Suppose $a \\gt b \\gt c \\gt d$. Consider the event that $A$ is the champion. In order for this to happen, $A$ must first defeat $D$ in the semifinals. Then, either $B$ wins their semifinal match and $A$ defeats $B$ in the finals, or $C$ wins their semifinal match and $A$ defeats $C$ in the finals. We can use a similar argument to compute the probability that $B$, $C$, or $D$ wins, and we obtain the following formulas:
\n\\begin{align}
\nP_A &= \\frac{a}{a+d}\\left(\\frac{b}{b+c}\\cdot\\frac{a}{a+b}+\\frac{c}{b+c}\\cdot\\frac{a}{a+c}\\right) \\\\
\nP_B &= \\frac{b}{b+c}\\left(\\frac{a}{a+d}\\cdot\\frac{b}{a+b}+\\frac{d}{a+d}\\cdot\\frac{b}{b+d}\\right) \\\\
\nP_C &= \\frac{c}{b+c}\\left(\\frac{a}{a+d}\\cdot\\frac{c}{a+c}+\\frac{d}{a+d}\\cdot\\frac{c}{c+d}\\right) \\\\
\nP_D &= \\frac{d}{a+d}\\left(\\frac{b}{b+c}\\cdot\\frac{d}{b+d}+\\frac{c}{b+c}\\cdot\\frac{d}{c+d}\\right)
\n\\end{align}One can check that $P_A+P_B+P_C+P_D=1$ for all values of $(a,b,c,d)$, since exactly one of these four outcomes must occur. The expected value of the quality of the champion is therefore
\n\\[
\nx(a,b,c,d) = a P_A + b P_B + c P_C + d P_D.
\n\\]Ultimately, we seek the average of $x(a,b,c,d)$ where each $a,b,c,d$ is chosen uniformly at random in $[0,1]$. The formulas above require $a\\gt b \\gt c \\gt d$, but by symmetry each of the $4!=24$ permutations of $a,b,c,d$ is equally likely to occur. Therefore, we can restrict our averaging to the case where $a\\gt b \\gt c \\gt d$ and multiply the result by $24$. This leads to the following formula for $\\bar x$, the average value of $x$.
\n\\[
\n\\bar x = 24 \\int_{a=0}^1 \\int_{b=0}^a \\int_{c=0}^b \\int_{d=0}^c x(a,b,c,d)\\,
\n\\mathrm{d}d\\,\\mathrm{d}c\\,\\mathrm{d}b\\,\\mathrm{d}a
\n\\]This integral is very difficult to evaluate by hand, so I used Mathematica, and found the result to be:
\n\\begin{align}
\n\\bar x
\n&= \\frac{2}{5} \\left(23-(29+2\\pi^2)\\log(2)+39\\log^2(2)-8\\log^3(2)-3\\zeta(3)\\right)\\\\
\n&\\approx 0.6735417813867959221089
\n\\end{align}Here, $\\zeta$ is the Riemann zeta function<\/a>. The constant $\\zeta(3)=\\sum_{n=1}^\\infty\\frac{1}{n^3}\\approx 1.2020569$ is known as Ap\u00e9ry\u2019s constant<\/a>. The reason this constant shows up is because we are repeatedly integrating products of rational functions, and it turns out that
\n\\[
\n\\zeta(3) = \\int_0^1\\int_0^1\\int_0^1\\frac{1}{1-xyz}\\,\\mathrm{d}x\\,\\mathrm{d}y\\,\\mathrm{d}z
\n\\]Fun fact: it is known that $\\zeta(3)$ is irrational, but it is still not known whether it is transcendental<\/a>!<\/p>\n

Numerical verification<\/h3>\n

I also obtained an estimate for $\\bar x$ by performing a Monte Carlo simulation of 10 million games of Riddler Football Playoff in Matlab. Here is my code:<\/p>\n

\r\nN = 1e7;\r\n\r\nquality = zeros(N,1);\r\nfor trial = 1:N\r\n   \r\n    x = sort(rand(4,1));\r\n    a = x(4); b = x(3); c = x(2); d = x(1);\r\n    \r\n    % F = winner of Semifinal 1 (A vs D)\r\n    if rand < a\/(a+d)\r\n        f = a;\r\n    else\r\n        f = d;\r\n    end\r\n    \r\n    % G = winner of Semifinal 2 (B vs C)\r\n    if rand < b\/(b+c)\r\n        g = b;\r\n    else\r\n        g = c;\r\n    end\r\n    \r\n    % X = winner of Finals (F vs G)\r\n    if rand < f\/(f+g)\r\n        quality(trial) = f;\r\n    else\r\n        quality(trial) = g;\r\n    end\r\nend\r\navg = mean(quality)\r\nsem = std(quality)\/sqrt(N);\r\nconf_interval = [avg-1.96*sem,avg+1.96*sem]\r\n<\/pre>\nRunning the code above took about 10 seconds and produced an estimate of $0.67353$, with a 95% confidence interval $[0.67339, 0.67367]$. So it looks like there is good agreement between the numerical value derived from the exact formula and the empirical Monte Carlo estimate.\n<\/p><\/div>\n<\/body>","protected":false},"excerpt":{"rendered":"This week\u2019s Riddler Classic is a probability question inspired by the ongoing World Cup. The Riddler Football Playoff (RFP) consists of four teams. Each team is assigned a random real number between 0 and 1, representing the \u201cquality\u201d of the team. If team $A$ has quality $a$ and team $B$ has quality $b$, then the … Continue reading “Riddler Football Playoffs”<\/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":[7],"tags":[28,8,2],"class_list":["post-3589","post","type-post","status-publish","format-standard","hentry","category-riddler","tag-calculus","tag-probability","tag-riddler"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3589","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/comments?post=3589"}],"version-history":[{"count":7,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3589\/revisions"}],"predecessor-version":[{"id":3599,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3589\/revisions\/3599"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=3589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=3589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=3589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}