Roulette Wheel Game

[IAS_LLC]

I don' understand. Why are you not accounting for the probability of a loss?

The expected value for the game is :
E = (-1*.02 +0.03*0.98)*100 = +0.94% /per roll.

Again, the expected value isn't really relevant since you may get wiped out by a single roll.

I'd never bet my entire livelihood on a single roll.

 

[nonlinear5]

The expected value for the game is :
E = (-1*.02 +0.03*0.98)*100 = +0.94% /per roll.

This is not quite right, either. Your calculated expectancy is positive, while the "intuitive" expectancy is negative (i.e. a near certainty of losing your entire account, given sufficiently many spins). The actual expectancy in this game is a function of the number of spins.

The OP wants to determine the degree of risk aversion of the traders, that is, a shape of the traders' utility function in term of its curvature. This is indeed an interesting topic, much discussed and written about. The problem is, the OP chose a rather esoteric game to explore this field. A much more illustrative game would be along the lines of, "given a game where you have 40% chance to make 5% return on your bet, and a 60% chance to lose 2% of your bet, how do you manage your position (bet) size relative to your account size". This formulation actually relates to real world trading, and the corresponding responses would indicate the responders' risk aversion levels in quantitave form, which is apparenly what the OP is looking for.

 

[Cswim63]

This is not quite right, either. Your calculated expectancy is positive, while the "intuitive" expectancy is negative (i.e. a near certainy of losing your entire account, giving sufficiently many spins). The actual expectancy in this game is a function of number of spins.

thank you

 

[IAS_LLC]

This is not quite right, either. Your calculated expectancy is positive, while the "intuitive" expectancy is negative (i.e. a near certainty of losing your entire account, given sufficiently many spins). The actual expectancy in this game is a function of the number of spins.

The OP wants to determine the degree of risk aversion of the traders, that is, a shape of the traders' utility function in term of its curvature. This is indeed an interesting topic, much discussed and written about. The problem is, the OP chose a rather esoteric game to explore this field. A much more illustrative game would be along the lines of, "given a game where you have 40% chance to make 5% return on your bet, and a 60% chance to lose 2% of your bet, how do you manage your position (bet) size relative to your account size". This formulation actually relates to real world trading, and the corresponding responses would indicate the responders' risk aversion levels in quantitave form, which is apparenly what the OP is looking for.

I agree with what you are saying, but my calculation IS technically correct for the classic expectancy calc. It's just irrelevant since the expectancy relies on repeated trials (said that previously).

 

[IAS_LLC]

Coded up a quick matlab script (I think its right.... but it took 2 minutes, so maybe its not) to simulate the game.

Set initial account size to $100, and the bet size to $100; and the profit target to $100; Set the Maximum number of rolls per monte carlo to (100k... outrageous)

Basically found you are going to go bust (not being able to place the min bit) 49.6% of the time, and you will hit the $200 profit target the rest of the time.

If you reduced the minimum bet to 20% of the initial account value and keep the profit target at $100....you'll only go bust 4% of the time
If 10%, you'll bust 0.1% of the time.......Risk Management is King.

Code:

clear all;
close all;
clc;

%% Define Input Params
NumMonteCarlos = 100e3;
InitAccountSize = 100;
MinBet = 20;
TargetAccountSize = 200;
MaxSpins = 100000;
WinProb = 0.98;
LossProb = 0.02;
LossPerc = -1.0;
WinPerc = 0.03;

%% Do it
Results = NaN(NumMonteCarlos,1);
AccountVal = NaN(NumMonteCarlos,1);
for iMC = 1:NumMonteCarlos
AccountVal(iMC) = InitAccountSize;
iRoll = 1;
rand('seed',iMC);
while( (iRoll <= MaxSpins) && ( AccountVal(iMC) >= MinBet) && (AccountVal(iMC) < TargetAccountSize) )
isWin = rand(1) > LossProb;

if(isWin)
AccountVal(iMC) = AccountVal(iMC) + WinPerc*MinBet;
else
AccountVal(iMC) = AccountVal(iMC) + LossPerc*MinBet;
end
iRoll = iRoll + 1;
end
iMC
end

 

[JB3]

Thanks for your participation. I get the sense that most people will not play this game unless the rule is changed that they don't lose their whole account if they hit 0 or 00. How about you lose 60% of your initial account if you hit 0 or 00? You playing?

I picked the roulette game because most people are familiar with it. It's probably not the best example because it's not the same as trading. Nevertheless, I'm just trying to convey the game aspect of trading.

 

[botpro]

Thanks for your participation. I get the sense that most people will not play this game unless the rule is changed that they don't lose their whole account if they hit 0 or 00. How about you lose 60% of your initial account if you hit 0 or 00? You playing?

I picked the roulette game because most people are familiar with it. It's probably not the best example because it's not the same as trading. Nevertheless, I'm just trying to convey the game aspect of trading.

It is possible to feed any (or say most) games, incl. roulette, from the official stock market data.
One just needs to define (and publish for the participants) the fair and realistic mathematical rules and formula...
Just a hint for some advanced ideas for game design... ;-)

 

[IAS_LLC]

Thanks for your participation. I get the sense that most people will not play this game unless the rule is changed that they don't lose their whole account if they hit 0 or 00. How about you lose 60% of your initial account if you hit 0 or 00? You playing?

I picked the roulette game because most people are familiar with it. It's probably not the best example because it's not the same as trading. Nevertheless, I'm just trying to convey the game aspect of trading.

Still a 2% chance I only survive 1 roll, and a 41% chance I go bust before I double up. I'll still sit this one out.

 

[kut2k2]

Still a 2% chance I only survive 1 roll, and a 41% chance I go bust before I double up. I'll still sit this one out.

OP doesn't realize what it means to overbet the Kelly fraction.

Better yet, here's the expectation when betting 60%:

((1 + .6*.03)^.98)*((1 - .6)^.02) = 0.99915766.

IOW, a return of −0.084234% on average. Not good.

 

[IAS_LLC]

If you lowered the minimum bet to 5%, I'd play this game. In 100e3 Monte Carlos, I doubled up every time (in the matlab code I pasted above).