The equation you need is:

P(x) = n! / [x! (n - x)!] p^x (1 - p)^(n - x)

where:

n = number of attempts (unknown)

x = number of successful attempts (known, but variable)

p = probability of success (1/6)

Since n is your unknown, this isn't going to be pretty. But since you are asked for "at least 2 6's" to be 0.6 (which means 2 or more, which could be up to 100 if n = 100, etc.) so if we add up all of the possible outcomes (2, 3, ..., n) we will get 0.6. We need something more static, so instead we look at the opposite cases (x = 0 and 1) which should then add up to 0.4.

So:

P(0) + P(1) = 0.4

Now that we have known values for x we can substitute the expressions in from the equation above into the second equation and find a value for n that will work. There is no good formula for the factorial of an unknown that I know of so this will need a little brute force:

P(x) = n! / [x! (n - x)!] p^x (1 - p)^(n - x)

we have a known "p", so:

P(x) = n! / [x! (n - x)!] (1/6)^x (5/6)^(n - x)

And we know we want P(0) and P(1), so:

P(0) = n! / [0! (n - 0)!] (1/6)^0 (5/6)^(n - 0) and P(1) = n! / [1! (n - 1)!] (1/6)^1 (5/6)^(n - 1)

0! = 1, 1! = 1, anything to the 0th power is 1, anything to the 1st power is just the base:

P(0) = n! / (n)! (1) (5/6)^n and P(1) = n! / (n - 1)! (1/6) (5/6)^(n - 1)

P(0) = n! / (n)! (5/6)^n and P(1) = n! / (n - 1)! (1/6) (5/6)^(n - 1)

Now we have n! / n!, which is 1 as everything cancel out, and n! / (n - 1)! is n, since:

n! = n(n - 1)!

And since we are dividing by (n - 1)! that cancels out leaving only the n:

P(0) = (5/6)^n and P(1) = n(1/6) (5/6)^(n - 1)

Now we want the sum of these two to be 0.4:

P(0) + P(1) = 0.4

(5/6)^n + n(1/6) (5/6)^(n - 1) = 0.4

Factoring out the (5/6)^(n - 1) from the left side, this leaves:

(5/6)^(n - 1)[(5/6) + n(1/6)] = 0.4

Now we can simplify this new binomial:

(5/6)^(n - 1)(5/6 + n/6) = 0.4

(5/6)^(n - 1)(5 + n)/6 = 0.4

Let's multiply both sides by 6:

(5/6)^(n - 1)(5 + n) = 2.4

Let's multiply both sides by (5/6) to get the exponent into "n":

(5/6)^n (5 + n) = 2.4 * (5/6)

(5/6)^n (5 + n) = 2

Here is where we need to do a little brute force. We know n must be an integer greater than 2. Let's try a number like 5 and 10 and compare the change between the two and how far away we are from 2 to see how much farther we need to go:

(5/6)^5 (5 + 5) = 2 and (5/6)^10 (5 + 10) = 2

0.40187 (10) = 2 and 0.1615 (15) = 2

4.0187 = 2 and 2.4225 = 2

So 10 is closer, and we're getting close to 2. Let's see what 12 and 14 do:

(5/6)^12 (5 + 12) = 2 and (5/6)^14 (5 + 14) = 2

0.112157 (17) = 2 and 0.077887 (19) = 2

1.906669 and 1.479663 = 2

We've overshot. Let's see what 11 looks like:

(5/6)^11 (5 + 11) = 2

0.13459 (16) = 2

2.15344

Note that the original problem says "at least 0.6", so it doesn't have to be exactly equal to 0.6, but can be greater.

So P(0) + P(1) needs to be at most 0.4. The first time that this is below 0.4 is when n = 12.

So you need to roll the die 12 times to have the probability of getting a six twice to be at least 0.6.

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The same process will work for the other two with difference numbers.