Is binomial distribution approriate

AI Thread Summary
The discussion centers on the probability distribution of rolling six dice, initially set at a constant probability of 1/6 for rolling a six. It explores a scenario where the probability of rolling a six increases continuously from 0 to 1/6 over time. The conclusion is that while each individual toss follows a binomial distribution, the overall sum of results from multiple tosses does not maintain a binomial distribution due to the changing probabilities. The mean and variance of the total number of sixes do not align with those expected from a binomial distribution. Therefore, the overall distribution cannot be classified as binomial.
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Suppose I have 6 die and toss them. The probability to have n 6's is binomially distributed with parameter 1/6.
Now suppose instead tossing the 6 die and having 1/6 probability for a 6 each dice's probability to show 6 grows continously in the time interval t=0 to t from 0 to 1/6. Can I then say that as before, the probability to have n 6's is binomially distributed with parameter 1/6?
 
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aaaa202 said:
Suppose I have 6 die and toss them. The probability to have n 6's is binomially distributed with parameter 1/6.
Now suppose instead tossing the 6 die and having 1/6 probability for a 6 each dice's probability to show 6 grows continously in the time interval t=0 to t from 0 to 1/6. Can I then say that as before, the probability to have n 6's is binomially distributed with parameter 1/6?

I think I know what you are trying to say, but let me re-word it first. You have 6 dice (not 'die'--a 'die' = one single cube). At any toss they each have the same probability p of showing a '6', but the probability p increases from 0 to 1/6 as we make more tosses. If ##p_k## is the probability of a die showing '6' on toss k, then the number ##N_k## of 6's on toss k is binomial with parameters ##(6,p_k)##. If we finally get a probability of 1/6 on the nth toss (that is, ##p_n = 1/6##), the total number of 6's altogether is ##X = \sum_{k=1}^{n} N_k##. This will NOT have a binomial distribution, because although the summands are independent, they are not identically distributed. In fact, you can easily work out the mean and variance of X and find that they are not related to each other as a binomial would give.
 
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