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Assume we start with 1.

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?

Thanks in advance!

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  • $\begingroup$ the answer depends on whether you can have negative balance, e.g. whether the game stops when you go broke $\endgroup$
    – Slug Pue
    Jan 3 '20 at 9:32
  • $\begingroup$ 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. $\endgroup$
    – roz
    Jan 3 '20 at 14:36
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    $\begingroup$ 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? $\endgroup$
    – amdopt
    Jan 3 '20 at 15:01
  • $\begingroup$ pretty sure op means gaining or losing 50% not half a dollar $\endgroup$
    – roz
    Jan 3 '20 at 15:18
  • $\begingroup$ 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. $\endgroup$
    – Chp
    Jan 5 '20 at 0:04
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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%.

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  • $\begingroup$ 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? $\endgroup$
    – Chp
    Jan 5 '20 at 0:08
  • $\begingroup$ 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. $\endgroup$
    – roz
    Jan 6 '20 at 1:03
  • $\begingroup$ 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. $\endgroup$
    – Chp
    Jan 7 '20 at 2:08
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    $\begingroup$ 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. $\endgroup$
    – roz
    Jan 8 '20 at 16:12

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