Expected number of flips until all coins show heads

[nolxiii]

Originally from the statistics forum but am told this is more of a calculus question.

I flip 10 coins, if any of the coins land on tails, all of the coins split into 10 new coins and I flip them all again. I keep doing this until a round where every single coin lands on heads. Can I expect to ever stop flipping coins as the number of flips goes to infinity? (and followup question: if so, on average, how many flips would it take me?)

I think we've managed to at least state the problem mathematically but am unsure how to go about deriving an answer.

From the other thread..

Ok, that's understandable.The probability that you stop flipping after $N$ or fewer tosses
$= \sum _{i=1}^n pr( \ stop\ after\ exactly\ i\ tosses)$
$= \sum_{i=1}^n pr( \ toss\ heads\ with\ each\ of\ xy^{i-1}\ coins$
$= \sum_{i=1}^n \big( \frac{1}{2} \big) ^{xy^{i-1}}$

Your ask the value of
$\lim_{n \rightarrow \infty}<br /> { \sum_{i=1}^n \big( \frac{1}{2} \big) ^{xy^{i-1}} }$
which is also denoted by $\sum_{i=1}^\infty \big( \frac{1}{2} \big) ^{xy^{i-1}}$

The general problem is how to deal with series like $\sum_{i=0}^\infty C^{D^{\ i}}$

If that question were asked in the Calculus section, someone would probably have some ideas!

 

[BvU]

If I understand you well, you want the probability of 10 heads in a row ? One in $2^{10}$ ?

 

[nolxiii]

If I understand you well, you want the probability of 10 heads in a row ? One in $2^{10}$ ?

No that is just the first round, if the first time I flip any of the coins land on tails, then the next round I will flip 100 coins, and so on.

 

[BvU]

I see. The number of coins multiplies by 10 every time you get a tail ? And all of them have to end up heads ?
So $2^{-10}$, then $2^{-100}, \quad 2^{-1000}$etc ?

 

[nolxiii]

Correct.

I flip 10 coins, if any of the 10 are tails, then I go and flip 100 coins. If any of those 100 are tails then I flip 1000 coins. If during any iteration all of the coins I flip in that iteration land on heads then I stop and my task is complete.

 

[nolxiii]

I suspect that I'm really asking if the series converges? And if it diverges I'd expect to get all heads an infinite number of times if I kept going, and if it converges then I'd expect to get all heads a finite (and possibly less than 1) number of times if I continued on forever?

Or is making the jump from the value of the series to a statement of probability unfounded/completely wrong?

 

[mfb]

You stop after 1 flip with probability 2^-10, after two flips with probability (1-2^-10)*2^-100 and so on. The total probability to stop converges to a value extremely close to 2^-10=1/1024, and below 1/1000.
The probability that you ever get "all heads" is also small (below 1/1000, and dominated by the first flip).

 

[nolxiii]

thank you!