We will assume the probability of spotting the creature on any given day is $p$ and is independent one day to the next. We are told the probability the expedition will fail is $\\frac{1}{2}$. This occurs when the creature is not observed for $n$ days. This event has probability $(1-p)^n$. Therefore, we have
\n\\[
\n(1-p)^n = \\frac{1}{2}
\n\\]We can solve this equation for $p$ and we find
\n\\[
\np = 1-2^{-1\/n}
\n\\]<\/p>\n

If Graydon sends $k$ letters (with $1\\leq k \\leq n-1$), this means that the creature was spotted on the $k^\\text{th}$ day, after not being spotted for the first $k-1$ days. The probability of this event is $p(1-p)^{k-1}$. If $n$ letters are sent, then either the creature was spotted on the last day (probability $p(1-p)^{n-1}$), or the creature was not spotted at all (probability $(1-p)^n$). Let the average fraction of days where a letter is received be $A$. This quantity is equal to the expected number of letters received divided by $n$. Therefore, we obtain:
\n\\[
\nA = \\frac{1}{n} \\sum_{k=1}^n k p(1-p)^{k-1} + (1-p)^n
\n\\]We can evaluate this sum<\/a> in closed form, and we obtain
\n\\[
\nA = \\frac{1-(1-p)^n}{n p}
\n\\]Substituting what we found for $p$, we obtain
\n\\[
\nA = \\frac{1}{2n(1-2^{-1\/n})}
\n\\]We are told that \u201c$n$ is very very large\u201d, so it makes sense to examine what happens as $n\\to\\infty$. Doing so, we observe that $A$ tends to a constant:<\/p>\n

With a bit of work, we can actually evaluate the limit exactly:
\n\\[
\n\\lim_{n\\to\\infty} \\frac{1}{2n(1-2^{-1\/n})}
\n= \\frac{1}{\\log (4)} \\approx 0.721348
\n\\]So, on average, David should expect to receive a letter on approximately 72.1% of the days.<\/p>\n

Alternative approach<\/h5>\n
A slightly faster way to get to the same answer is to realize that if $(1-p)^n=\\frac{1}{2}$ and $n$ is \u201cvery very large\u201d, then it must be true that $p$ is \u201cvery very small\u201d. Specifically, if we recall the fact that
\n\\[
\n\\lim_{n\\to\\infty} \\left(1+\\frac{x}{n}\\right)^n = e^x
\n\\]Then we can rearrange:
\n\\[
\n(1-p)^n = \\left(1 + \\frac{(-np)}{n}\\right)^n
\n\\]So if $n\\to\\infty$, we conclude that $e^{-np} \\to \\frac{1}{2}$, or that $np \\to \\log(2)$. Therefore, we can return to our original expression for $A$ and substitute $(1-p)^n \\to \\tfrac{1}{2}$ and $np\\to \\log(2)$ and we find:
\n\\[
\nA = \\frac{1-(1-p)^n}{n p} \\to \\frac{1}{\\log(4)}
\n\\]which is the same thing we found using the first approach.\n<\/p><\/div>\n
<\/p>\n<\/body>","protected":false},"excerpt":{"rendered":"
This week\u2019s Riddler Classic is a question about large numbers of attempts at a very unlikely thing. Graydon is about to depart on a boating expedition that seeks to catch footage of the rare aquatic creature, F. Riddlerius. Every day he is away, he will send a hand-written letter to his new best friend, David … Continue reading “Spotting a rare creature”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":3491,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[7],"tags":[39,8,2],"class_list":["post-3468","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-limits","tag-probability","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t