Main content

### Course: Precalculus > Unit 8

Lesson 8: Decisions with probability# Using probabilities to make fair decisions example

We can determine whether or not probabilities are being used to make a fair decision. In this example, we look at whether different outcomes have the same probability or not when we roll two dice to make a decision. Created by Sal Khan.

## Want to join the conversation?

- Why are 3 and 4 in the first row selected？(18 votes)
- This is an error as those 3 and 4 from the first row shouldn't be selected. The only way to get a 3 or 4 from 2 die is getting (1&2;2&1) and (1&3; 3&1; 2&2).

The answer should be 5/36 chance (or 0.138889%) for Jordan to vacuum and 1/6th chance (or 0.166667&) for Miguel to vacuum. Therefore, Miguel has a higher chance of vacuuming their apartment (by 1/36th or 0.127778%).

TLDR: No, their is a higher probability that Miguel will vacuum.(25 votes)

- At about2:47, 3 and 4 in the first row was selected, and the only way that could be right is for the other dice to roll a 0, which it can't, since the dice only has the numbers 1-6 on it.

So, the correct answer is A:

Because Miguel has a`"No, there is a higher probability that Miguel vacuums."`

probability of vacuuming, whie Jordan only has a**6/36**probability of vacuuming**5/36**(7 votes) - Why is everybody talking about Miguel and Jordan? The names of the people in this video are Roberto and Jocelyn.(6 votes)
- It seems like the video got updated and the names as well as the scenario in the problem changed.

Previously it was a dice roll about whether Miguel or Jordan will vacuum.

However it seems that the video had an error and it got updated as a result.(3 votes)

- If I am not mistaken, in that problem probability of Miguel doing the work is 6/36 and jordan doing the work is 5/36(6 votes)
- Yeah Jordan’s probability is 5/36(5 votes)
- Can Sal please remake this video to show the correct answer? Or start making columns with rows in different colors that are easier to see, tee hee!(5 votes)
- this is indeed wrong as Sal is using the row of the first die to include in the count...(1 vote)
- Is it possible to calculate these probabilities without drawing the graph and visualizing it?(1 vote)
- yes.

You just have to find out how many combinations of two numbers between 1-6 add up to the target number, then divide it by 36 to get the decimal probabilty.(1 vote)

- what if I want to go backwards?(1 vote)
- Who are Miguel and Jordan? The names are Roberto and Jocelyn, also what do people mean "3 and 4 in the first row are selected"?(1 vote)

## Video transcript

- [Instructor] We're told
that Roberto and Jocelyn decide to roll a pair
of fair six-sided dice to determine who has to
dust their apartment. If the sum is seven,
then Roberto will dust. If the sum is 10 or 11,
then Jocelyn will dust. If the sum is anything
else, they'll roll again. Is this a fair way to decide who dusts? Why or why not? So pause this video and see if you can figure this out
before we do it together. All right, now let's do this together. So what I wanna do is
make a table that shows all of the different scenarios for rolling two fair six-sided dice. So let me make columns for roll one. So that is when you get a one. This is when you get a two. This is when you get a three. This is when you get a four. This is when you get a five. And then, this is when you get a six. And then here, let's do the other die. So this is when you get a one. This is when you get a two. This is when you get a three. This is when you get a four. This is when you get a five. And then, this is when you get a six. So one way to think about it is this is roll one or let me write it this way, d1 and d2. This could be a one, a two,
a three, a five or a six. And this could be a one,
a two, a three, a four, a five or a six. Now what we could do is
fill in these 36 squares to figure out what the sum is. Actually, let me just do that and I'll try to do it really fast, one + one is two. So it's three, four, five, six, seven. This is three, four,
five, six, seven, eight. This is four, five,
six, seven, eight, nine. This is five, six, seven, eight, nine, 10. This is six, seven, eight, nine, 10, 11. Last but not least, seven, eight, nine, 10, 11 and 12, took a little less time than I suspected. All right, let's think
about this scenario. If the sum is seven,
then Roberto will dust. So where is the sum seven? So we have that ones, twice, three times, four, five, six. So six out of, so six of these outcomes result in a sum of seven. And how many possible equally
likely outcomes are there? Well, there are six times six equally possible outcomes or 36. So six out of the 36 or this is another way of saying there's a 1/6 probability
that Roberto will dust. And then, let's think
about the 10s or 11s. If the sum is 10 or 11, then Jocelyn will dust. So 10 or 11, so we have one, two, three, four, five. So this is only happening
five out of the 36 times. So in any given roll, it's a higher probability that Roberto will dust than Jocelyn will dust. And of course, if neither of these happen, they are going to roll again. But on that second roll, there's a higher probability
that Roberto will dust than Jocelyn will dust. So in general, this is not fair. There's a higher probability
that Roberto dusts. So this is our choice.