How to quantify the confidence intervals of the ratio of two binomial probabilities of getting exactly k successes in n trials

Question

I have a question about how to quantify the confidence intervals of the ratio of two binomial probabilities of getting exactly k successes in n trials.

Consider I have a binomial distributed random variable X ~ B(n,p). When my n is 10 and my p is 0.03 then the chance of getting 1 hit in 10 trials, 0 hits, and their ratio is: \begin{align} Pr(k=1) &= 0.22806 \\[5pt] Pr(k=0) &= 0.73742 \\[5pt] \frac{Pr(k=1)}{Pr(k=0)} &= 0.3092784 \end{align}

But when I will do a study like throwing a dice 100 x 10 times where 100 are the repetitions of the experiment and 10 are my trials per repetition -with the probability of success for a single roll = 0.03 like above- I will not always get one six out of 10 in 22 of the trials and not zero times six out of 10 in 74 trials. There will be a variance that is probably changing (becoming lower) with the times I repeat this (here 100). Because of that also my quotient Pr(k=1)/Pr(k=0) will vary probably depending on how often I will carry out my 10 times dice rolling.

Is there a way to estimate confidence intervals of my Pr(k=1)/Pr(k=0) distribution probably depending on the size of my repetitions of the 10 times dice rolling experiment? And will this change when I change my n (here 10) to a higher amount, e.g., 50 trials per repetition, so that Pr(k=1) and Pr(k=0) will change.

I want to check if my quotient is originating in a binomial distribution with Pr(k=1;n,0.03)/Pr(k=0;n,0.03).

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@Zahava Kor I cannot comment to my question, so I try to answer your comment here:

Yes, I know that the expected ratio is a single number, but what will differ is the observed ratio. Consider the example above where I repeat this experiment 100 times. I will not always get one six out of 10 in 22 trials and not zero times six out of 10 in 74 trials although this would be expected due to my Pr(k=1) and Pr(k=2) so what I want is to have confidence intervals of these expected outcomes of 22/74 (22 trials out of 100 show one six out of ten due to Pr(k=1)). But when I get values of 24 for Pr(k=1) and 69 for Pr(k=0), I get a quotient of 24/69. I want to know if this quotient which is a single number is still possible due to a binomial distribution or lies outside the confidence intervals of that quotient for a given n and k. And that this quotient is on a p<0.05 significance level higher or lower than possible due a binomial distribution. When I get e.g a quotient of 0.8 for the n and k in the example it will be highly unlikely that this is based on a binomial distribution as the expected value will be Pr(k=1)/Pr(k=0)= 0.3092784.

I hope that this makes my concern a little clearer.

Answer 1

Whatever you do - don't use the ratio between the counts to estimate the ratio between the probabilities - it doesn't work this way. For one - the numerator and denominator are dependent random variables. Even worse - the ratio of counts does not have a distribution at all, because the denominator can be 0 and you cannot divide by 0. The best thing you can do is treat the N x n Bernoulli (0/1) experiments as one random variable with distribution Bin (Nn,p), count the number of successes out of the total of Nn experiments, and find a confidence interval for p (using the Central Limit Theorem - i.e., the Normal approximation). You get an interval [a,b] for p. If you want to find an interval for, say, np/(1-p), c ≤ np/(1-p) ≤ d, simple algebra will give you c and d based on a and b. Notice that no ratios of random variables are calculated here!