100 coins problem 100 coins in a circle, clockwise labeled 1-100. Starting at   coin 1, you flip each one moving clockwise. If a coin lands tails, you keep it, if it lands heads, you remove it. After a long time, only one coin will remain. What would you expect its label is
The problem sounds pretty tricky but my initiation is that since each coin has an equal chance to become a tail or head, this would really become a uniform distribution so with that said the expected value for uniform distribution would be (100 + 1) /2 = 50.5
is this correct ? 
 A: Break the flippings into rounds around the table. Further, to make things easier mathematically, pretend that we flip every coin, including the eliminated ones, every time around. We just ignore the result. Further, we keep going until all coins have flipped a head. The "winner" is the final coin to flip a head. (We may know it will be the winner much earlier, but we will still stubbornly flip it until it finally shows a head.)
Every coin has an equal chance of surviving a round. The probability of any particular coin surviving $n$ rounds is $(1/2)^n$. the probability of not surviving $n$ rounds is therefore $$1 - (1/2)^n = \frac {2^n - 1}{2^n}$$
In order for coin $i$ to be the winner in round $n$, three conditions must hold:


*

*Coin $i$ survives $n-1$ flips.

*Every coin greater than $i$ does not survive $n-1$ flips.

*No coin survives $n$ flips.


The probability of $i$ surviving $n-1$ flips but not $n$ is $(1/2)^{n}$. The probability of every coin greater than $i$ not surviving $n-1$ flips is $$\left(\frac {2^{n-1} - 1}{2^{n-1}}\right)^{100 - i} = \left(\frac {2^n - 2}{2^n}\right)^{100 - i}$$
The probability of every coin less than $i$ not surviving $n$ flips is: $$\left(\frac {2^{n} - 1}{2^{n}}\right)^{i-1}$$
So the total probability of $i$ winning in round $n$ is:
$$\left(\frac 1{2^n}\right)\left(\frac {2^{n} - 1}{2^{n}}\right)^{i-1}\left(\frac {2^n - 2}{2^n}\right)^{100 - i} = \frac{(2^n-1)^{i-1}(2^n - 2)^{100-i}}{2^{100n}}$$
Which means that the total probability of $i$ winning is $$\sum_{n=1}^\infty \frac{(2^n-1)^{i-1}(2^n - 2)^{100-i}}{2^{100n}}$$ and of course the expected index is 
$$\sum_{i=1}^{100}\sum_{n=1}^\infty \frac{i(2^n-1)^{i-1}(2^n - 2)^{100-i}}{2^{100n}}$$
Since it is very late, I'll leave it there.
