Hello chemicalforums users,

I come to you all today with a probability concept that I am having a bit of trouble understanding.

First Scenario:

Using tetrahedral (4-sided dice) determine the chance of at least one dice landing on a "4" when you throw six dice.

My solution:

For this problem, I came up with the equation 1 - (3/4)^6 to determine the probability of ~82%. I used the fraction of (3/4) because, since the tetrahedral dice has four sides, there is a 75% chance that the number four will not be rolled. And, since I am rolling six dice, I raised (3/4) to the sixth power.

So, I understand the above problem to a certain degree, but I was a bit lost when I came across this second scenario:

Second scenario: What is the chance of rolling at least one double when you toss twelve tetrahedral dice?

In solving this problem, I came across this equation:

1 - (3/4)^12 - 12 * (3/4)^11 * (1/4) = 0.842

The question I have is WHY do I use the above equation to solve the tetrahedral dice problem? Can somebody explain to me the elements of the equation so I would sufficiently be able to come up with my own formula for, hypothetically, a trial that measured the chance of throwing at least three common numbers out of 18 tetrahedral dice? If you could explain each element in the above equation and its purpose, that would be perfect. Also, if you have an additional example that you could provide to increase my understanding further, I would be very appreciative.

Thank you,

Steve