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Suppose you are playing a game where you are betting dollars and if you flip a coin and it is heads, then you win that amount, but if it's tails, you lose that amount. You use the strategy that you start by betting $\$1$. If you lose, you bet $\$2$. If you lose, you bet $\$4$ dollars. So in round $n$, you are betting $2^{n-1}$ dollars. However, if you win, you stop playing.

What is the net winning after we win on the $n$th coin flip? How do we find the expected value of our winnings in the $i$th round and our expected net winnings after the $n$th round?

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closed as off-topic by Ian Miller, colormegone, choco_addicted, Watson, Shailesh Apr 8 '16 at 5:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ian Miller, colormegone, choco_addicted, Watson, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi, welcome to Math.SE. Please indicate what you have tried and where you are stuck. This will help people better tailor their answer to your background and situation. It will also demonstrate that you are interested in your question and not just looking for someone to do your homework for you - Math.SE is not a homework site. $\endgroup$ – Ian Miller Apr 8 '16 at 4:53
  • $\begingroup$ Have you done any research yet yourself? This is a very commonly discussed betting system and a quick internet search would turn up lots on it - en.wikipedia.org/wiki/Martingale_(betting_system) $\endgroup$ – Ian Miller Apr 8 '16 at 4:55
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Suppose that you win on the $n$th turn, then your net gain is $$2^{n-1}-(1+2+\dots+2^{n-2})=2^{n-1}-(2^{n-1}-1)=1$$

This betting strategy is called a martingale, and is the origin of the term martingale in probability theory. The downside of course is that the game can last arbitrarily long, so you need an unlimited amount of money to employ it.

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  • $\begingroup$ Thank you. What about the expected values? $\endgroup$ – user329672 Apr 8 '16 at 4:56
  • $\begingroup$ The expected value is zero: after $i$ steps, you've lost $2^i-1$ dollars with probability $2^{-i}$, and won a dollar with probability $1-2^{-i}$. $\endgroup$ – carmichael561 Apr 8 '16 at 5:01
  • $\begingroup$ Then is the expected net winnings after the nth round also 0? $\endgroup$ – user329672 Apr 8 '16 at 5:03
  • $\begingroup$ Yes. Your net gain is only 1 only if you win on that turn. $\endgroup$ – carmichael561 Apr 8 '16 at 5:04
  • $\begingroup$ Got it.. Thanks $\endgroup$ – user329672 Apr 8 '16 at 5:05

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