Hello everyone,

I know this equation can be solved by numerical methods, using Excel, but have no idea how to work the program.

(95/2611)*exp(-12.6*x) + (85/2611)*exp(-8.7*x) + (54/2611)*exp(-6.68*x) + (64/2611)*exp(-5.06*x) + (53/2611)*exp(-3.64*x) + (62/2611)*exp(-2.42*x) + (66/2611)*exp(-1.4*x) + (2132/2611)*exp(x) = 1

For those curious, x represents a risk of ruin constant. I'm trying to find my risk of ruin for 45-man turbo SNGs. I want to solve for the constant directly to compensate for the skew positive nature of these tournaments--as opposed to assuming the outcomes are normally distributed about my win rate.

Thanks!

x = 0.080463525

You can represent x by a cell and guess values until it comes out to 1. That's basically what the solver does. If you go under Tools->Solver it's pretty self-explanatory. You make a cell 1 by changing another cell, and you can set the precision under options. I used goal seek at first, and the result wasn't very accurate, so I switched to the solver once I had a pretty good initial guess.

Even easier, just copy that equation exactly as you typed it into the box at wolframalpha.com and click the = button. It will solve it for you. No fuss, no muss.

And how did you derive that equation for risk of ruin? What are the details? Which of course is always the next thing to ask to be safe about the numerical answer anyway.


You can always plot it by the way in excel (in 0,1 range) to visually see what a good starting guess is and then go as BruceZ suggested or try the Mathematica version.


Also recognizing the answer ought to be small you can also do it using Taylor expansion (level 2) around 0. This gives a quadratic for x that you can solve and get something like 0.064 that you can then use as a new Taylor expansion point for the exponentials and obtain instantly another quadratic that is giving 0.0803. A third time is not needed at least to 2 digits accuracy but can be done for fun too. Of course none of this is needed, i am only suggesting how to kill time in an airport (with no interesting people around) if you have only paper and a calculator (lol) or a new laptop with no internet connection in a bus ride with no wifi. Finally you can also rederive some derivative approximation solutions for general functions F in the bus ride (assuming its not a bumpy one!)

http://en.wikipedia.org/wiki/Newton%27s_method

These days even some cheap $15 calculators have the ability to program functions into them so that you can then numerically get the answer by trial and error in less than 1 min after inputing the monstrous long expression of course first.

(kind of a mathematician's hunt in the wild with primitive weapons survivalist challenge)

Thanks for the help guys!

The risk of ruin (RoR) model that I am using is the one proposed by Bill Chen & Jerrod Ankenman in The Mathematics of Poker.

They present the RoR function as R(x) = exp(-αx) with the property that 1 = <exp(-αx)> which allows for the solving of the risk of ruin constant (α) directly.

The equation I asked how to solve:

(95/2611)*exp(-12.6*x) + (85/2611)*exp(-8.7*x) + (54/2611)*exp(-6.68*x) + (64/2611)*exp(-5.06*x) + (53/2611)*exp(-3.64*x) + (62/2611)*exp(-2.42*x) + (66/2611)*exp(-1.4*x) + (2132/2611)*exp(x) = 1

...was me trying to find my own risk of ruin constant based on a 2,611 game sample of my play at the $0.50 & $1.50 45-man turbo SNGs on PokerStars.

Now that I have that value, I will use R(x) = exp(-αx) to calculate my risk of ruin for bankrolls of various sizes. I want to develop a bankroll schedule that allows me to move up in stakes as quickly as possible, while keeping my risk of ruin very low.

Thanks again!

Let me understand better since i dont have the book. You play 0.5 and 1.5 45 player turbo SNGs and what is the prize structure and method to construct risk of ruin here? Dont you need to provide also your bankroll and your actual winrate and volatility (or much better the actual probabilities for various prizes) . Where are those in your expression?

Your risk of ruin depends on how much you start with and your probabilities to win various prizes. How can you know those probabilities? Are these coefficients your actual past records? I guess i am asking a little bit of clarification on how the equation is constructed.

He's using the same method described in sections 3-4 of this paper. That paper also appears in the book Optimal Play, Mathematical Studies of Games and Gambling which is a collection of very interesting mathematical papers which I'm sure you would enjoy. In his expression, exp(-x) is lambda. Each exponential corresponds to a lamda^X where X is a payout based on his past performances, and he is averaging these based on how many of each payout he won. The x he is solving for is alpha which is -ln(lamda) which can then be used to comptute a risk of ruin for a given bankroll B as exp(-alpha*B).

Nice, thanks. Yes i was thinking of a generalized risk of ruin problem with multiple probabilities for various steps back and forth. Lets see what this paper does and whether the full exact problem has a closed form solution (even if involving higher polynomial roots) in a manner derived at say Feller moreover the added complexity. At least the special version applied to poker tournaments has always the same back step and only the ones ahead are of variable depth and probability. I was wondering if the full problem has an exact solution (some recursive expressions that lead to characteristics etc) or if an approximation is all one can hope for. I will go over it and see what he does.

The risk of ruin formula that we normally use for ring games based on our win rate E and standard deviation sigma

ror = exp(-2EB/sigma^2)

is what they refer to as the equivalent coin tossing model, and I give a simple derivation of this here. That breaks down for highly skewed games like tournaments, so for those games, or games with multiple payouts in general, the method they present should be exact, but since there can be cases where we don't go exactly broke, you have to compute an effective bankroll by averaging over all possible sequences of wins and losses. It's not obvious to me how we can always do that easily, though often this isn't necessary for an accurate approximation. The rest of the paper comes up with various approximations of this based on consideration of the moments. Here is another paper that considers these moments.