the biggest bluff [not a book review]

It came as a surprise to me that the book reviewed in the book review section of Nature of 25 June was a personal account of a professional poker player, The Biggest Bluff by Maria Konnikova.  (Surprise enough to write a blog entry!) As I see very little scientific impetus in studying the psychology of poker players and the associated decision making. Obviously, this is not a book review, but a review of the book review. (Although the NYT published a rather extensive extract of the book, from which I cannot detect anything deep from a game-theory viewpoint. Apart from the maybe-not-so-deep message that psychology matters a lot in poker…) Which does not bring much incentive for those uninterested (or worse) in money games like poker. Even when “a heap of Bayesian model-building [is] thrown in”, as the review mixes randomness and luck, while seeing the book as teaching the reader “how to play the game of life”, a type of self-improvement vending line one hardly expects to read in a scientific journal. (But again I have never understood the point in playing poker…)

Les Enfants Rouges

An happenstance reservation at Les Enfants Rouges led to a great and unique meal. The restaurant is located in Le Marais, north of Paris City Hall, and the chef Daï Shinozuka brings a precision in the cooking and presentation of traditional French food that makes each dish a masterpiece. Above the “neo bistro-fare” lauded by The New York Times. But definitely part of the new and exciting food scene in Paris!

  

cyclic riddle [not in NYC!]

In the riddle of this week on fivethirtyeight, the question is to find the average number of rounds when playing the following game: P=6 players sitting in a circle each have B=3 coins and with probability 3⁻¹ they give one coin to their right or left side neighbour, or dump the coin to the centre. The game ends when all coins are in the centre. Coding this experiment in R is rather easy

situz=rep(B,P)
r=1
while (max(situz)>0){
 unz=runif(P)
 newz=situz-(situz>0)
 for (i in (1:P)[unz<1/3]) 
  newz[i%%P+1]=newz[i%%P+1]+(situz[i]>0)
 for (i in (1:P)[unz>2/3])
  newz[i-1+P*(i==1)]=newz[i-1+P*(i==1)]+(situz[i]>0)
 situz=newz
 r=r+1}

resulting in an average of 15.58, but I cannot figure out (quickly enough) an analytic approach to the problem. (The fit above is to a Gamma(â-1,ĝ) distribution.)

In a completely unrelated aparté (aside), I read earlier this week that New York City had prohibited the use of electric bikes. I was unsure of what this meant (prohibited on sidewalks? expressways? metro carriages?) so rechecked and found that electric bikes are simply not allowed for use anywhere in New York City. Because of the potential for “reckless driving”. The target is apparently the high number of delivery e-bikes, but this ban sounds so absurd that I cannot understand it passed. Especially when De Blasio has committed the city to the Paris climate agreement after Trump moronically dumped it… Banning the cars would sound much more in tune with this commitment! (A further aparté is that I strongly dislike e-bikes, running on nuclear power plants,  especially when they pass me on sharp hills!, but they are clearly a lesser evil when compared with mopeds and cars.)

Le Monde puzzle [#1019]

A gamey (and verbose) Le Monde mathematical puzzle:

A two-player game involves n+2 cards in a row, blue on one side and red on the other. Each player can pick any blue card among the n first ones and flip it plus both following ones. The game stops when no blue card is left to turn. The gain for the last player turning cards is 20-t, where t is the number of times cards were flipped, with gain t for its opponent. Both players aim at maximising their gain.

1. When n=4 and all cards are blue, can the first player win? If not, what is the best score for this player?

2. Among all 16 configurations at start, how many lead to the first player to win?

3. When n=10 and all cards are blue, how many cards are flipped an odd number of times for the winning configuration?

The first two questions can easily be processed by an R code like the following recursive function:

liplop <- function(x,n,i){
  if (max(x[1:n])==0){
    return(i)
  }else{
    sol=NULL
    for (j in (1:n)[x[1:n]==1]){
      y=x;y[j:(j+2)]=1-y[j:(j+2)]
      sol=c(sol,20-liplop(y,n,i+1))}
    return(max(sol))}}

Returning

> liplop(rep(1,6),4,0)
[1] 6

Meaning the first player cannot win, by running at most six rounds. Calling the same function for all 4⁴=16 possible configurations leads to 8 winning ones:

[1] 0 0 0 1
[1] 0 0 1 1
[1] 0 1 0 1
[1] 0 1 1 1
[1] 1 0 0 0
[1] 1 0 1 0
[1] 1 1 0 0
[1] 1 1 1 0

Solving the same problem with n=10 is not feasible with this function. (Even n=6 seems out of reach!)