You are flipping n fair coins, putting aside those that come up heads after each flip. What's the expected number of rounds? This question is related to https://www.quora.com/You-are-flipping-100-coins-putting-aside-those-that-come-up-heads-after-each-flip-How-many-flips-do-you-expect-to-make-before-running-out-of-coins-to-flip

At each stage, you flip all your coins and only keep the ones that are tail. You stop when you ran out of coins. Some Quora answers mention that the expected number of rounds is logarithmic with respect to $n$. Any suggestions on how you would prove this fact?
Since the probability of getting a head is $1/2$ after every round the number of coins will be probably halved. So let's call $C_j$ the number coins at the $j$-th round:

$$C_j=\frac{n}{2^j}$$

The game ends when:

$$C_j&lt;1$$
$$2^j&gt;n$$
$j&gt;\log_2(n)$

So the expected value $x(n)$ searched is:

$$x(n) \sim \log_2(n) $$

:)
Since the probability of getting a head is $1/2$ after every round the number of coins will be probably halved. So let's call $C_j$ the number coins at the $j$-th round:

$$C_j=\frac{n}{2^j}$$

The game ends when:

$$C_j&lt;1$$
$$2^j&gt;n$$
$j&gt;\log_2(n)$

So the expected value $x(n)$ searched is:

$$x(n) \sim \log_2(n) $$

:)