# What's the expected value of a repeated game with 50% chance to win 0.5 and 50% to lose 0.5?

In the first bet the expected value of remained balance is 1.5 * 0.5 + 0.5 * 0.5 = 1 For N times, is it still 1 according to E(XYZ)=E(X)E(Y)E(Z)? But 1.5^50 * 0.5^50 is not 1.

If the game is repeated N times, what's the expected value of remained balance in the end?

• the answer depends on whether you can have negative balance, e.g. whether the game stops when you go broke Jan 3 '20 at 9:32
• you can not reach a negative balance in this game since 1.5*x > 0 for all x > 0 and 0.5*x > 0 for all x > 0, but in the limit your bankroll will go to zero.
– roz
Jan 3 '20 at 14:36
• Does the winning or losing of '0.5' constitute winning or losing [0.5 * bank roll] or just an absolute 0.5 regardless of the bankroll? Jan 3 '20 at 15:01
• pretty sure op means gaining or losing 50% not half a dollar
– roz
Jan 3 '20 at 15:18
• thanks guys. I mean gaining or losing 50%. Since the expectation of return is 0 every time, shouldn't the overall expected value of return be 0 too according to E(XYZ)=E(X)E(Y)E(Z)? And this made the remained balance 1, which is false.
– Chp
Jan 5 '20 at 0:04

On average half the time you will win 0.5 times your current bankroll and half the time you will lose 0.5 times your current bankroll. Over N plays your expected growth will be (0.5)^(N/2)(1.5)^(N/2) and you will tend to lose money over time and in the limit since 0.5*1.5 = 0.75 < 1. This happens because gaining and losing 50% are not equivalent. Think about starting with 1 dollar and then losing 50%. In order to get back up to a dollar you have to gain 100%, not 50%.

• Thanks for your answer. I'm still confused about the formula E(XYZ)=E(X)E(Y)E(Z). The expected return is 0 every time. Does it mean that the overall expected return is 0 too?
– Chp
Jan 5 '20 at 0:08
• I don't know exactly what it is you are asking now. The EV over the ensemble and the EV of time across repeated plays will be different since this is a multiplicative wealth process. So it depends if you take the expectation over time or over space.
– roz
Jan 6 '20 at 1:03
• Sorry for the ambiguity. Let's say computing the expected return after 100 rounds of game. Since E(return i) = 1, E(total return)=E(return1*return2*return3*****)=E(return1)*E(return2)**** = 1. However it should be a very small number according to your answer above. I don't know where I get wrong here.
– Chp
Jan 7 '20 at 2:08
• I think this resource will help you: ergodicityeconomics.files.wordpress.com/2018/06/… . On page 5 at the start of the first chapter the author describes a game ery similar to the one you have described. If you read through the first chapter I think your issue is clearly explained in detail.
– roz
Jan 8 '20 at 16:12