Probability of Throwing at least 1 "6" vs Double 6 in Dice Rolls

[karambos]

TASK: Show that throwing at least one "6" within 4 throws of a die is more likely than throwing a "double 6" within 24 throws of two dice.

SOLUTION: I can do the first result two ways. Firstly, I can use the complement of "four dice all showing something other than six" which would be 1 - ((5/6)^4) = 51,7%. Or I can use the long-winded version of

Binomial[4, 1]*((5/6)^3)*((1/6))^1) +
Binomial[4, 2]*((5/6)^2)*((1/6))^2) +
Binomial[4, 3]*((5/6)^1)*((1/6))^3) +
Binomial[4, 4]*((5/6)^0)*((1/6))^4)

= 51,7%

which translated into words is roughly: "the chances of selecting 1 from 4 multiplied by the chances of three dice showing not six multiplied by the chances of one die showing six PLUS the chances of selecting two from 4 etc etc"

I can show the second part by using the complement of "no six or exactly one six" which is 1 - ((5/6)^12) - (Binomial[12,1])*(5/6)^11)*(1/6)

MY QUESTION: if I were insane enough (which I am, believe me ) to want to solve the second part using the "long-winded" binomial version, how would I do it?

I'm gratefu for any help

 

[TenaliRaman]

P(getting a double 6) = 1/36
P(in 24 throws , no double throws) = (35/36)^24
P(getting atleast 1 double throw in 24 throws)
= 1-(35/36)^24
approx .491 or 49.1%

Hence the result

-- AI

 

[tacman]

To solve the second part using the "long-winded" binomial version, you would need to use the binomial coefficient formula:

Binomial[n, k] = n! / (k! * (n-k)!)

where n is the total number of trials (24 in this case) and k is the number of successes (2 in this case, since we are looking for a double 6).

So the calculation would be:

Binomial[24, 2] * ((5/6)^22) * ((1/6)^2) = 0.025%

This is the probability of getting exactly 2 sixes in 24 throws. To get the probability of getting at least one double 6, we need to add the probability of getting exactly 3 sixes, 4 sixes, and so on up to 12 sixes.

So the final calculation would be:

Binomial[24, 2] * ((5/6)^22) * ((1/6)^2) + Binomial[24, 3] * ((5/6)^21) * ((1/6)^3) + ... + Binomial[24, 12] * ((5/6)^12) * ((1/6)^12) = 0.025% + 0.001% + ... + 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000