Using probability to make fair decisions

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 want to 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: 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 it for 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: this is roll one, or let me write it this way: die one and die two. This could be a one, a two, a three, a four, a five, or a six, and this could be a one, a two, a three, a four, a five, or 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 plus 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, ten. This is six, seven, eight, nine, ten, eleven. Last but not least, seven, eight, nine, ten, eleven, and twelve. Took a little less time than I suspected.

All right, let's think about this scenario. If the sum is 7, then Roberto will dust. So where is the sum 7? 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 7.

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 one-sixth 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. And of course, if neither of these happen, they're 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.