 Hello and welcome to the session. In this session we discuss how to find the expected payoff for a game of chance. Now we know that expected value of a random variable is defined as the weighted average of all possible values that this random variable can take on and it is a measure of central redundancy of random variable. Now we know that in statistics weighted average is equal to summation of w i x i whole upon summation of w i where w i represent the weights and x i represent the data points. Now when calculating expected values in probability the weights that is w i are just the probabilities of the values x i. In this case the sum of the weights in the denominator is equal to 1 because the probabilities will sum to 100% if you are considering the complete probability distribution. Hence the expected value becomes summation of p i x i where i varies from 1 to n and here we know that sum of the probabilities is equal to 1. All you can say if capital X is a random variable with possible values denoted by x i and p of x i denotes probability distribution of random variable capital X is equal to x i then expected value of random variable capital X is given by even capital X is equal to summation x i into p x i the whole where i varies from 1 to n. Now the expected value of the game is the sum of the expected payoffs of all the consequences or we can say expected value is the sum of the possible payoffs each multiplied by its probability. Now let us discuss an example. Here we are given that 500 tickets are sold at $1 each for a charity effort. Tickets are to be drawn at random and cash prices are given as follows. One price of $50, two prices of $20, three prices of $10 and we have to find that what is the expected value of this example if you buy one ticket. Now let us start with the solution. Here we are given that 500 tickets are sold at $1 each and we have one price of $50. So probability of this one price out of these 500 tickets is equal to 1 upon 500. This means that you have 1 by 500 chance of winning $49 as this one price is of $50 and cost of one ticket is $1. So the person has a chance that is 1 by 500 chance of winning $50 minus $1. That is $49. Now it is given that there are two prices of $20. Now again we know that total number of tickets is equal to 500. So you have 2 by 500 chance of winning $19. That is $20 minus $1. Then there are three prices of $10. This means you have 3 by 500 chance of winning $10 minus $1. That is $9. Now we know that probability of an event not happening is equal to 1 minus probability of an event happening. And here we have found probability of winning one price, two prices and three prices. So probability of not winning any of the prices will be 1 minus probability of winning one price that is 1 upon 500 plus probability of winning two prices that is 2 upon 500 plus probability of winning three prices that is 3 upon 500 the whole. So this is equal to 1 minus 6 upon 500 which is equal to 500 minus 6 whole upon 500 and this is equal to 494 upon 500. So probability of not winning any of the prices is 494 upon 500. This means you have 494 upon 500 chance of winning minus $1. That is losing $1 which is the cost of ticket. Now let us make a payoff table for the above obtained results. Here let the random variable capital X represents the amount 1 and P of X be its probability. Now in this table we will denote the possible values of random variable capital X by Xi and their probabilities by P of X equal to P of random variable capital X is equal to Xi. Now for one price the amount 1 is $39 and its probability is 1 upon 500 then for two prices amount 1 is $19 and its probability is 2 upon 500 similarly for three prices amount 1 is $9 and its probability is 3 upon 500 then for not winning any of the prices amount 1 is minus $1 and its probability is 494 upon 500. Now we know that expected value of random variable capital X is equal to summation of Xi into P of Xi the whole where i varies from 1 to n. So here expected value is equal to $49 into 1 upon 500 plus $19 into 2 upon 500 plus $9 into 3 upon 500 plus of minus $1 into $494 upon 500 solving this is equal to $49 plus $38 plus $27 minus $494 upon 500 which is equal to minus $380 upon 500 that is equal to minus $0.76. So the expected value of this example if you buy one ticket is equal to minus $0.76. So in this session we have discussed how to find the expected payoff for a game of chance and this completes our session hope you all have enjoyed the session.