I'm preparing for some interviews and I'm stuck on a question which is rather simple, I'm just trying to get my head around how we redefine the sample space.
Given 4 coin flips, you get 1$ for each head you get. What's the expected value of the game (this is trivial = 2), now suppose you can flip an even number of coins after you see your coin flips, what's the expected value of the game now?
I know that if you see 0 heads then you definitely want to flip all 4 again, but how do you decompose the new probability distribution given the results? Also, we also want to flip the 2 T's we see if we have 2H2T, but then again how do we decompose the results?
would appreciate some intuitive response to this.
Edit:
If I have 0 heads, I would flip all 4:
Initially: $P(x=0) = \frac{1}{16}$, after flipping, the expected value is 2, so does this mean that now $P(x=2) = \frac{6}{16} + (\frac{1}{16}\frac{1}{16} + \frac{1}{16}\frac{4}{16})$? I'm kind of stuck as to how knowing the expected value of the second flip changes the initial distribution.
Best Answer
Just go case by case.
Sketch:
We see one of $\{0,1,2,3,4\}$ Heads initially, with easily derived probabilities.
$0$: probability $\frac 1{16}$. Then you retoss all $4$, giving you an expected value of $2$.
$1$: probability $\frac 4{16}$. Then you retoss two of the Tails, giving you an expected value of $2$ (the one you start with plus $1$ from the two new tosses). Note: you could retoss all $4$ if you like, as that also gives you an expected value of $2$.
$2$: probability $\frac 6{16}$. Then you retoss the two Tails, giving you an expected value of $3=2+1$ (the $2$ you had initially, plus $1$ from the new two.
$3$: probability $\frac 4{16}$. Then you do nothing and keep the $3$. If you tossed $2$ you;d still have an expected value of $3$ while if you tossed $4$ your expectation would drop to $2$.
$4$: probability $\frac 1{16}$. Obviously you do nothing and keep the $4$.
Thus the new expected value is $$\frac 1{16}\times (1\times 2+4\times 2+6\times 3+4\times 3+1\times 4)=\frac 1{16}\times (2+8+18+12+4)=\cdots $$ $$\cdots =\frac {44}{16}=\boxed {2.75}$$
Note: as a sort of sanity check, let's suppose you could retoss as many coins as you liked. Then the only way you get $T$ is if the coin comes up Tails twice in a row (a probability $\frac 14$ event). Thus, in this case the expected value would be $\frac 34\times 4 = 3$. Of course it makes sense that the expected value of your game ($2.75$) would be less than this but greater than $2$ (the value you'd get if you had no chance to retoss).