How to compare binomial distributions with different number of trial?

Question

I want to know which event has the best score (or luck or odd)

$p$ (prob of success) is equal with all event. There are givne set of events, that is (number of try, number of success)

$$\mathrm{event1} (n_1, x_2) \sim \mathcal{B}(n_1,P)$$ $$\mathrm{event2} (n_2, x_2) \sim \mathcal{B}(n_2,P)$$ $$\dots$$ $$\mathrm{event99} (n_{99}, x_{99}) \sim \mathcal{B}(n_{99},P)$$

The number of trials,success is all different.

And how compare probability of each cases? for example, ranking NO.1 = event33 / ranking NO.2 = event44 / ... ranking NO.99 = event11

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In the same trial, the more success In the same success, the less trial

Q1. I try cdf of Binomial distribution. (quantile) But I don't know it is okay. Because Binomial distribution has diffrent distribution with different number of trial.

Q2. so, I try2 Normal distribution approximation. But N is not big enough.

Q3. and I don't know if I should use the negative binomial distribution. In that cases, the prob of binomial dist is bed.. there are too many 1

Thank you

Answer 1

If you want probabilities of specific events, you can use the binomial PDF function. For example, if $X \sim \mathsf{Binom}(5, .4),$ then $P(X = 2) = {5 \choose 2}(.4)^2(.6)^3 = 0.3456.$

10*.4^2*.6^3
[1] 0.3456

You can use the procedure dbinom in R to compute this easily:

dbinom(2, 5, .4)
[1] 0.3456

Notice that this is quite different from getting 4 successes in 10 trials, when $p = 0.4:$

dbinom(4, 10, .4)
[1] 0.2508227

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If you want confidence intervals for $p$ given $X = 2$ successes in $n = 5$ trials, then there are various methods: You can read about them in the Wikipedia article on binomial confidence intervals. I will show you two of them:

A 95% Agresti-Coull confidence interval in this case would use $\hat p = (X+2)/(n+4) = 4/9$ to get the interval $\hat p \pm 1.96 \sqrt{\hat p(1-\hat p)/(n+4)}.$ This computes to the interval $(0.119, 0.769),$ which does include $p = 0.4.$

n = 5;  x = 2
hat.p = (x+2)/(n+4)
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/9)
[1] 0.1198065 0.7690824

Another style of confidence interval is the Jeffries interval, based on a beta distribution: $(0.094, 0.791),$ which also includes $p = 0.4.$

qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.09439033 0.79058334

Note: For larger $n$ the two types of intervals would tend to be shorter and to agree more closely. For example, consider these two styles of confidence intervals for 20 successes in 50 trials:

Agresti-Coull

x = 20; n = 50
hat.p = 22/54
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/54)
[1] 0.2763554 0.5384594

Jeffries:

qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.2729524 0.5382564