Probability of getting two / four of the same color out of three possibilites Basic probability question, with random assortment and unlimited numbers, there are three lollipops with colors red, yellow, and green. How many times must you pick a lollipop from the jar to guarantee two of the same color? Four of the same color?
Is the answer as simple as 4 for getting two of the same color and 10 for four of the same color? Or for two would the probability be (1/3) * (1/3) so 9? And for four of the same color (1/3) * (1/3) * (1/3) * (1/3) so on average 81? I know technically there is no guarantee but I think the question is asking the average.
 A: The answers are 4 and 10, but as André Nicolas pointed out, your reasoning is not correct. Let us first begin with seeing how many times we need to draw to get two of the same color. The best case scenario would be if, on our first 2 draws, we got 2 of the same color. If we think about the worst case scenario within 3 draws, we see that our first 3 draws would include red, yellow, and green. If any of those were different, we would have 2 of the same color, and our condition would be met. If we've reached 3 draws and still don't have 2 of a kind, then we must have one of each color. This means that no matter which color we draw next, we must already have one of that color in our hand, meaning that we will make 2 of a kind. That means that 4 is the minimum number of draws it takes to guarantee that we will have drawn 2 of a kind, and it is an absolute guarantee.
This same logic can be applied for 4 of a kind. Our best case scenario is drawing 4 in a row of the same color. Our worst case scenario is 9 draws, giving us 3 of a kind of each color (R, Y, G, R, Y, G, R, Y, G). Like before, no matter which color we draw next, we will make 4 of a kind. Once again, this is an absolute guarantee.
