The Angry Statistician

As I've mentioned previously , a common method used in sports for estimating game outcomes known as log5 can be written \[p = \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\] where \(p_i\) is the fraction of games won by team \(i\) and \(q_i\) is the fraction of games lost by team \(i\). We're assuming that there are no ties. What's the easiest way to derive this estimate? Here's one argument. Assume team \(i\) has a probability \(p_i\) of beating an average team (a team that wins half its games). Now imagine that this means for any given game the team has some "strength" sampled from [0,1] with median \(p_i\) and that the stronger team always wins. Thus, the probability that team 1 beats team 2 is \[ p = \int_0^1 \int_0^1 \! \mathrm{Pr}(p_1 > p_2) \, \mathrm{d} p_1 \mathrm{d} p_2 .\] This looks complicated, but but with probability \(p_1\) team 1 is stronger than an average team and with probability \(p_2\) team 2 is stronger than an average team. From this perspecti...

It's time for a workstation upgrade; here's what I've assembled. The massive case has plenty of space for additional drives to store the extreme amount of data that nearly every sport is now collecting together with the associated video footage. The case, power supply and motherboard allow up to 3 additional video cards to use as GPU units as your analytical needs demand (and budget can handle). Cooler Master Cosmos II - Ultra Full Tower Computer Case An absolutely massive case - plenty of room for an E-ATX motherboard, large power supply, multiple video cards and hard drives. PC Power & Cooling 1200W Silencer MK III Series Modular Power Supply An exceptionally large power supply, but platinum rated and plenty of room to spare allows it to operate nearly silenty, plus it'll handle any additional video cards you'll add later for GPU computing. ASUS Rampage IV Black Edition LGA 2011 Extended ATX Intel Motherboard Space for 4 video cards,...

As I mentioned in the previous article "The Good, the Bad and the Weird: Duels, Truels and Utility Functions" , a classic probability puzzle involves a 3-way duel (called a "truel" ). A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be? There's a subtle issue involved in these types of problems in that we don't know how each participant values each outcome. If we allow duelists to deliberately miss there are \(2^3-1=7\) possible outcomes; each person may or may not be shot and at least one person will not be shot. Even if deliberate missing isn't allowed, there are still 3 possible outcomes. A, for example, could conceivably value B winning m...

A man, a plan, a cat, a master, a minae, a lunula, a tora, a vela, a cola, a belga, a tola, an uta, a dona, a tiara, an aura, a kata, a mana, a nipa, a cabana, a law, a harem, a taces, a liard, a gnat, a mut, a dawed, a bura, a nim, a levator, an eh, a trams, a sall, a vacates, an up, a garb, a keets, a serin, a sored, a patamar, a gater, a lees, a strep, a lama, a pat, a lame, a haded, a wag, a naan, a kalian, a fer, a yaws, a remit, a keep, a loom, a neep, a ha, an ai, a retailor, a cabals, a subah, a slits, a pated, a callet, a xu, a faena, a tups, a sati, a rapider, a reps, a rapid, a korat, a dens, a sled, a tical, a pipages, a rats, a seeks, a defi, a wasts, a peracid, a nomas, a rosarian, a bran, a nonet, a gama, a sac, a scrota, a lab, a cain, an aid, a statin, a reel, a tiff, a haj, a talas, a nares, a mares, a laics, an allot, a til, an arena, a nits, a palace, a fanum, a yarer, a daub, a basts, a masas, a called, a slay, a galoots, a senega, a ton, a span, a mum, a dracaenas...

In the excellent (and highly recommended) book "Fifty Challenging Problems in Probability with Solution" , Frederick Mosteller poses "The Three-Cornered Duel": A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be? This is problem 20 in Mosteller's book, and it also appears (with an almost identical solution) in Larsen & Marx "An Introduction to Probability and Its Applications" . Mosteller's solution: A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the...

A bar bet as presented in the YouTube video  The HARDEST Puzzle Yet!  involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be \[ 6+6-6=6 .\] I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say \( N \), which involves using at most three different mathematical operations (although some of these may be used many, many times). If \( 0 \leq N \leq 2 \) we have the solutions \begin{eqnarray} (0! + 0! + 0!)! &= 6 \\ (1! + 1! + 1!)! &= 6 \\ 2+2+2 &= 6. \end{eqnarray} If \( N\geq 3\), concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result \( x \) with \( 3 \leq x < 9\).  If we now take the greatest integer \(\lfloor x \rfloor\) we have an integer \( n \) with \( 3 \leq n \leq 8 \). ...

Sir Arthur Eddington posed this difficult logic puzzle to the readers of Caliban's ( Hubert Phillips ) newspaper puzzle columns, and famously (or infamously) provided a solution using Dirac matrices . Sir Eddington's zoo puzzle: I took some nephews and nieces to the Zoo, and we halted at a cage marked Tovus Slithius, male and female.  Beregovus Mimsius, male and female.  Rathus Momus, male and female.  Jabberwockius Vulgaris, male and female.  The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no! It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right. As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr Tove identified the other animal of the same species a...