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/
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