Please help me - probability problem

Ringo · #1 03-10-2005, 07:56 AM

I've been reading about standard deviation, and am stuck trying to figure something out. Imagine a game, where one 6 sided dice is used. As the "house", if a 3,4,5 or 6 roll, I win, and a 1 or a 2 will pay out at even money. I know it`s not a very good game, but bear with me!

I was trying to work out how many games, with constant stakes, it would take to ensure a profit for the house - ie, how many games it would take to make a 99% chance (ie, 3 standard deviations) of the house coming out ahead. So..

.. I've been approaching the problem from this angle. I know I want to solve the SD formula for "n" (number of trials), so have been doing

SD = sqrt[N.(2/3).(1/3)]

..but, what goes on the left of the equation? To solve for N, there needs to be something. The expectation for the house is 0.66, and would I be right in saying that to ensure a win for the house, 3 SD`s must be less than (0.66-0.5)?

This type of problem can be applied to counting cards, can it not? Ie, if I only place bets when I have a >10% edge, whats the chance of me being behind after X trials, and how many trials it would take to ensure (within 3 SD`s) a profit. etc.

Can anyone help me out here? [img]/images/graemlins/smile.gif[/img]

Ringo

elitegimp · #2 03-10-2005, 01:52 PM

The Standard Deviation tells you how far from the mean you should be. So you want [mean] - 3*SD > 0. In this case, if you play N times, you expect to win 2N/3 of those times... So you want SD < 2N/9, or Sqrt[N]*Sqrt[2]/3 < 2N/9, or Sqrt[N] > 3/Sqrt[2], so N > 9/4. That seems extremely low, so I'm gonna go out on a limb and say I probably messed something up [img]/images/graemlins/smile.gif[/img]

Paul2432 · #3 03-10-2005, 02:17 PM

There are two steps to solve this problem. First determine the mean and SD. Each game has two possible outcomes. +1 or -1. The mean is 0.66 x 1 - 0.33 x 1 = 0.33. (not 0.66) SD is simply the square root of the variance.

Variance is 0.66 x (1-0.33)^2 + 0.33 x ((-1-0.33)^2 = 0.88

SD = 0.94

The second step uses the formula for standard error:

SE = SD / sqrt(N)

99% corresponds to 2.33 SD or SE. Our win rate is 0.33. 0.33/2.33 = 0.14. So

0.14 = 0.94 / sqrt(N)

Solving,

N = 45

That is if you play 45 games you will have less than 1% chance of being behind.

Paul