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I am modifying a question asked in MSE here.

Flip four coins. For every head, you get $\$1$. You may reflip at most two coins after the four flips. Calculate the expected returns.

In this case, would the answer be the same as reflipping one coin, which is $$\frac{79}{32}?$$

gt6989b
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Idonknow
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  • @MatthewDaly If I got four heads in the first try, then I will not reflip any of them and get $$4$. – Idonknow Jan 21 '20 at 14:55
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    This is the same as flipping six coins, and getting $1 for each head up to $4. – Thomas Andrews Jan 21 '20 at 14:57
  • I get an answer of $\frac{45}{16}$. Arrived at as $\frac{(4+0)\cdot 1+(3+\frac{1}{2})\cdot 4+(2+1)\cdot 6+(1+1)\cdot 4+(0+1)\cdot 1}{2^4}$. The method employed should be clear from the numbers used. – JMoravitz Jan 21 '20 at 14:57
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    As for "would the answer be the same as reflipping one coin?" very obviously not. By giving the player the option to flip an additional coin, you will strictly increase their expected return, seen by the fact that given the same scenario to both players, having flipped two heads and two tails, the player who can only reflip $1$ coin will increase their score by $\frac{1}{2}$ on average while the player who can flip two will increase their score by $1$ on average. – JMoravitz Jan 21 '20 at 15:00
  • @ThomasAndrews If this is the case, why wouldn't the linked question be solved in the same way, that is, treat the problem as flipping five coins? – Idonknow Jan 21 '20 at 15:00
  • That should be one way to solve the original question. @Idonknow It's not the only way – Thomas Andrews Jan 21 '20 at 15:02
  • @Idonknow who says it wouldn't be solved in the same way? Note: $\frac{0+1\cdot 5 + 2\cdot 10+3\cdot 10+4\cdot 5+4\cdot 1}{2^5} = \frac{79}{32}$ – JMoravitz Jan 21 '20 at 15:02
  • Oh, my approach gives the wrong answer if flipping six gives HHHTTH because we only get the one after the four. Actually my method works in the original problem, but fails in the six-coin version. – Thomas Andrews Jan 21 '20 at 15:05
  • @ThomasAndrews hmm... My method earlier does not agree with that answer... The odd case out would seem to be the sequence of coinflips $HHHT$ and choosing to reflip one coin and getting a tail. In my interpretation we are not allowed to reflip that reflipped coin as our second reflip. Thus, the sequence of six flips $HHHTTH$ does not count as having gotten a score of four in my interpretation. If we are allowed to reflip the coin that was reflipped as our second reflip, then they would agree. – JMoravitz Jan 21 '20 at 15:06
  • Yeah, i just realized that case. @JMoravitz – Thomas Andrews Jan 21 '20 at 15:07

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This seems similar to your earlier question. Without reflips your expected gain is 2. You have 1/4 chance of flipping just one T, in which case you reflip one coin, and so your expected gain increases by $\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$. You have 11/16 chance of getting at least two T, in which case you reflip two coins, and your expected gain increases by 11/16. So total expected gain $$2+\frac{13}{16}=\frac{45}{16}=\frac{90}{32}$$

almagest
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