What is a rolling calendar?

What is the probability of rolling at least two fives or greater when rolling four dice?

• You roll four dice. Each die has six sides. The desired outcome is a 5 or greater (so 5 or 6), therefore 2 sides of each die are desirable. I know the probability of rolling a 5 or greater on a single die is 2/6, or 1/3. I also know to calculate the probability of rolling *at least* one die that's 5 or greater with four dice is to multiply together the probability of *not* getting a 5 or greater for each die (so 4/6x4/6x4/6x4/6 = 256/1296, or 0.197530864, or ~19.75%) then subtract this total from 1 (so 1 - 0.197530864 = 0.802469136, or ~80.24%). And further more, I know how to calculate the probability that *all* four dice will show a 5 or greater, which is accomplished by multiplying together the probability of rolling a 5 or greater for each die (so, 2/6x2/6x2/6x2/6 = 16/256, or 0.0625, or ~6.25%). In fact, I even know how to calculate the probability of *exactly* rolling two dice that show a 5 or greater when rolling four dice. This is done by multiplying the probability of rolling a 5 or greater for 2 of the dice (2/6x2/6) as well as multiplying the probability of *not* rolling a 5 or greater for the other 2 dice (4/6x4/6) and then multiply by the number of dies being rolled (4), so it would be 4x2/6x2/6x4/6x4/6 = 0.197530864 or ~19.75%. But what I can not figure out is how to find the probability of rolling *at least* two dice which are 5 or greater when you are rolling four dice! To clarify, that means rolling only one die with a 5 or greater roll is *not* a desired outcome (...at least two...), but rolling two, three, or four dice with a 5 or greater *is* a desired outcome! Again, please note that the phrase is "at least" and *not* the word "exactly". I already know how to calculate the probability for exact roles. If you could show me a detailed description of how this probability question is solved - complete with mathematical formula - that would be greatly appreciated! I am not looking for simply an answer but an explanation to the method of solving it! THANKS!

• Answer:

  The first step is knowing how to calculate for "exactly". Your method is slightly off. "This is done by multiplying the probability of rolling a 5 or greater for 2 of the dice (2/6x2/6) as well as multiplying the probability of *not* rolling a 5 or greater for the other 2 dice (4/6x4/6) and then multiply by the number of dies being rolled (4)" It's just the last "and then" that's wrong. You don't multiply by #dice, you multiply by the #ways to get exactly 2 successes. It could be on the 1st and 2nd roll, 1st and 3rd, 2nd and 3rd, etc. In total there are (4 C 2) = 6 combinations. So P(exactly 2) = 29.63% This probability distribution is called the Binomial Distribution, and is a special case of the Multinomial distribution. Wikipedia has a good page on both. For P(at least 2) there are two methods: P(at least 2) = P(exactly 2) + P(exactly 3) + P(exactly 4) or P(at least 2) = 1 - [P(exactly 1) + P(none)] It's not such a hassle for this problem, but if you were talking about 100 rolls and wanted P(at least 50), then using either of those formulas would be a pain, so a computer or advanced calculator comes in handy. Excel, Wolfram, TI calculators have built-in binomial distribution functions. Or you can just express it in summation form and let it evaluate the sum (without you having to type any + signs or a bunch of terms). Answer: P(at least 2) = 40.74%