Let's roll one dice two times, and ask some questions about our rolls. In this case, let's see what the probability as a whole is, of getting 4, 5, or 6 on the first roll, and a 1, 2, 3, or 4 on the second roll.

As in last lessons, we'll set

Note here that we have two separate results,

Note a key word in the question, the "and": "getting a 4, 5, or 6 on the first roll**and** a 1, 2, 3, or 4 on the second roll. "And" in probability means to multiply any separate and independent probabilities, so in the end, we multiply

As in last lessons, we'll set

`N`

equal to the number of rolls we'd like to make. Next, we'll set up a "roll for-loop" for each roll. The first one study what the probability is to roll a 4, 5, or 6 on dice #1. The second will look for the probability of rolling a 1, 2, 3. A dice roll will be simulated with `math.random(1,6)`

.
Note here that we have two separate results,

`d1`

and `d2`

. `d1`

will be the chance of rolling a 4, 5, or 6. `d2`

will be the chance of rolling a 1, 2, 3, or 4.
Note a key word in the question, the "and": "getting a 4, 5, or 6 on the first roll

`d1`

and `d2`

to find the final answer.
Type your code here:

See your results here:

We can apply some common sense to verify the computer's answer. Keep in mind, for a fair dice, each number is equally likely to show after a roll. Thus, each number has a 1 in 6 chance of showing, or a probability of $1/6$.

The chance of rolling a 4, 5 or 6, is $3\times 1/6$ or $3/6$ or $1/2$ or $0.5$. So, `d1`

in the above code should result in around this number. For large values of `N`

, like $1,000$ or so, $0.5$ should result.

The chance of rolling a 1, 2, 3 or 4 us $4\times 1/6$ or $4/6$ or $2/3$ or 0.66. So, `d2`

in the above code should result in around this number. For large values of `N`

, like 1,000 or so, $0.66$ should result. Note the repeated look for a "4" between the two rolls has nothing to do with the fact that the dice is "fair" and all numbers have a 1 in 6 chance of coming up.

Lastly, we address the "and" by multiplying `d1`

and `d2`

for the final answer to our question.