Previously, we've covered some simple probability distributions. Now in this section, I'd like us to move on and consider the concept of expectations. So, remember what we said about decision making under uncertainty. We have to make decisions now not knowing exactly what's going to happen in the future. So, the best we can do today is form an expectation about what we anticipate would happen in the future. So, let's revisit the case of rolling a fair die. So, previously, we saw the topiram and graphical form of that probability distribution, i.e., all possible outcomes of that experiment and assigned to each value its probability of occurrence. So, in advance of rolling the die, what would we expect to be the outcome? So, up to this point we've just looked at the distribution, the probability distribution as a whole, and now, I'd like us to focus on one particular attribute of that distribution i.e., what is the expectation? On average what score on the die do we think is going to occur? So, we're now going to consider the concept of a mean, an expectation, an average. We can use these terms synonymously, i.e., interchangeably, to represent what is the expected score on this die. So, we see those six possible values one, two, three, four, five, six, so, what we're going to do, is to take a probability weighted average, i.e., we can consider each possible value of the variable one, two, three, four, five, six, and assign to each of those a weight reflecting its probability of occurrence. Now, in the case of a fair die, of course these weights are going to be equal, because we said for a fair die each of those six scores was equally likely to occur. So, the way to extract, or indeed calculate this expectation of X, the expected score on a die, is to take each value of the variable, multiply it by its probability of occurrence, and add them all together. So, what is the expected score when we roll a fair die? It will be one times a sixth plus two times a sixth etc., up to six times a sixth. Now, if you want to type those values into your calculator, you will come up with an expectation of 3.5. So, what this is saying is that you're expected score when you roll a fair die is three and a half. Now, of course, we know you would never roll 3.5 when you roll that dice. It was not a member of the sample space. It was not a possible outcome of this experiment. So, how should we interpret this concept of an expectation? Well let's imagine you did this sort of relative frequency experiment. You took your fair die and you rolled it. Not once, but a very large number of times. Let's keep the numbers simple. Let's assume you roll this fair die, let;s say 600 times. Then, one would anticipate approximately, 100 times out of those 600, one would be the uppermost space, roughly a 100 times it would be a two, three, four, five, and six. So, if you now consider the average score across those 600 rolls of the die, you would get a value of approximately 3.5. I say approximately because it's pretty unlikely you would roll exactly 100 ones, twos, threes, fours, fives, and sixes. So the expected value, this probability weighted average that we had calculated, I'd like you to view this as a long-run average, such that if you roll this die a very large number of times, the 3.5, we would call it a measure of central tendency, a measure of location i.e., the value that we tend to get on average, even though the 3.5 itself is not an attainable value by itself. Let's consider another example, let's go back to tossing the fair coin. So, there we said, the outcomes, heads and tails were not numeric, but we could easily apply some simple binary coding let's say one for heads and zero for tails. And assuming this fair coin, there may have a 0.5 or a 50% chance of each of those outcomes. So, what would be the expectation for the score on the die here. Well, we take each value of that variable, zero and one in this case, multiply it by its corresponding probability of occurrence, in this case, this equal weights of 0.5 and 0.5. So, zero times 0.5 plus one times 0.5, will give us no 0.5. So, there is our expectation for this simple probability distribution. Now, you might argue 0.5 has no sort of real world meaning in the sense of tossing this fair coin, and I'd perhaps agree with you on that point. But nonetheless, that methodology of calculating an expected value remains the same as in the case of the fair die. So, let's consider one more example. Suppose you're feeling lucky and you head down to the casino, and let's suppose you're interested in roulette. So you go to the roulette table, and the casino offers the following game. Namely, you bet one pound, one dollar, one euro whichever currency you prefer such that if when you spin the ball on the roulette wheel, if a red number comes up, you get that one pound stake back, but you also get winnings of a pound. If you don't get a red number coming up, you lose that stake of a pound. Now roulette wheels do vary in different parts of the world. For our example here, we will consider a roulette wheel with 18 red numbers, 18 black numbers, and one green zero. So, thinking back to our previous session on some simple probability distributions, let's assume this is a fair roulette wheel, such that when you spin it and the ball bounces around, it's equally likely to land on any of those numeric slots on the roulette wheel. So, if we have 18 red numbers, 18 black numbers, and don't forget that green