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).

@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.

  • $\begingroup$ There is something wrong in the way you describe your problem. There is no distribution to the ratio Pr(k=1)/Pr(k=0), as a simple algebraic calculation shows that it is a constant parameter equal to np/(1-p). Of course it changes with n and with p, but not with your experiments. You need to be more clear as to what you want to check. $\endgroup$
    – Zahava Kor
    Commented Dec 21, 2017 at 16:06
  • $\begingroup$ Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. $\endgroup$ Commented Dec 21, 2017 at 18:07
  • $\begingroup$ @Zahava Kor I have added some clarification in my q! thanks a lot $\endgroup$
    – MriRo
    Commented Dec 21, 2017 at 18:26
  • $\begingroup$ @Zahava Since $p$ is a parameter and $n$ is known, you have shown that this question is equivalent to asking for a confidence interval on the odds ratio $p/(1-p)$ based on a binomial experiment. There's nothing wrong with that $\endgroup$
    – whuber
    Commented Dec 21, 2017 at 18:36
  • 1
    $\begingroup$ Sure - I misunderstood the question. I thought that the OP only has the ratios between the counts and not the counts themselves (this is the impression I got from reading the question). $\endgroup$
    – Zahava Kor
    Commented Dec 21, 2017 at 21:03

1 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!

  • $\begingroup$ would it be possible to do this on ratios only, without knowing the N. Or is this mathematically not possible? $\endgroup$
    – MriRo
    Commented Dec 25, 2017 at 20:54

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