What is the probability that at least two of the dice land on a number greater than $4$ when three six-sided dice are rolled?

We roll three fair six-sided dice.

(a) What is the probability that at least two of the dice land on a number greater than $$4$$?

(b) What is the probability that we roll a sum of at least $$15$$?

(c) Now we roll three fair dice $$n$$ times. How large need $$n$$ be in order to guarantee a better than $$50\%$$ chance of rolling a sum of at least $$15$$, at least once?

Of course, the size of the sample is $$6^3=216$$. For (a), my thinking is that 2 options (5 and 6) on the first die, 2 options on the 2nd die, 6 options on the 3rd die, multiplied by 3 orders of the dices resultings; So $$2\cdot2\cdot6\cdot3 = 72$$. Which gets me $$\frac{72}{216}$$. However I'm not sure if I'm correct. For (b) and (c), I am not sure how to even approach the question.

PART A

Separate into 2 cases, 2 dice and 3 dice greater than 4.

2 dice greater than 4 $$=3\cdot(2\cdot2\cdot4)=48$$. The three represent the possible choices of the 2 dice from 3 dice.

3 dice greater than 4 $$=2\cdot2\cdot2=8$$.

$$P(A)={48+8\over216}=\frac7{27}$$

You are counting 16 extra combinations.The 72 combinations you counted, combinations like (6,6,6),(5,5,6) are repeated 3 times...similarly a total of 8 combinations are repeated 3 times.So as dice have no orders you have to neglect the 16 extra combinations that you considered.(72-16 = 56)

(a) The probability of one die landing on a number greater than 4 is $$2/6 = 1/3$$. The probability of one die not landing on a number greater than 4 is thus $$2/3$$

1. The probability of two dice landing on a number greater than 4 is $$\frac 13 \frac 13 \frac 23 + \frac 13 \frac 23 \frac 13 + \frac 23 \frac13\frac13 = 3\cdot \frac{2}{27} = 2/9$$
2. The probability of three dice landing on a number greater than 4 is $$\frac13\frac13\frac13 = 1/27$$

It follows that the result for a is $$\frac 29 + \frac {1}{27} = 7/27$$

(b) Effectively what you need to do is write 15 as a sum of three numbers you can obtain on your dice. It would be fairly easy but time consuming to construct a 3D table in which you sum all the 216 possible cases and find which ones sum to 15.

Instead, it's easier to see which "biggest" numbers you can combine to get 15.

$$3\cdot6 = 18$$, thus you take $$2\cdot6 + x = 15$$ . You find x, consider the number of combinations and you keep decreasing the left sum until you can no longer achieve 15 or you repeat your combination.

1. 6+6+3 = 15 and there are 3 cases in which this combination occurs. Probability for this is $$\frac 16\frac 16\frac16 3 = \frac {3}{216}$$
2. 6+5+4 = 15 and there are 3 cases in which this combination occurs. Probability for this is $$\frac {1}{216}3 = \frac {3}{216}$$
3. 6+3+6 = 15 and there are 3 in which this combination occurs. Probability for this is $$\frac {3}{216}$$
4. 6+2+7 - oops, 7 isn't on a die, so decrement the left-most number and repeat.
5. 5+5+5 = 15 and there is only 1 case in which this combination occurs. Probability for this is $$\frac 16\frac 16\frac16 = \frac {1}{216}$$

If you consider all the other possible combinations, you will see that you cannot get 15 anymore without considering cases that you have already covered above. Thus the result for b should be $$\frac {3+3+3+1}{216}=10/216$$

(c) This is an example of a binomial distribution. You first need to establish the probability of the desired case on one die - getting at least a 15. You have 4 cases there: 15, 16, 17 and 18.

1. You can get 15 in 10 cases (b).
2. You can get 16 as 6+6+4 in 3 cases and 6+5+5 in 3 cases.
3. You can get 17 as 6+6+5 in 3 cases.
4. You can get 18 as 6+6+6 in 1 case.

Thus the total number of desirable cases on one die is 10+6+3+1 = 20 and the probability for them is $$20/216$$ and let's call this p (little p).

$$p=20/216$$

Let's say P (big p) is the answer for c. Then you can do the opposite. 1-P = the probability of NOT getting 15 at least once = the probability of getting it 0 times out of n attempts. From this it follows: $$P>.5\\ -P<-.5\\ 1-P<1-.5\\ 1-P<.5\\ 1-P = \neg P$$

I don't know if you're familiar with the binomial distribution in probability, but it can be shown that if you want to get your desirable case k times out of n attempts, where each attempt has a probability of p and is performed independently of the previous, then the probability for that will be: $$B(k,n,p) = \binom{n}{k}p^k(1-p)^{n-k}$$

In your case: $$\neg P = B(0, n, p) = \binom {n}{0}(20/216)^0(1-20/216)^{n-0} = (1-20/216)^n$$ and combining this with the inequality from above: $$1 - P = \neg P < .5 \\ \neg P - .5 < 0 \\ (1-20/216)^n - .5 < 0$$ If you make a graph of this equation, you will find it intersects the x-axis at n=7.133..

Since you want the first value of x for which the last inequality holds true, that cannot be 7, since the slope of the equation is negative and the left side is greater than 0 for n=7, but instead it must be the next whole value - 8.

Thus, the answer for c is $$8$$

• For (a) case 1, it should be $\frac13\frac13\frac23$ as you calculate the probability exactly 2 of them greater than 4. Commented Feb 1, 2020 at 7:56
• That's a fair point. My bad. Commented Feb 1, 2020 at 7:59
• Michelle seems to have forgotten about her question. Commented Feb 3, 2020 at 12:56