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?

  • $\begingroup$ It partly depends on what counts as flip. If each individual coin counted, it would be $4\times 2=8$. $\endgroup$
    – Henry
    Oct 14, 2021 at 8:48
  • $\begingroup$ Oh I think what the problem intends is that you should flip all coins except for the heads. So at first flips you get 1 head out of 4, then next time you should flip 3 coins. That counts as one flip. $\endgroup$
    – cycla
    Oct 14, 2021 at 8:51

1 Answer 1


I am not completely sure about my understanding of the question (please check).

My understanding described with $n$ replacing $4$ to make things more general:

Let there be $n$ coins. The first flip is a flip of all coins and if $k$ head appears then we put these coins aside so that the next flip will be a flip of $n-k$ coins. The process ends if all coins are put aside.

To be found is then $\mu_4$.

This can be done on base of the following equalities:

  • $\mu_4=1+\frac4{16}\mu_1+\frac6{16}\mu_2+\frac4{16}\mu_3+\frac1{16}\mu_4$
  • $\mu_3=1+\frac38\mu_1+\frac38\mu_2+\frac18\mu_3$
  • $\mu_2=1+\frac24\mu_1+\frac14\mu_2$
  • $\mu_1=1+\frac12\mu_1$

This is based on: $$\mu_n=\sum_{k=0}^nP(n-k\text{ heads appear by throwing }n\text{ coins})(1+\mu_k)$$where $\mu_0:=0$.

It is handsome to start with the last one and find $\mu_1,\mu_2,\mu_3,\mu_4$ respectively.

For the second question I have no answer mainly because I do not understand what they mean (what exactly is a "wand"?).

Edit on second problem:

The wand (a new word for me) reduces things to the question: is the number of tails that appear even or odd? Then for every $n\geq1$ we have the equality:$$\mu_n=1+\frac12\mu_1$$

Consequently $\mu_n=2$ for every $n$.

  • $\begingroup$ Actually this is what I thought the answer would be. Using the recursive method. What do you think about the problem with the second method: which is just adding the expected value for leftovers after each toss? So for 4 tosses, it would be, 1 toss: 2 heads as expected values, then for 2 tails, you toss one more, than you will likely get 1 head so one tail left, so add 2 tosses more (since this is expected number of tosses for 1 coin). That being the answer: 4 tosses. $\endgroup$
    – cycla
    Oct 14, 2021 at 9:21
  • $\begingroup$ For the second part, the wand is just an explanation of flipping two tails, to two heads. So if you get HHTT, you could make it to HHHH without additional flips. $\endgroup$
    – cycla
    Oct 14, 2021 at 9:22
  • 1
    $\begingroup$ From the fact that the second method gives different answers than the first it can be concluded that the second does not work properly. Further it is based on "intuition" only and a proof of it lacks. $\endgroup$
    – drhab
    Oct 14, 2021 at 9:35

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