Biased Dice

Date: 02/18/99 at 08:18:10
From: sharron
Subject: Probability of biased dice

My question is:

If one die is biased so as to make 1, 2, 3, 4, and 5 three times more 
likely than 6, and the other die is biased so 1 is three times as 
likely as for the other die, how would you begin to work out the 
possible outcomes of the two dice adding up to 7?

In addition, I understand how to work out probabilities on unbiased 
dice but need to understand how to work out the table of outcomes. Can 
you help me with that too?

Thanks,

Sharron

Date: 02/18/99 at 08:38:41
From: Doctor Mitteldorf
Subject: Re: Probability of biased dice

Start by calculating the probabilities for each die separately. For the 
first die, 1,2,3,4, and 5 all have probabilities proportional to 3, 
while 6 has a probability proportional to 1. So add up the 5 threes and 
the 1 to get 16, and that's the number you use as a denominator to 
_normalize_ the probabilities. The probability of 1 = 3/16, of 2 = 
3/16, etc., but the probability of 6 = 1/16. Check that these 
probabilities now add up to 1. (How do you know that these 
probabilities DO add up to 1?)
 
Follow the verbal directions to do the same calculation for the other 
die. Continue by listing all the possibilities: 36 different ones that 
look like this:

First die         Second die
1                   1
2                   1
3                   1
4                   1
5                   1
6                   1
1                   2
2                   2
3                   2

Then assign probabilities to each throw. You do this by multiplying the 
probability for the left column by the probability for the right 
column. For example, the first row will be 3/16 times whatever number 
you get for the right column.

Now pull out the rows that add up to 7:  6, 1 and 5, 2, etc.  Add 
those 6 probabilities together to get your answer.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/