Using probabilities to make fair decisions example

- [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.