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  #1  
Old 11-19-2004, 12:22 PM
e_fermat e_fermat is offline
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Default Bankroll theory applied to SNG\'s

I've read several posts and recommendations where people have suggested that they try to accumulate X number of buy-ins for their bankroll before moving up a level. I've seen numbers ranging from 15 or 20 on the low end to 40 or 50 on the higher end. Now, I may be ignorant as I'm still fairly new to poker and am only a marginal winning player but as someone from a blackjack/sports betting background, I was surprised that I haven't seen any references to the application of the Kelly Criterion for bankroll requirements for SNGs. (My search did find some discussion applied to limit HE, such as: Approximate Kelly Bankroll for LHE )

I think Kelly is even more applicable for the SNG setting than even limit ring games since each SNG is discreet event where you can estimate the probability of finishing ITM and the payouts and maximum losses are known in advance. So if the formula for Kelly is:

b=(p*(c-1))/(c-1)

where:
p is the probability of finishing ITM. (0<p<1)
c is the gross payoff (a multiple of stake) in case you win. (c>1)
b gives the maximum fraction of your current bankroll that should be used to buy into a single SNG

Then for someone who finishes ITM 35% of the time and has an equal distribution of 1,2,3 place finishes, then p=0.35 and we'll use c=2.73 for simplicity, a second place finish where vig is 10% (obviously this is conservative given a 1st place finish is worth more proportionately more than a second or third place finish).

Solving for b we get b=0.0257

Therefore, for a full Kelly 35% ITM player, they should risk no more than 2.57% of their bankroll on a given buy-in. A more conservative half Kelly player would risk no more than 1.29% of their bankroll. Translated to buy-ins, this equates to 39 buy-ins at full Kelly and 78 buy-ins at half Kelly.

For various playing, this would mean having a minimum bankroll at full Kelly (and half Kelly) of:

$10+1 = $429 ($858)
$50+5 = $2145 ($4290)
$100+9 = $4251 ($8502)

Any comments on this? Obviously, this is just the basic application of the Kelly Criterion and there are many other factors to consider since there are only a limited number of dollar amounts to play SNG's at. For instance, your bankroll fraction may indicate optimal play at the $75 level but there are no $75 SNG's. Also, it may not be practical to move around levels frequently since the gameplay is different at different levels, unlike blackjack for instance. I dunno, I thought I would just throw this out to see if anyone else had considered this approach as a way to reduce risk of ruin...
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  #2  
Old 11-19-2004, 12:46 PM
tigerite tigerite is offline
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Default Re: Bankroll theory applied to SNG\'s

I think your formula is wrong mate.. even if it's (p*(c-1))^2 / (c-1) I get 0.211925? Am I missing something?
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  #3  
Old 11-19-2004, 12:56 PM
Marcotte Marcotte is offline
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Default Re: Bankroll theory applied to SNG\'s

[ QUOTE ]
b=(p*(c-1))/(c-1)


[/ QUOTE ]

the right hand side of this formula reduces to 'p'. (ie. (c-1) cancels out).
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  #4  
Old 11-19-2004, 01:27 PM
stupidsucker stupidsucker is offline
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Default Re: Bankroll theory applied to SNG\'s

I am not certain where this formula came from, but it goes against what we have all believed for some time now.

Perhaps we have all been wrong, but.. RoR (risk of ruin) is rather small playing with only 30 buyins up till at least the 50s.

I am a big BR fan, and I think that most players are effected mentaly based on their BR. I feel way more comfortable when I have 50 buy ins online and more to back it up in the bank. I rarely have, but it feels good.
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  #5  
Old 11-19-2004, 02:08 PM
e_fermat e_fermat is offline
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Default Re: Bankroll theory applied to SNG\'s

Sorry, there should not be a bracket in the numerator on the c-1. The numerator should be just p*c-1. I think this addresses Tigerite's difference in calculated result.
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  #6  
Old 11-19-2004, 02:13 PM
hurlyburly hurlyburly is offline
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Default Re: Bankroll theory applied to SNG\'s

Solve the stuff in parantheses first, then divide (blue, red, green).

b=(<font color="red">p*</font>(<font color="blue">c-1</font>))<font color="green"> / </font>(<font color="blue">c-1</font>)


Thanks for sharing that formula, that is the first approach I've seen that has true merit, and can be applied to results at different levels equally to determine hourly rate expectations and variance. I've never seen any math backing up the 20x guideline so I never quite trusted it.

