, that means it's a bad investment if you have log utility.

(X-Post from NVG thread) Here are the results with 4% rake included for each game type, for 1%, 2%, 3%, 4% and 5% ROI players:



So:

1% ROI player needs P(1st) = 0.350694
2% ROI player needs P(1st) = 0.354167
3% ROI player needs P(1st) = 0.357639
4% ROI player needs P(1st) = 0.361111
5% ROI player needs P(1st) = 0.364583

1% ROI player's optimal bankroll fraction for 3-handed "winner take all" SNG = 0.0053185 (~188 buyins)
2% ROI player's optimal bankroll fraction for 3-handed "winner take all" SNG = 0.0106388 (~94 buyins)
3% ROI player's optimal bankroll fraction for 3-handed "winner take all" SNG = 0.0159576 (~63 buyins)
4% ROI player's optimal bankroll fraction for 3-handed "winner take all" SNG = 0.0212764 (~47 buyins)
5% ROI player's optimal bankroll fraction for 3-handed "winner take all" SNG = 0.0265952 (~38 buyins)

1% ROI player's optimal bankroll fraction for 3-handed "Spin and Go" SNG = 0.00047093 (~2123 buyins)
2% ROI player's optimal bankroll fraction for 3-handed "Spin and Go" SNG = 0.00140016 (~714 buyins)
3% ROI player's optimal bankroll fraction for 3-handed "Spin and Go" SNG = 0.00302398 (~331 buyins)
4% ROI player's optimal bankroll fraction for 3-handed "Spin and Go" SNG = 0.00533822 (~187 buyins)
5% ROI player's optimal bankroll fraction for 3-handed "Spin and Go" SNG = 0.00816306 (~123 buyins)

1% ROI player's optimal bankroll fraction increase ratio = 0.0053185/0.00047093 = ~11.3x
2% ROI player's optimal bankroll fraction increase ratio = 0.0106388/0.00140016 = ~7.6x
3% ROI player's optimal bankroll fraction increase ratio = 0.0159576/0.00302398 = ~5.3x
4% ROI player's optimal bankroll fraction increase ratio = 0.0212764/0.00533822 = ~4.0x
5% ROI player's optimal bankroll fraction increase ratio = 0.0265952/0.00816306 = ~3.3x

I'll see if I can re-do this with the top 3 prize-levels split 80/10/10 tomorrow (got killer backache from sitting infront of computer today and off to chill for a bit now...).

Juk

I'll post in here when I've got the result (prolly tomorrow or Monday now though).

Also, big thanks again for the help! I've tried to work out how to do this once before (from a VERY badly translated book on sports-betting...), but couldn't quite get it.

Juk

Yes, the differentiation idea is correct for multiple outcomes too (i.e. you only need to solve an equation f'(x)=0, with the left-hand side being a sum of fractions of the form p/(x+a); it will give the same answer as finding the maximum of the logarithmic sum, but will speed the computation up). I was going to write an Excel sheet for this (to use its solver) a few days ago (had done it for Full Tilt JP SnGs, but not updated it to include the stars.com Spin & Go payout structure). Thanks for the job!

Also, rake, rakeback and life expenses can be accounted for by changing the amount lost when losing a tourney, e.g. with 4% rake and 25% rakeback (45% rakeback minus life expenses that are worth 20% of rake, in the example of a single 3x Supernova), it will be 0.99 BIs instead of 1 BI (because the rakeback with deducted expenses, which is 1%=0.01 of the BI here, is accrued independently of the tourney outcome).

^ Oops, excuse my faulty brain... Life expenses matter in risk of ruin calculations, but not in Kelly management - they're constant (don't depend on which stake is played and contribute equally to the utility function being maximised at all points), while rakeback and winnings do.

So, in my above example, the money lost when losing a Spin & Go is 0.982 BI (because rakeback is 0.018 BI).

Also, the set of stakes offered is discrete, so it would be more interesting to determine 'cutoff bankrolls', e.g. the bankroll size point above which playing $30s makes more sense with $15s (note that the ROI at the $30s will most likely be smaller than at the $15s). To that extent, just solve an inequality whose left-hand side is the EV of the logarithm of the roll size (p1*ln(1+b1x) + p2*ln(1+b2x) + ...) with payouts and probabilities (based on the ITM) at the $30s, and the RHS is a corresponding expression for the $15s.

The continuous Kelly criterion also makes sense - it helps determine whether a player should sell action on specified terms, and if yes, how many %s of it; but to those who have an allergy to being staked or aren't chasing a VIP level and don't see sense in entering a less soft higher buy-in level, it's rather useless.

Of course these approaches can be combined: the LHS can be written for a fixed BR but a variable % of action sold at the $30s (i.e. if one has 200 BIs for the $30s, we can write 0.005*y, where y is the fraction of the action to remain unsold, from 0=0% to 1=100%, instead of x - the fraction of the roll to be risked), and the RHS - for playing $15s on one's own (where we'll write 0.005/2=0.0025 instead of x in this example because the BI is twice smaller).

E.g. for 3-handed non-lottery tourneys with 4% rake, 45% RB, 35.5% ITM at the $30s, 36% ITM at the $15s, when the player has a $375 roll:

0.64*ln(1-0.982*0.08y)+0.36*ln(1+(2.88-0.982)*0.08y) = 0.645*ln(1-0.982*0.04)+0.355*ln(1+1.898*0.04)

There are 2 roots - y~0.03 and y~0.71, which means that the optimal % is somewhere in between. If there was only one root, then playing 100% at the $30s on one's own would be optimal and we'd then have to consider selling action at the $60s instead.

To find the optimal y, we can differentiate the LHS by y and equate it to 0, pick the root that is between 0 and 1.

-0.982*0.64/(1-0.982*0.08y) + 1.898*0.36/(1+1.898*0.08y) = 0
y~0.3675
i.e. leave 36.75% to oneself and sell 63.25%.

^ Sorry, I've confused the ITM numbers

0.645*ln(1-0.982*0.08y)+0.355*ln(1+(2.88-0.982)*0.08y) = 0.64*ln(1-0.982*0.04)+0.36*ln(1+1.898*0.04)

No real roots! I.e. it's better to play the $15s.

Now let's assume we have a $1500 roll.

0.645*ln(1-0.982*0.02y)+0.355*ln(1+(2.88-0.982)*0.02y) = 0.64*ln(1-0.982*0.01)+0.36*ln(1+1.898*0.01)

y~1.085>1. But alas we can't sell our action short, and again, playing $15s is better.

And it appears that, with any bankroll size, selling action at the $30s is not a viable option - it's optimal to either play $30s on one's own or play $15s on one's own!

It happens because of the disparity in ROIs - even at the Kelly roll for the $30s (which is ~$1389), playing the $15s with the bigger ROI is slightly preferable!

The cutoff bankroll is ~$1620 (i.e., above it, $30s should be played, but the relative BR growth rate will still not be as big as if the roll was $1389):

0.645*ln(1-1.964x)+0.355*ln(1+3.796x)=0.64*ln(1-0.982x)+0.36*ln(1+1.898x)
has roots x=0 (obv) and x~0.009262
15/0.009262~1619.52