Okay I have one more question, what are the odds than in three separated picks the same group is chosen twice in a row (regardless the person chosen)?
This problem is similar, but actually quite a bit more complex to solve, so you'd need a different method of approaching it. There are three separate picks, but only two of them need to be the same. We're given the further stipulation that the two groups that are the same must be picked back-to-back. So, just reasoning it out, we have two scenarios which satisfy these conditions:
Group 1 and Group 2 are the same
Group 2 and Group 3 are the same
To make the concept simpler, let's consider flipping coins. We're going to flip three coins, and we want to find the probability of getting two of the same face in a row. If we want coins 1 and 2 to both be heads, how many possible outcomes are there? Notice that it doesn't matter whether the first coin is heads or tails. So there are two outcomes for the first coin. But then the second coin has to the same as the first. Regardless of what face the first coin was, only one outcome will work. And, going further, we also don't care what face the third coin is. Now overall, how many outcomes does the first scenario have? Well, there are two outcomes for the first coin, one for the second, and two again for the third, so we have:
2 * 1 * 2 = 4 outcomes
Applying similar logic for the second scenario, when the second and third coins are the same, we note that the first coin doesn't matter, nor does the second, but the third coin must be the same as the second, so we have:
2 * 2 * 1 = 4 outcomes
There are 4 outcomes in the first scenario, and 4 more in the second scenario, for a total of 8 outcomes which satisfy the given conditions. However, we're not quite done. Consider the possibility that all three coins are the same. If that's the case, then those possibilities were counted in
both of our scenarios. That means we counted them twice, and need to subtract off that many possibilities. So, then, how many ways can it occur that all three coins are the same? The first coin tells us which face we're looking for, and then each subsequent coin must match that face, so we have:
2 * 1 * 1 = 2 outcomes
Taking our 8 outcomes from before and subtracting the 2 outcomes where the scenarios overlap, we have 6 scenarios which satisfy our conditions. And since probability is just favorable outcomes divided by total outcomes, we have: 6 / (2 * 2 * 2) or 6/8, which is 0.75
Now can you apply a similar reasoning to your original problem where there are five groups to pick from?
