Le Monde mathematical puzzle launched a competition to celebrate its 1000th puzzle! A fairly long-term competition as it runs over the 25 coming puzzles (and hence weeks). Starting with puzzle #1001. Here is the 1000th puzzle, not part of the competition: Alice & Bob spend five (identical) vouchers in five different shops, each ...

After a long lag (due to my missing the free copies distributed at Paris-Dauphine!), here is a Sudoku-like Le Monde mathematical puzzle: A grid of size (n,n) holds integer values such that any entry larger than 1 is the sum of one term in the same column and one term ...

“I was just thinking that there might be a magic trick to simulate directly from this distribution without having to go for less efficient methods.” In a simple question on X validated a few days ago [about simulating from x²φ(x)] popped up the remark that the person asking the ...

When I saw this title among the CRC Press novelties, I immediately ordered it as I though it fairly exciting. Now that I have gone through the book, the excitement has died. Maybe faster than need be as I read it while being stuck in a soulless Schipol airport and ...

As I discovered (!) the Annals of Applied Statistics in my mailbox just prior to taking the local train to Dauphine for the first time in 2017 (!), I started reading it on the way, but did not get any further than the first discussion paper by Pengsheng Ji and Jiashun Jin on ...

Still about this allocation riddle of the past week, and still with my confusion about the phrasing of the puzzle, when looking at a probabilistic interpretation of the game, rather than for a given adversary’s y, the problem turns out to search for the maximum of where the Y’...

The [then current now past] riddle of the week is a sort of multiarmed bandits optimisation. Of sorts. Or rather a generalised knapsack problem. The question is about optimising the allocation of 100 undistinguishable units to 10 distinct boxes against a similarly endowed adversary, when the loss function is and the distribution ...

A recent question on X validated ended up being quite interesting! The model under consideration is made of parallel Markov chains on a finite state space, all with the same Markov transition matrix, M, which turns into a hidden Markov model when the only summary available is the number of ...

A quick puzzle on The Riddler this week that enjoys a quick solution once one writes it out. The core of the puzzle is about finding the average number of draws one need to empty a population of size T if each draw is uniform over the remaining number of ...

Nicolas Chopin (CREST) just posted an entry on Statisfaction about the comparison of truncated Normal algorithms run by Alan Rogers, from the University of Utah. Nicolas wrote a paper in Statistics and Computing about a simulation method, which proposes a Ziggurat type of algorithm for this purpose, and which I ...

The Riddler of this week has an extinction riddle which summarises as follows: One observes a population of N individuals, each with a probability of 10⁻⁴ to kill the observer each day. From one day to the next, the population decreases by one individual with probability K√N 10⁻⁴ What is the ...

In answering yet another question on X validated about the numerical approximation of the marginal likelihood, I suggested using an harmonic mean estimate as a simple but worthless solution based on an MCMC posterior sample. This was on a toy example with a uniform prior on (0,π) and a “likelihood” equal ...

Luca Martino pointed out to me my own and forgotten review of a 2012 paper of his, “On the Generalized Ratio of Uniforms as a Combination of Transformed Rejection and Extended Inverse of Density Sampling” that obviously discusses a generalised version of Kinderman and Monahan’s (1977) ratio-of-uniform method. And further points ...

The last riddle from the Riddler [last before The Election] summed up to find the probability of a Binomial B(2N,½) draw ending up at the very middle, N. Which is If one uses the standard Stirling approximation to the factorial function, log(N!)≈Nlog(N) – N + ½log(2πN) the ...