4 you are playing a game at a casino which has the feature...

You are playing a game at a casino which has the feature that your probability of winning is 0.48, of losing is 0.52 (no draws). Each time you bet $1, you get $2 if you win, $0 if you lose (so your net gain is +$1 or -$1). After you have played n times, we will say that you are a net winner if your total (or average) net gain is positive (i.e. you win more than 50% of the time), net loser if your total or average net gain is negative. (e.g. if you play three times, win twice and lose once, your net gain is +$1+$1-$1 = +$1 > 0, so you are a net winner). a) Find the mean and variance of your net gain, or 'payoff', on each (single) bet. b) Find the mean and variance of your average payoff on n bets. c) State the asymptotic distribution of your average payoff on n bets and use it to find the approximate probability of being a net winner after playing 10 times, and after playing 200 times. d) Interpret c) using the weak law of large numbers.

Transcript

00:01 So in this question, we have the probability of winning is 0 .48.

00:06 Probability of losing is 0 .52.

00:09 So the payoff, so we have x is 1 or minus 1.

00:15 That's our payoff.

00:17 And the probability of x is 0 .48, 0 .52.

00:26 So after n times, after you've played n times, you're a winner if your net gain is positive or a loser if your net gain is negative.

00:41 So we want to find the variance of, the mean and variance of your net gain on each bet.

00:49 So that is the expectation value of x, which is 0 .48 minus 0 .52, which is minus 0 .04.

01:01 We want to find the variance.

01:02 So the expectation of x squared is 1 times 0 .48.

01:09 Plus 1 times 0 .52 because we square it.

01:12 So that's just going to be 1.

01:15 So that means that the variance of x is 1 minus 0 .04 squared.

01:23 So 1 minus 0 .04 squared is variance of x is 0 .994.

01:33 So there we go.

01:34 We've got our expected value and our variance.

01:38 Part b, what about the payoff on nbets? well, the average payoff on n -bats, x -bar n, is the sum from i equals 1 to n x -i divided by n.

01:53 So where x -i is identically distributed with x, and they're all independent.

02:04 So what's the expected value of x -bar n? well, that's just 1 over n, sum from i equals 1 to n, of the expected value of x -i, which is exactly going to be the expected value of x -i, which is exactly going to be the expected value of x because these all have the same expected value.

02:22 There's n copies of them you divide by n, so this is going to be minus 0 .04.

02:33 But what about the variance? well the variance of x bar n is going to be the sum from i equals 1 to n variance of 1 over n x i because they're all independent.

02:46 We can split the variance into this sum.

02:49 So that means we're going to have to divide by n squared and then sum from i equals 1 to n variance of xi.

02:58 So this just gives us 1 over n times the variance of x.

03:04 So this is going to be variance of x bar n is equal to 0 .9984 divided by n.

03:19 Okay so we want the asymptotic distribution of your average payoff on n bets.

03:26 Well we can use the we can use the central limit theorem.

03:34 It tells us that asymptotically, x bar n is normally distributed with mean expectation of x bar n, which is minus 0 .04.

03:50 And variance, which is the variance of x bar n, 0 .9984 divided by n.

03:59 So we want to find it to find the approximate probability of being a net winner after playing 10 times...

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