Zero

Posted by Richard Reid on June 27, 1997 at 19:00:43: In Reply to: Re: Law of Independent Trials: Is It Flawed? posted by Pete Moss on June 27, 1997 at 11:55:26:

Pete Moss wrote:
Next question: You've seen exactly 400 rolls, and no 7 has come up. What odds would be fair on the proposition that the next roll will be a seven?

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Pete

Here's my take on your question. On pp. 127-130, Epstein is using the Chi-square test. The purpose of the Chi-square test is to compare any 3 or more groups or classes of things. So, for example, it can be used to compare the outcomes of a sample of die rolls to an expected outcome. Notice that his example is explaining the detection of a bias. I see no problem with his method and believe he is using the Chi-square test correctly.

However, you are correct in saying that Epstein's treatment should not be used to answer the type of question you have posed. Your question is different, but in a subtle way. It is worded such that one is not trying to determine whether each of the numbers on a die shows up 1/6th of the time. Rather, the question one is trying to determine the probability of a 7 appearing on the next roll, given that the last 400 rolls have been non-seven. There is a very real albeit subtle distinction here.

As Don (and others) have said, if the dice are fair, then the answer is 1/6th.

We can determine whether the dice are fair or not by using the zI Test, where the expected number (E) of occurrences for the sample of 400 non-seven occurrences can be determined by

E = Px * n, where Px = the fractional proportion of sevens to non-sevens (ie 1/6) and n = 400.

E= 400/6 ~ 67

We can then determine z from the formula:

z = (|E-x| - c) / sqrt(E*Py), where Py = 1-Px, and x = the actual number of 7s rolled, c = a correction factor (normally = 0.2)

Therefore, z = ( | 67- 0| - 0.2 ) / sqrt(67*(1-5/6))

z = 66.5 / sqrt(10) = 66.2/3.16 ~ 21

z is the difference between the actual and expected values expressed in terms of standard deviations. The larger z is, the less likelihood of our non-significant difference assumption being correct. With a z = 2.58 there is a 1% probability that there is no significant difference. With z = 21, there is an extremely minute probability that there is no significant difference. In layman's terms, it would be very safe to say that the dice are not fair!!

Now, given that the dice are not fair, then what is the probability of a seven appearing on the next roll?

Well, the first thing we must do is calculate a new mean. Since there have been no sevens rolled, the new mean is 0/400 = 0. Therefore, our best estimate of the probability of a seven appearing on the next roll is "zero."

Sincerely,
Richard Reid



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