nolxiii
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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..
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..
Stephen Tashi said: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!