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Game 1: Flip a fair coin $100$ times. Heads, you win $2$ gold coins. Tails, you lose $1$ gold coins. What is the fair price of this game? Calculate expectation.

Game 2: You start with $50$ gold coins and play Game $1$. Is $\mathbb{E}$[game $1$] = $\mathbb{E}$[game $2$]? Is there a different and is it a significant difference?

I think the meaning of this question is, for Game 1, you can finish the game by flipping coin 100 times anyway. However, for Game 2, if you lose all your gold coins, you cannot continue the game, which means the total number of flips may not be 100. Hence, the expectations should have slightly difference. Any ideas?

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  • $\begingroup$ What are the properties of expected value? Review that and you will have your answer. $\endgroup$
    – yiyi
    Mar 4, 2013 at 1:53
  • $\begingroup$ For game 2, do you flip each coin once? Or do you flip (some other coin) 100 times and win/lose just like game 1? It sounds like you are just 50 coins ahead in game 2. $\endgroup$
    – Ross Millikan
    Apr 16, 2013 at 13:32

3 Answers 3

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Hint: what is the expected value of one coin flip in game 1? You multiply the results by their probability and add. Then the expected value of 100 flips is just 100 times that.

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The linearity of the expected value.

$E[x+c] = E[x] + c$

This is what you need to know.

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  • $\begingroup$ I do not understand how E[x+c]=E[x]+c relates to the original question. Could you please explain? What is c here? $\endgroup$
    – user72835
    Apr 16, 2013 at 12:27
  • $\begingroup$ Did you read the link? Do you know what $E\left[x\right]$ means? $\endgroup$
    – yiyi
    Apr 17, 2013 at 10:14
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I know what $E[x]$ means but what I cannot figure out how it related to Game 2. If $c=50$ and $E[x]$ is the expectation for the first Game or Game 1 and you are saying that $E[Game 2] = E[Game 1 + 50]=E[Game 1] + 50$ then I do not think it is right. for example, we have 2 and we get 2 if it is a head or loose 1 if it is a tail. We stop if we tried 3 coin flipping or when we do not have money.

$HHH = 2*3=6\\ HHT = 2*2-1=3\\ HTH = 2*2-1=3\\ HTT = 2-1*2=0\\ THH = -1+2*2=3\\ THT = -1*2+2=0\\ TTH = -1*2=-2 \text{we stop here after two tails because we don't have money for the 3rd flip.}\\ TTT = -1*2=-2 \text{ we stop here after two tails because we don't have money for the 3rd flip.}$

So expectation for the Game 2: E[Game2]=1/8[6+3+3+0+3+0-2-2]=11/8=1.375. Ok, I can add 2$ to E[Game2] to say how much money I have on average at the end of Game 2. E[Game1]= 3*1/2*2- (3-3*1/2)*1=1.5 SO, E[Game2+2]=E[Game2]+2 but it is not equal to E[Game1]+2

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  • $\begingroup$ I don't think people believed that you would stop game 1 if your partial result went negative, so we saw game 2 as the same as game 1 with an extra 50 in your pocket. A starting reserve of 50 is enough that you will almost never run out, so the important question is what is the impact of running out on the expectation of game 1? If you can do all 100 flips regardless of partial results, the expectation of game 1 is clearly +50. $\endgroup$
    – Ross Millikan
    Apr 23, 2013 at 13:26
  • $\begingroup$ You have to be careful with dollar signs, as this site uses them to set $\LaTeX$ formatting. To get one without messing stuff up you have to escape them with dollar-backslash-dollar-dollar or just ignore them. $\endgroup$
    – Ross Millikan
    Apr 23, 2013 at 13:30
  • $\begingroup$ if you flipped the tail 50 times in a row, then 1 head + 52 times in a row (1 head can be anywhere starting with the first flip and finishing on 51 flip) then you stop the game and so on... $\endgroup$
    – user72835
    Apr 23, 2013 at 13:35

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