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If you can toss 4 coins and get 1 dollar for 1 head, 2 dollars for 2, 3 for 3 and 4 for 4. How much are you willing to pay for this game? If you are given an option that you can choose to toss the 4 coins again after the first toss, how much are you willing to pay now?

Attempt:

1) This is binomial distribution with p=1/2. Expected value is thus np = 4*0.5 =2.

2) I will reflip if I get 0 or 1 head. Therefore, my re-flip probability is P(0 head ) + P(1 head) = $(0.5)^4 *4C0 + (0.5)^4 * 4C1 =0.3125$. If I reflip, I would get $2 as the expected value. On the other hand, I wouldn't flip if I have 2, 3, 4 head, which give me the expected value of (0.5)^4*(2*4C2 + 3*4C3 + 4*4C4 )= 1.75.

Thus, the total amount would be 2*0.3125 + 1.75 = 2.375

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    i think the question isn't clear enough. – 2017-02-19
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    Your attempt at the first part is incorrect. Check again your calculations for the expected value. Note that the probability of getting two heads in four coin tosses is not equal to the probability of getting three heads or four heads, and the probabilities are not equal to $1/6$. Remember your binomial distribution. Also, do not confuse coins with dice. – 2017-02-19
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    @ JMoravtiz, just realize that. I have edited it. – 2017-02-19
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    Now you have the correct value for the first part. For the second part, do you have any intuition on when would be a good idea to choose to try again? If you get three heads the first time, do you want to toss again and possibly get less? If you get only one head the first time, do you want to toss again and possibly get more? What is the probability distribution for the amount of money you get if you have the option to reflip the coins? (*remember you get 4 heads either by flipping four heads initially or by flipping badly and choosing to reflip and getting four heads second time*) – 2017-02-19
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    Can you prove your intuition for when it is a good idea to reflip is correct? What is the expected value of this new random variable? – 2017-02-19
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    Allow me some to think through this. I think I should flip again if I get something below 2, since this is the expected value. – 2017-02-19

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Let $X$ be the amount of money you win from playing the game. There are $2^4=16$ possible outcomes of the 4 tosses so $\mathbb{P}(X=1)=4/16$, $\mathbb{P}(X=2)=6/16$, $\mathbb{P}(X=3)=4/16$, and $\mathbb{P}(X=4)=1/16$. Now $$\mathbb{E}[X]=\sum_{n=1}^4n\mathbb{P}(X=n)=4/16+12/16+12/16+4/16=2$$ As for part 2, use the total expectation theorem. Let $A$ be the event that your first toss is a heads. $$\mathbb{E}[X]=\mathbb{E}[X|A]\mathbb{P}(A)+\mathbb{E}[X|A^c]\mathbb{P}$$ $\mathbb{E}[X|A^c]$ is still 2. Given that $A$ occurs there are now $2^3=8$ possibilities corresponding to the last three flips. Now $\mathbb{P}(X=1)=1/8$, $\mathbb{P}(X=2)=3/8$, $\mathbb{P}(X=3)=3/8$, and $\mathbb{P}(X=4)=1/8$. Ill leave you to crunch the numbers.

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    I think the answer for part 2 is not 2.5 – 2017-02-19
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    Yes, I fixed my mistake. – 2017-02-19