Define the following random variables:<\/p>\n

$m$: number of candy corn kernels in the bag\n<\/li>

$K$: whether we draw $k$ peanut butter cups in a row (true\/false)\n<\/li><\/ul>\n
We are asked to calculate the expected number of corn kernels in the bag given that we draw $k$ peanut butter cups in a row. In other words, we want to find $\\mathbb{E}(m\\mid K)$.<\/p>\n
An easier quantity to calculate is the probability that we draw $k$ peanut butter cups in a row given that there are $m$ candy corn kernels in the bag. This is given by:
\n\\begin{align*}
\n\\mathbb{P}(K\\mid m)
\n&= \\frac{n}{n+m}\\cdot \\frac{n-1}{n+m-1}\\cdots\\frac{n-k+1}{n+m-k+1} \\\\
\n&= \\frac{n!(n+m-k)!}{(n-k)!(n+m)!} \\\\
\n&= \\binom{n}{k}\\binom{n+m}{k}^{-1}
\n\\end{align*}Using Bayes\u2019 rule<\/a>, we can express the desired expectation in terms of the conditional probability above:
\n\\begin{align*}
\n\\mathbb{E}(m\\mid K)
\n&= \\sum_{m=0}^\\infty m\\cdot \\mathbb{P}(m\\mid K)
\n= \\sum_{m=0}^\\infty m\\cdot \\frac{\\mathbb{P}(K \\mid m)\\mathbb{P}(m)}{\\mathbb{P}(K)}
\n\\end{align*} Now substitute: $\\mathbb{P}(K) = \\sum_{m=0}^\\infty \\mathbb{P}(K\\mid m)\\mathbb{P}(m)$ and obtain:
\n\\begin{align*}
\n\\mathbb{E}(m\\mid K)
\n&= \\frac{\\sum_{m=0}^\\infty m\\cdot \\mathbb{P}(K \\mid m)\\mathbb{P}(m)}{\\sum_{m=0}^\\infty \\mathbb{P}(K \\mid m)\\mathbb{P}(m)} \\\\
\n&= \\frac{\\sum_{m=0}^\\infty m\\cdot \\mathbb{P}(K \\mid m)}{\\sum_{m=0}^\\infty \\mathbb{P}(K \\mid m)}
\n\\end{align*}where in the last step, we used the fact that $\\mathbb{P}(m)$ is the same for all $m$ so we canceled it from the numerator and denominator. Substituting our expression for $\\mathbb{P}(K\\mid m)$, we obtain:
\n\\begin{align*}
\n\\mathbb{E}(m\\mid K)
\n&= \\frac{\\sum_{m=0}^\\infty m\\binom{n}{k}\\binom{n+m}{k}^{-1}}{\\sum_{m=0}^\\infty \\binom{n}{k}\\binom{n+m}{k}^{-1}}
\n= \\frac{\\sum_{m=0}^\\infty m\\binom{n+m}{k}^{-1}}{\\sum_{m=0}^\\infty \\binom{n+m}{k}^{-1}}
\n%= \\frac{n-k+1}{k-2}
\n\\end{align*}
\nTo simplify this, define the following quantity:
\n\\[
\nf(k,a,b) := \\sum_{i=a}^b \\binom{i}{k}^{-1}
\n\\]and observe that we have:
\n\\[
\n\\mathbb{E}(m\\mid K) = \\frac{\\sum_{j=1}^\\infty f(k,n+j,\\infty)}{f(k,n,\\infty)}
\n\\]It turns out we can evaluate $f$, and that it has the nice closed-form expression:
\n\\[
\nf(k,a,b) = \\frac{k}{k-1}\\left( \\binom{a-1}{k-1}^{-1}-\\binom{b}{k-1}^{-1}\\right)
\n\\]This formula can be proved by induction and is related to something called the German tank problem<\/a>.<\/p>\n
With this formula in hand, we can take the limit $b\\to\\infty$ and see that:
\n\\[
\nf(k,a,\\infty) = \\frac{k}{k-1}\\binom{a-1}{k-1}^{-1}
\n\\]Substituting into our expression for the desired expectation, we obtain:
\n\\begin{align*}
\n\\mathbb{E}(m\\mid K)
\n&= \\frac{\\sum_{j=1}^\\infty f(k,n+i,\\infty)}{f(k,n,\\infty)} \\\\
\n&= \\frac{\\sum_{j=1}^\\infty \\frac{k}{k-1}\\binom{n+j-1}{k-1}^{-1}}{\\frac{k}{k-1}\\binom{n-1}{k-1}^{-1}} \\\\
\n&= \\binom{n-1}{k-1}\\sum_{j=1}^\\infty \\binom{n+j-1}{k-1}^{-1} \\\\
\n&= \\binom{n-1}{k-1} f(k-1,n,\\infty) \\\\
\n&= \\binom{n-1}{k-1} \\frac{k-1}{k-2} \\binom{n-1}{k-2}^{-1} \\\\
\n&= \\frac{(n-1)!}{(n-k)!(k-1)!} \\frac{k-1}{k-2} \\frac{(n-k+1)!(k-2)!}{(n-1)!} \\\\
\n&= \\frac{n-k+1}{k-2}
\n\\end{align*}<\/p>\n
Therefore, our final answer is:<\/p>\n
$\\displaystyle
\n\\mathbb{E}(m\\mid K) = \\frac{n-k+1}{k-2}
\n$<\/span><\/p>\n
I undoubtedly found the most complicated possible solution for this problem\u2026 So if somebody can show me a more elegant solution (perhaps a counting argument?) then I would be much obliged!\n<\/p><\/div>\n<\/body>","protected":false},"excerpt":{"rendered":"
This week\u2019s Fiddler is about rounding! You are presented with a bag of treats, which contains $n \\geq 3$ peanut butter cups and some unknown quantity of candy corn kernels (with any amount being equally likely). You reach into the bag $k$ times, with $3 \\leq k \\leq n$, and pull out a candy at … Continue reading “Halloween Puzzle”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"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":[42],"tags":[6,43,8],"class_list":["post-4149","post","type-post","status-publish","format-standard","hentry","category-the-fiddler","tag-counting","tag-fiddler","tag-probability"],"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