probability rolling a dice 5 times

Question

I can't solve this problem:

What is the probability that, when rolling a dice 5 times, the number of times when you get a 1 or 2 is greater than the number of times when you get a 6.

any help?

Answer 1

This is essentially a counting problem: in the $6^5=7776$ possible outcomes, how many satisfy the condition "the number of rolls resulting in 1 or 2 is greater than the number of rolls resulting in 6"? The probability is this number divided by $6^5$.

We can find this number by filling five initially empty cells with the possible numbers.

Hint: We can find a formula for number of ways of rolling exactly $s$ 6s and $n$ 1s or 2s:

• What is the number of ways of filling exactly $s$ 6s in those $5$ empty cells?

• After the above, what is the number of ways of choosing exactly $n$ empty cells (to be filled with 1s and 2s in the next step)?

• After the above, what is the number of ways of filling the chosen $n$ cells with 1s and/or 2s?

• After the above, what is the number of ways of filling the remaining empty cells with 3s, 4s, or 5s?

Once we have the formula, we just do the bookkeeping:

$$\begin{array}{cc|c} \text{no 6s} & \text{no 1s and 2s} & \text{nr ways this can occur} \\ \hline 0 & 1 & ??? \\ 0 & 2 & ??? \\ 0 & 3 & ??? \\ 0 & 4 & ??? \\ 0 & 5 & ??? \\ 1 & 2 & ??? \\ 1 & 3 & ??? \\ 1 & 4 & ??? \\ 2 & 3 & ??? \\ \end{array}$$

Answer 2

@Stones Yes, but the table is not quite right. If there are three or more 1s or 2s, then it does not matter how many 6s there are. So N5 the number of ways of getting five 1s or 2s is 32. Similarly, N4 the number of ways of getting four is $5\times 2^4\times 4=320$ and N4 the number of ways of getting three is $10\times 2^3\times 4^2=1280$. If we get just one 1/2 then we cannot get any 6s, so N1 is $5\times 2\times 3^4=810$.

That leaves the trickier case of two 1s/2s. If there are no 6s, then we have N2a as $10\times 2^2\times 3^3=1080$. If there is one 6, then we have N2b as $10\times 3\times 2^2\times 1\times 3^2=1080$. Adding those up, we get 4602, so the probability is 767/1296 = 0.5918 approx.

[The coefficients 1, 5, 10 are just the binomial coefficients, because we are looking at the number of ways of picking 5, 4, 3 things from 5. The extra 3 for N2b comes because there are 3 ways of choosing which of the three results not a 1 or 2 is the 6.]

Answer 3

It can be solved using ordinary generating functions:

Since there are three groups of numbers of our interest, take three formal variables for the ogf.

Let $x$ indicate 1 or 2, $z$ indicate a 6, and $y$ the other three numbers of the dice.

Hence, the required ogf is: \begin{align*} \left(2\, x + 3\, y + z\right)^5 \end{align*} We need to consider cases with the variables, though.

Grouping the coefficients for which the degree of $x$ $>$ the degree of $y$, the sum turns out to be $4602$, and the probability is \begin{align*} \mathbb{P} &= \frac{4602}{6^5} = \frac{767}{1296} \approx 0.5918209876543 \end{align*}