 So one of the things that makes probability difficult is that, except for a very, very, very, very, very, very, very, very few cases, computing a probability is challenging. So last time we looked at the problem of computing the probability that the sum of two six-sided dice add to ten. And when we first solved this problem, we computed a probability of one-eleventh. As a result, we supported one of the fastest growing industries in the world, gambling. And this is one of the industries that thrives when people don't know probability and statistics. So let's see what the problem is with our conclusion. The first important thing to recognize is that we computed this probability under the assumption that our sample space consisted of equally likely outcomes. But remember, most sample spaces do not consist of equally likely outcome. And so we might pause and ask ourselves, why doesn't this sample space consist of equally likely outcomes? Well, let's think about that. To get a two on two dice, both dice need to be ones. On the other hand, I can get a five by having one die show a one and the other a four, or one die showing a two and the other a three. And what that means is that some of these numbers, like five or seven or nine, can be obtained in many different ways, while others, like two or twelve, can only be obtained by one very specific roll of the dice. And so that means our sample space does not consist of equally likely outcomes. And this leads to the following problem. We can compute probabilities if we have a sample space of equally likely outcomes. And so this leads to the following problem. Given a random experiment, find a sample space of equally likely outcomes. And you might wonder how this is possible, because aren't the outcomes the possible results of the random experiment? And the answer is yes, but we might be able to describe those results in different ways. So let's analyze our problem. Suppose you roll the dice one at a time. Now the first die can land one, two, three, four, five, or six, and it seems that these are all equally likely outcomes. Certainly if one of these came up more often than the other, you'd have a die that was considered loaded, and people would be very upset with you if you should play with it. By the same argument, that second die can also land one, two, three, four, five, or six, and these also seem to be equally likely outcomes. This suggests our sample space might consist of the ordered pairs x, y, where x is the number shown in the first die, and y is the number shown on the second die. The idea being that since these individual numbers are equally likely to occur, then our ordered pairs are also going to be equally likely to occur. Let's formalize that question. Suppose you roll two dice and record the outcome as x, y, where x is the number shown in the first die and y is the number shown in the second die. What's the sample space and does it consist of equally likely outcomes? So we could try to list all of our outcomes and let's try to be somewhat systematic about this. If x is the number shown on the first die, then that first number could be anything between one and six. Now if the first die shows a one, that second die could also show a number between one and six. So our outcomes could be 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, and 1, 6. Similarly, if our first die shows a 2, that second die could be a 1, 2, 3, 4, 5, or 6. And similarly for all the others. We see there are 36 outcomes in S. Now we made a claim that this sample space consists of equally likely outcomes, but let's see if we can make a good argument for that. So a good way to reason in mathematics and life is to ask yourself, what are the natural consequences? So what if our outcomes are not equally likely? So that one outcome, say 5, 3, is more likely than another outcome, for example 2, 1. It seems that in order for this to happen, this would require that 5 is more likely to show on the first die than 2, or that 3 is more likely to show on the second die than 1. But the outcomes on the individual dies are supposed to be equally likely. So it doesn't seem like this is a possibility. So S seems to consist of equally likely outcomes. So let's get back to our question about the probability of rolling a 10 on two dice. So the first thing we did is we actually found a sample space of what we might consider to be equally likely outcomes. Now we need to find the event E that corresponds to rolling a 10 on two dice. So how can we get a 10? Well, maybe our first die will roll a 1 and our second die will roll a 9. Oh, wait, that can't happen because our dice could only show 1 through 6. So that means the lowest the first die can be as a 4 and a second die would have to be a 6. And that corresponds to the outcome 4, 6. Our first die might be a 5 and the second die could also be a 5. And that corresponds to the outcome 5, 5. Our first die could be a 6 and the second die could be a 4. That corresponds to the outcome 6, 4. Our first die could be a 7. Wait, that's not possible either. And so the event that corresponds to rolling a 10 consists of the three outcomes 4, 6, 5, 5, 6, 4. And so the probability of this event is going to be the number of outcomes in our event over the number of outcomes in S. And so that's 3 over 36.