zero, that would mean we have 37 equally likely outcomes when we spin this roulette wheel. So let's come up first of all with a simple probability distribution, and from then, I'd like us to calculate the expected value and think about the real world meaning of this as well. So, our experiment we are roll or spin the ball around the roulette wheel, and consider what the outcome is. So, if we take things from the perspective let's say of the casino, sometimes called the house. So, what will happen? When the person stakes the pound, dollar, the euro, one of two things will happen. The ball will spin, and if that ball lands on a red number, then the customer wins a pound from the casino. And of course that would reflect a loss from the house's or the casino's perspective. So, let's say in this experiment one possible outcome is the loss of a pound, minus one from the casino's perspective if a red number comes up. Of course, the other possibility is that it's not a red number, which means either one of the 18 black values or the green zero, and in this case the casino gets to keep the stake of the customer and hence it makes if you like a profit of plus one pound, dollar or euro. So, if we write out this probability distribution, from the casinos perspective, there were two possible outcomes, loss of minus one, if it is a red number which occurs, or a profit of plus one if it's a black number or the green zero. So, considering this sort of classical approach to probability we saw previously, whereby every possible outcome in the experiment is equally likely. So, in this case, these are the 37 different values on the roulette wheel. And we're imagining if it's a fair game, that that ball is equally likely to land on any of them. Of course, 18 of those 37, agree with the event of a red number occurring. So, the probability that the house, the casino loses this pound, will be 18 over 37. Remember the little n over capital n. Number of outcomes agreeing with the event of interest, over the total number of equally likely outcomes in the game. Of course, to complete this probability distribution somewhat trivially, if there were 18 red numbers, every of all of the other values are either black numbers, or that green zero, which there are a total of 19 of those possible values. So, the probability that the house wins that pound from the customer will be 19 over 37. So, let's now work out the expected winnings or potentially losses which are accrued to the casino. So, we take each value of this random variable i.e., the winnings or losses to the house, multiply by the corresponding probability of occurrence and add these together. So, there's a loss of minus one occurring with a probability of 18 over 37. So, we do minus one times 18 over 37, plus one the winnings over of a pound times 19 over 37, the probability that that event occurs. Now, I will allow you to use a calculator to work this out but I checked this earlier, and the expectation you would get is 0.027. So, just like with the toss of a roll of a fair dice where we had an expected score of 3.5 that was not attainable, similarly here, the expected outcome, this 0.027, could never occur in a single play of this roulette game. Because we know the only possible outcomes is that the casino either wins a pound or it loses a pound. But remember the interpretation now, of viewing the expectation as a long run average. So, what this means that for every pound that is staked by a customer on this roulette game, the casino can expect to win 2.7 pence. That's a 2.7% gain of all of the money which is staked by the customers. So, true in any one play of this game, maybe the customer is lucky, a red number comes up and the casino makes a loss. But of course, we would anticipate this is going to be happening 18 out of 37 plays or 18 plays out of every 37 of this game. Of course on average, 19 out of every 37 plays of the game will be good for the casino and it wins some money. So, in the long run, of course the casino sees a positive expectation here. Now, this is no coincidence, the casino operators are not stupid. They design these games such that on average, in the long run that the house always wins. Of course, the casino would want customers to win on occasions. If every time someone made a gamble in the casino, they lost money, if there was never any winnings on the customer's part, who would ever go to the casino if it was a certainty that you're going to lose money? So, the casino welcomes these success stories on occasions where some people do win lots of money. But of course for every winner there are going to be typically more losers and on average, in the long run, the casino comes out ahead. So, nice section here for us to consider the concept of an expectation. May also like to relate this back to that Monty Hall problem. You have a two thirds chance of success if you switch the door, only a one third chance of success if you stick with your original choice. So, again thinking in terms of the long run, as a strategy, if you switch each time and one is expected to win, two thirds of the time and lose only one third of the time. So, in the long run you would win far more often than you lose. So, remember there are no certainties typically in life other than there's death and there's taxes. So, if we can think about forming expectations, we would tend to base the decision-making under uncertainty but use these expectations really at the heart of our decision making process.