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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

[sjb]

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

Have you verified this is correct, as it does not seem right to me. By the pigeon-hole principle you must have a double:

1st roll is irrelevant
2nd roll either matches the first (and so a double and stop) or doesn't (probability 3/4)
3rd roll either matches 1 or 2 (so stop) or doesn't (probability 1/2),
4th matches 1, 2, or 3 (so stop), or doesn't (1/4).
5th roll must match one of the previous 4 if they were all different.
6-12th irrelevant.

[Corribus]

Yes, if you have more dice than you have possible output values per die, it's impossible NOT to roll at least one double.