[Ganchrow]

I saw Peep mention to you a simple formula that can be used to figure out your chnaces of going broke. Do you know anything about that?

I assume you're referring to this statement by Peep:

If we knew that we could win say 30% of our bets at average odds of 9/2, [the formula] let us know what our chance of going broke was if we bet ____ percent of our bankroll.

The simplicity of the formula to which he's referring, however, is rather muted by its general lack of utility.

The risk-of-ruin approximation (it's only an approximation because it ignores the higher-order moments of a betting distribution and then appeals to the Central Limit Theorem, which will only be true asymptotically) is given by:

R ≈ e^{-2μ(x) / σ2(x)}

where R represents our risk-of-ruin, μ our expectation, and σ our standard deviation (the latter two both given in units of initial bankroll and as a function of bet size, x).

Note that even as an approximation this is only really applicable for constant μ and σ. While this assumption remains robust across multiple classes bets of varying size, it won't hold for a player sizing his bets as a percentage of bankroll. Furthermore this formulation refers to a player's probability of going broke before the End of Time, rather than of going broke before reaching a certain goal. As such, it's of no use to the recreational bettor (whose probability of eventually going broke approaches 100%). We see this by noting that when μ < 0, the exponent goes positive and R would be > 1, a logical impossibility.

So for a bet size of x (as a fraction of initial bankroll) we have:

R ≈ e^{-2μ / (x * σ2)}

Recall that with binary outcome wagers, if p represents the win probability, d the decimal payout odds, and x the bet size, then we have:

μ(x) = x * (p*d - 1)
σ²(x) = x² * (d - μ/x - 1) * (1 + μ/x)
σ²(x) = x² * d² * (1-p) * p

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So in Peep's example of a 9/2 dog winning at a 30% clip we'd have for a unit bet size:

μ = 11/2 * 30% - 1 = 65%
σ² = (11/2)² * 70% * 30% = 6.3525
e^{-2μ / σ2} = e^{-2 * 65% / 6.3525} ≈ 81.494%

So our risk-of-ruin approximation would be given by

R ≈ 81.494% ^1/x

So for a player betting 5% of bankroll, risk-of-ruin would be 81.494%^1/5% ≈ 1.669%, while for a player betting 25% of bankroll it would be 81.494%^1/25% ≈ 44.106%.

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Note that we can restate the above formula as follows:

R = e^{-2μ / (x * σ2)}
ln(R) = -2μ / (x * σ²)
x = -2μ / ( ln(R) * σ²)
x = -2*(p*d - 1) / ( ln(R) * d² * (1-p) * p)

Which tells us our maximum bet size if we're willing to accept a given risk-of-ruin.

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So using the above example of a 9/2 dog winning at 30%, if we're willing to accept a risk-of-ruin no greater than R*:

x ≤ -20.464% / ln(R*)

So for R* = 5%, we'd have:

x ≤ -20.464% / ln(5%) = 6.831%

Implying we'd need to select a bet size of no more than 6.831% of initial bankroll in order to keep our risk-of-ruin at 5% or less.

[gamble4heisman]

Ganchrow,

i was looking back through old threads and i had a few questions about this one. i get lost when you combine steps in the following:

μ = 11/2 * 30% - 1 = 65%
σ2 = x2 * (11/2)2 * 70% * 30% = 6.3525.

i don't understand where the 11/2 comes from when the formulas above in the main post are considered. i am looking to calculate a risk of ruin given a certain winning percentage and percentage of bankroll risked and also to calculate an idea bet size given a desired risk of ruin and winning percentage at stated average decimal odds. thanks.

G4H

[Ganchrow]

Fractional payout odds of 9/2 were given in the initial question. As decimal odds = fractional odds + 1, so decimal odds = 9/2 + 1 = 11/2.

[gamble4heisman]

thanks for the help, i think now i'm at the last part and am of course confused again. 'σ2 = x2 * (11/2)2 * 70% * 30% = 6.3525' when i solved with a value of x that i thought correct to use (fraction of bankroll wagered) i got an undefineable answer. i took your answer and solved for x and got 1. i'm pretty confident x is 1 but i'm curious as to why, can you explain? thanks.

G4H

[Ganchrow]

My above post is a bit unclear.

x represents bet size, expressed as a fraction of total bankroll.

When I wrote R = e^{-2μ / (x * σ2)}, I referring to μ and σ² off a unit-sized bet.

So:

R = e^{-2μ / (x * σ2)}

R = e^{-2*(p*d-1)/(x*d2*(1-p)*p)}

In other words, you should ignore the initial x² in the Peep example (since removed from the above post) as it was erroneous.

[gamble4heisman]

Thanks!! I got it now, did all the calculations. I am going to make a spreadsheet for different winning percentages and odds percentages now. thanks for the help on an old thread topic, i've learned a lot from you. thanks again.

G4H