I've been trying to figure out this question using the recursive method.
What is the expected value of number of flips to get 4 coins to all land on heads, where once a coin lands on heads you dont have the reflip it, Also, if you had a wand that could flip any pair of identically facing coins with unlimited uses what would the new expected value be?
For the first question, my attempt is as follows:
For one coin case, expected flip is 2, which is clear. For 2 coins, you flip them first. There are 3 possible scenarios: 1/4 chance of having 0 head, 1/2 chance of having 1 head, 1/4 chance of having 2 heads.
If we set $a$ as the expected number of tosses needed for 2 coins, then we could set the equation as this:
$a = 1 + (1/4)*a + (1/2)*2 + (1/4) * 0 $ so that $a$ would be 8/3.
However, my intuition tells me that: by using simpler method: if we toss 2 coins, that expected number of head is 1, assuming the fair coin. And then since 1 coin is left, we add 2 since expected tosses for getting head with one coin is 2. So the total expected number would be $1 + 2 = 3$
So I'm confused which approach is right, and if one is wrong, why would it be?