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BGonline.org Forums
Tables to Improve Your Game
Posted By: Bob Koca In Response To: Tables to Improve Your Game (Joe Russell)
Date: Tuesday, 16 March 2010, at 1:17 a.m.
I think you are applying a rule that is meant for something that has a geometric distribution which is not the case here. What is meant by odds of success averaging something here? The "success" is actually two different successes. This has less variability meaning there will be more instances close to the mean of 6, meaning less of a chance of being done in just a few rolls. Another consideration is the skew. The geometric distribution has more right skew which makes the median less than the mean by a greater amount.
As an example suppose we roll a single die and we see how many rolls it takes to roll a 4. This is a geometric distribution with parameter 1/6 (the prob of success at any given trial). The expected value of this distribution is 6.
Now suppose that instead we need to roll either a 1 or a 2, and then do it again. This is the sum of two independent geometric distributions each with parameter 1/3. The expected total number of rolls is 3 + 3 = 6. This is similar to the backgammon example we are talking about but without the possibility of the 16 roll doing two things at once.
For the first situation taking exactly n turns means you have n-1 failures and then a success for a probability of (5/6)^(n-1) * (1/6) Not that to go from one probability to the next just multiply by (5/6)
For the second situation taking exactly n turns means you have exactly one success in the first n-1 trials and then a success. There are n-1 arrangements e.g. SFFFS or FSFFS or FFSFS or FFFSS each with the probability (2/3)^(n-2)*(1/3)^2 for a probability of (n-1)(2/3)^(n-2)*(1/3)^2 Note that to go from one number to the next multiply by (n/n-1)(2/3). As n gets larger we will be multiplying by a value close to 2/3. 2/3 is less than the 5/6 so eventually these values decrease more quickly than for the geometic situation.
P(takes exactly 1 roll) .167 .000
P(takes exactly 2 rolls) .139 .111
P(takes exactly 3 rolls) .116 .148
P(takes exactly 4 rolls) .096 .148
P(takes exactly 5 rolls) .080 .132
P(takes exactly 6 rolls) .067 .110
P(takes exactly 7 rolls) .056 .088
P(takes exactly 8 rolls) .046 .068
P(takes exactly 9 rolls) .039 .052
P(takes exactly 10 rolls) .032 .039
P (takes exactly 11 rolls) .027 .029
P(takes exactly 12 rolls) .022 .021
For taking exactly n rolls with n > 11 the first situation is more likely. For n small the first situation is also more likely. Thus the first situation has much more variability and is much more right skewed. The more right skew there is the greater the difference between the mode and the mean.
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