## a funny mistake

Posted in Statistics with tags , , , , , , , , , , , on August 20, 2018 by xi'an While watching the early morning activity in Tofino inlet from my rental desk, I was looking at a recent fivethirthyeight Riddle, which consisted in finding the probability of stopping a coin game which rule was to wait for the n consecutive heads if (n-1) consecutive heads had failed to happen when requested, which is

p+(1-p)p²+(1-p)(1-p²)p³+…

or $q=\sum_{k=1}^\infty p^k \prod_{j=1}^{k-1}(1-p^j)$

While the above can write as $q=\sum_{k=1}^\infty \{1-(1-p^k)\} \prod_{j=1}^{k-1}(1-p^j)$

or $\sum_{k=1}^\infty \prod_{j=1}^{k-1}(1-p^j)-\prod_{j=1}^{k}(1-p^j)$

hence suggesting $q=\sum_{k=1}^\infty \prod_{j=1}^{k-1}(1-p^j) - \sum_{k=2}^\infty \prod_{j=1}^{k-1}(1-p^j) =1$

the answer is (obviously) false and the mistake in separating the series into a difference of series is that both terms are infinite. The correct answer is actually $q=1-\prod_{j=1}^{\infty}(1-p^j)$

which is Euler’s function. Maybe nonstandard analysis can apply to go directly from the difference of the infinite series to the answer!