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The questions reads:

There are $37$ numbers, from $0$ to $36$. Each number has an equal chance of turning up. Zero is green in color and odd numbers are in black and even numbers are in red. If you place $\$1$ on red (black) you get $2$ dollars if you are right and $\$0$ if you are wrong. For each bet there are $19$ losing numbers and $18$ winning numbers.

Suppose you place $\$1$ bets $18$ times on red. What is your expected gain or loss after $18$ tries?

My working:

Expected value of one game = $2(18/37) + -$1(19/37)

Expected value of 18 games = ($2(18/37) + -$1(19/37))^18

Is this right?

Part 2:

Suppose I use a dollar for each color bet? How many games can I play if I have an initial capital of $18?

Not sure how to start on this...

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It's the right way to go about it, but there are two bugs:

  • The value of a win is $1$, not $2$.
  • The expected outcome after $18$ rounds is $18$ times the expected expected outcome after $1$ round, by linearity of expectation.
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  • $\begingroup$ I don't understand what "I use a dollar for each color bet" means. $\endgroup$ – Rebecca J. Stones Sep 23 '13 at 13:53
  • $\begingroup$ You place a dollar bet on either color. $\endgroup$ – lakesh Sep 23 '13 at 13:54
  • $\begingroup$ Are you asking about the expected stopping time? I'll need to think a bit about that. My guess would be that it's the minimium $k$ such that $k\times \mathbb{E}[\text{outcome after 1 game}] \leq -18$. $\endgroup$ – Rebecca J. Stones Sep 23 '13 at 14:07
  • $\begingroup$ since a dollar for each colour bet, do i need to multiply the outcome of each game *2? $\endgroup$ – lakesh Sep 24 '13 at 17:23
  • $\begingroup$ I'm not sure what you mean, but the change in the total money will either be +1 or -1 after betting $1. $\endgroup$ – Rebecca J. Stones Sep 24 '13 at 17:26
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We can think of the game as a $1$-dimensional random walk where we take a step $+1$ (when we win $\$1$) with probability $p := \frac{18}{37}$ and a step $-1$ (when we lose $\$1$) with probability $1-p = \frac{19}{37}$.

The expected profit after $n$ plays is

$$(2 p - 1) \, n = \left(\frac{36}{37} - 1\right) n = -\frac{n}{37}$$

which is negative, as expected. If $n = 18$, then the expected loss is $\frac{18}{37}$ dollars, which is approximately $50$ cents.

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