Probability of occurrence vs. Probability of x successes

Question

I've been reading for two hours and I'm having a hard time getting the grasp of these two concepts.

I'm developing a tool for a game, in which I'm given n dice and a success threshold (ie. roll at least a 4 means p=0.5), and I want to return the probability of reaching the threshold at least k number of times.

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If I'm not mistaken, probability of x successes goes by:

$$P(n,k,p) = \binom{n}{k} * p^k * (1-p)^{n-k}$$

So, 2 dice, at least 1 success, roll 4+ [1]

$$P(2,1,0.5) = 0.5$$

However, intuitively I know that it doesn't make sense, because you had two chances with the dice to roll a 4+. The probability of occurrence should be [2]:

$$P(n,k,p) = 1 - p^n$$

$$P(2,1,0.5) = 1 - 0.5^2 = 0.75$$

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Maybe in the former I'm counting not at least k successes but exactly k? Why would the numbers make more sense to me when $k>1$? What's the concept I am missing?

[1] http://goo.gl/NmwRWO (wolframalpha.com)

[2] http://goo.gl/V56v6s (wolframalpha.com)

Answer 1

You are right that in the first case you computed the probability of exactly $k$ successes. In general, if you want to compute the probability of at least $k$ successes, you have two options.

1. Compute the probability of exactly $j$ successes for every $j$ with $k \le j \le n$. Add all those probabilities together.
2. Compute the probability of exactly $j$ successes for every $0 \le j < k$, add all those probability together, then subtract that sum from $1$.

Your second answer correctly gave the probability of at least $1$ success because you were taking the second method above: you computed the probability of exactly 0 successes, and subtracted that from 1.

Notice that once $n$ becomes large, if $k$ is not close to either $0$ or $n$, then either method takes a long time, because you're adding a bunch of probabilities. One application of normal distributions is that they can be used to quickly give good approximations for large values of $n$.