Been searching for something like this for a long time!
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  #7  
Old 11-19-2004, 02:34 PM
e_fermat e_fermat is offline
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Default Re: Bankroll theory applied to SNG\'s - Link

For those interested in the full background to Kelly, this is a great resource:

Thorp
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  #8  
Old 11-19-2004, 02:36 PM
adanthar adanthar is offline
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Default Re: Bankroll theory applied to SNG\'s

[ QUOTE ]
Then for someone who finishes ITM 35% of the time and has an equal distribution of 1,2,3 place finishes

[/ QUOTE ]

I just attempted to calculate it for a 40% ITM player with a 35-30-35 distribution of finishes and my results seem...odd.

c=(.35x4.55)+(.30x2.73)+(.35x1.82) [the second decimal point is approximate] or c=3.0485.

b=(.40*3.0485-1))/2.0485 = .1071

So, for our hypothetical 40% ITM hero, a full Kelly suggests a whopping 9 buyins. A half Kelly would make it 18.

Is my math wrong anywhere?
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  #9  
Old 11-19-2004, 02:54 PM
Paul2432 Paul2432 is offline
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Location: Bryn Mawr, PA USA
Posts: 374
Default Re: Bankroll theory applied to SNG\'s

[ QUOTE ]
I think Kelly is even more applicable for the SNG setting than even limit ring games since each SNG is discreet event where you can estimate the probability of finishing ITM and the payouts and maximum losses are known in advance. So if the formula for Kelly is:

b=(p*c-1))/(c-1) corrected

where:
p is the probability of finishing ITM. (0&lt;p&lt;1)
c is the gross payoff (a multiple of stake) in case you win. (c&gt;1)
b gives the maximum fraction of your current bankroll that should be used to buy into a single SNG

Then for someone who finishes ITM 35% of the time and has an equal distribution of 1,2,3 place finishes, then p=0.35 and we'll use c=2.73 for simplicity, a second place finish where vig is 10% (obviously this is conservative given a 1st place finish is worth more proportionately more than a second or third place finish).

Solving for b we get b=0.0257

Therefore, for a full Kelly 35% ITM player, they should risk no more than 2.57% of their bankroll on a given buy-in. A more conservative half Kelly player would risk no more than 1.29% of their bankroll. Translated to buy-ins, this equates to 39 buy-ins at full Kelly and 78 buy-ins at half Kelly.

For various playing, this would mean having a minimum bankroll at full Kelly (and half Kelly) of:

$10+1 = $429 ($858)
$50+5 = $2145 ($4290)
$100+9 = $4251 ($8502)

Any comments on this?

[/ QUOTE ]

I think your approach is reasonable but some sloppiness with the numbers is giving you misleading results.

First, a player with 35% in the money finishes equally divided between 1st, 2nd and 3rd is barely a winning player. This player will have an ROI of only 1.06. Most winning players here have an ROI of around 20-30%.

The second problem, is that your approximation of c=2.73 is way off. Using your numbers c=3.03 (in fact if c=2.73 then the player is not even a winning player).

If we redo the numbers so that:

1st = 15%
2nd = 15%
3rd = 10%

we get:

p=0.4
c=3.5

Plugging those numbers into the Kelly equation we get:

b=0.16

Hence you need a BR of ~ 6 buy-ins at full kelly.

I'm not sure this is entirely accurate because the use of the average value for c.

Here is a good link for a derivation of the kelly formula

Kelly Derivation

Paul
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  #10  
Old 11-19-2004, 03:12 PM
Paul2432 Paul2432 is offline
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Posts: 374
Default Re: Bankroll theory applied to SNG\'s

Your math is fine.

The OP had some errors in his math. See my other post in this thread.

Paul
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