# PMF of throwing a die 4 times

We throw a fair 6-sided die independently four times and let $$X$$ denote the minimal value rolled.

1. What is the probability that $$X \ge 4$$?

2. Compute the PMF of $$X$$.

3. Determine the mean and variance of $$X$$.

My attempt:

1. $$(1/2)^4$$ because that means each of the $$4$$ rolls, you either get a $$4, 5$$ or $$6$$.

2. for $$X=1: (1/6)(6/6)(6/6)(6/6)$$
for $$X=2: (1/6)(5/6)(5/6)(5/6)$$
for $$X=3: (1/6)(4/6)(4/6)(4/6)$$
for $$X=4: (1/6)(3/6)(3/6)(3/6)$$
for $$X=5: (1/6)(2/6)(2/6)(2/6)$$
for $$X=6: (1/6)(1/6)(1/6)(1/6)$$

3. I can calculate this once I know I did the PMF correctly

Did I do (1) and (2) correctly?

• For (1) you need $\left(\frac{1}{2}\right)^4$ – Daniel Mathias Feb 7 at 19:29
• ad a) But you have to roll 4 times. So $P(X=4)=\left(\frac{1}{2}\right)^4$ And $P(X=5)=\left(\frac{1}{3}\right)^4$ And al last $P(X=6)=\left(\frac{1}{6}\right)^4$. To get $P(X\geq 4)$ you have to sum up the probabilities. – callculus Feb 7 at 19:31
• @calculus that should be $P(X\ge4)$ and $P(X\ge5)$ – Daniel Mathias Feb 7 at 19:34
• @DanielMathias No, since $X$ denote the minimal value rolled – callculus Feb 7 at 19:35
• @callculus for $X=4$, one of the rolls must be 4. But $\left(\frac{1}{2}\right)^4$ includes the events that all rolls are greater than 4. – Daniel Mathias Feb 7 at 19:38

We throw a fair 6-sided die independently four times and let $$X$$ denote the minimal value rolled.

What is the probability that $$X \ge 4$$?

For each roll, the value must be greater than three. This event has probability $$\frac{3}{6}=\frac{1}{2}$$. As this must occur on each of $$4$$ rolls, we have: $$P(X\ge4)=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$

Compute the PMF of $$X$$.

$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$

$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$

$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$

$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$

$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$

$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$

1) is now correct after your edit based on Daniel's comment.

2) is incorrect since a) it doesn't sum to 1 and b) $$X = 4, 5, 6$$ doesn't sum to what you got in 1).

My approach to this question is to break it down into a bunch of simpler problems as such:

For $$P(X=1)$$, let's consider the $$4$$ dice as dice $$1, 2, 3, 4$$. There are $$4$$ possibilities, the result can have $$1,2,3,$$ or $$4$$ ones in it.

• $$1$$ one: $${4 \choose 1} (\frac{1}{6})^1(\frac{5}{6})^3$$
• $$2$$ ones: $${4 \choose 2} (\frac{1}{6})^2(\frac{5}{6})^2$$
• $$3$$ ones: $${4 \choose 3} (\frac{1}{6})^3(\frac{5}{6})^1$$
• $$4$$ ones: $${4 \choose 4} (\frac{1}{6})^4(\frac{5}{6})^0 = (\frac{1}{6})^4$$

The sum of the above will give you the $$P(X=1)$$.

Similarly, you can solve for the rest. Make sure that the end results sums to $$1$$ and $$P(X=4) + P(X=5) + P(X=6) = (\frac{1}{2})^4$$.