triple ruin

An almost straightforward riddle from The Riddler involving a triple gambler’s ruin: Dawn competes against three players Alessandra, Berenike, and Chinue, with probabilities of winning one round ¾, ½, and ¼, respectively, until the cumulated score reaches ±15, ±30, and ±45, for the first, second, and third games. What is Dawn’s optimal sequence of adversaries?

First, a brute force R simulation shows that the optimal ordering is to play the three adversaries first weakest, third strongest and middle fair:

ord=function(p){
  z=2*(runif(1)<p[1])-1
  while(abs(z)<15)z=z+2*(runif(1)<p[1])-1
  y=2*(runif(1)<p[2])-1
  while(abs(z+y)<30)y=y+2*(runif(1)<p[2])-1
  x=2*(runif(1)<p[3])-1
  while(abs(z+y+x)<45)x=x+2*(runif(1)<p[3])-1 
  return(x+y+z>0)}
mcord=function(p,T=1e2){
  for(t in 1:T)F=F+ord(p)
  return(F/T)}
comp=function(T=1e2){
  return(c(mcord(c(.5,.55,.45),t),
    #mcord(c(.5,.45,.55),t),#1-above
    mcord(c(.55,.5,.45),t),
    #mcord(c(.45,.5,.55),t),#1-above
    mcord(c(.55,.45,.5),t)
    #mcord(c(.45,.55,.5),t)))#1-above
    ))}

where I used probabilities closer to ½ to avoid estimated probabilities equal to one.

> comp(1e3)
[1] 0.051 0.038 0.183

(and I eliminated the three other probabilities by sheer symmetry). Second, checking in Feller’s bible (Vol. 1, XIV.3) for the gambler’s ruin probability, a simple comparison of the six orderings confirms this simulation.

Take thrice thy money, bid me tear the bond

A rather fun riddle for which the pen&paper approach proved easier than coding in R, for once. It goes as follows: starting with one Euro, one sequentially predicts a sequence of four random bits, betting an arbitrary fraction of one’s money at each round. When winning, the bet is doubled, otherwise, it is lost. Under the information that the same bit cannot occur thrice in a row, what is the optimal minimax gain?

Three simplifications: (i) each bet is a fraction ε of the current fortune of the player, which appears as a product of (1±ε) the previous bets (ii) when the outcome is 0 or 1, this fraction ε can thus be chosen in (-1,1), (iii) while the optimal choice is ε=1 when the outcome is known, i.e., when both previous are identical. The final fortune of the player is thus of the form

(1±ε)(1±ε’)(1±ε”)(1±ε”’)

if the outcome is alternating (e.g., 0101 or 0100), while it is of the form

2(1±ε)(1±ε’)(1±ε”)

if there are two identical successive bits in the first three results (e.g., 1101 or 0110). When choosing each of the fractions ε, the minimum final gain must be maximised. This implies that ε=0 for the bet on the final bit  when the outcome is uncertain (and ε=1 otherwise). In case of an alternating début, like 01, the minimal gain is

min{(1±ε)(1±ε’)(1+ε”),2(1±ε)(1±ε’)(1-ε”)}

which is maximised by ε”=1/3, taking the objective value 4(1±ε)(1±ε’)/3. Leading to the gain after the first bit being

min{4(1±ε)(1+ε’)/3,2(1±ε)(1-ε’)}

which is maximised by ε’=1/5, for the objective value 8(1±ε)/5. By symmetry, the optimal choice is ε=0. Which ends up with a minimax gain of 3/5. [The quote is from Shakespeare, in the Merchant of Venice.]

Le Monde puzzle [#1088]

A board (Ising!) Le Monde mathematical puzzle in the optimisation mode, again:

On a 7×7 board, what is the maximal number of locations that one can occupy when imposing at least two empty neighbours ?

Which I tried to solve by brute force and simulated annealing (what else?!), first defining a target

targ=function(tabz){
  sum(tabz[-c(1,9),-c(1,9)]-1.2*(tabz[-c(1,9),-c(1,9)]*tabz[-c(8,9),-c(1,9)]
      +tabz[-c(1,9),-c(1,9)]*tabz[-c(1,2),-c(1,9)]
      +tabz[-c(1,9),-c(1,9)]*tabz[-c(1,9),-c(8,9)]
      +tabz[-c(1,9),-c(1,9)]*tabz[-c(1,9),-c(1,2)]>2))}

on a 9×9 board where I penalise prohibited configuration by a factor 1.2 (a wee bit more than empty nodes). The perimeter of the 9×9 board is filled with ones and never actualised. (In the above convoluted products, the goal is to count how many neighbours of the entries equal to one are also equal to one. More than 2 is penalised.) The simulated annealing move is then updating the 9×9 grid gridz:

temp=1
maxarg=curarg=targ(gridz)
for (t in 1:1e3){
  for (v in 1:1e4){
    i=sample(2:8,1);j=sample(2:8,1)
    newgrid=gridz;newgrid[i,j]=1-gridz[i,j]
    newarg=targ(newgrid)
    if (log(runif(1))<temp*(newarg-curarg)){
      gridz=newgrid;curarg=newarg}}
temp=temp+.01}

and calls to the procedure always return 28 entries as the optimum, as in

     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    1    0    1    0    1    0    1
[2,]    0    1    1    0    1    1    0
[3,]    1    1    0    1    0    1    1
[4,]    0    0    1    0    1    0    0
[5,]    1    1    0    1    0    1    1
[6,]    0    1    1    0    1    1    0
[7,]    1    0    1    0    1    0    1

As it happens, I had misread the wording of the original puzzle, which considered a dynamic placement of the units on the board, one at a time with two free neighbours imposed.

Le Monde puzzle [#1086]

A terse Le Monde mathematical puzzle in the optimisation mode:

What is the maximal fraction of the surface of a triangle occupied by an inner triangle ABC where Abigail picks a summit A on a first side, Berenice B on a second side, and then Abigails picks C on the third side, towards Abigail maximising and Berenice minimising this surface?

Which I first tried to solve by pen & paper, completing another black block for the occasion, as coding the brute force R version sounded too painful:

leading me to conclude that, for a rectangle triangle (although the result sounds independent of this feature), the optimum was the middle triangle, weighting one-fourth of the original surface. Reprogramming the question in the plane to Angkor produced the same output, modulo my approximation of the triangle continuum with a 200×200/2grid:

RSS Read Paper

I had not attended a Read Paper session at the Royal Statistical Society in Errol Street for quite a while and hence it was quite a treat to be back there, especially as a seconder of the vote of thanks for the paper of Andrew Gelman and Christian Hennig. (I realised at this occasion that I had always been invited as a seconder, who in the tradition of the Read Papers is expected to be more critical of the paper. When I mentioned that to a friend, he replied they knew me well!) Listening to Andrew (with no slide) and Christian made me think further about the foundations of statistics and the reasons why we proceed as we do. In particular about the meaning and usages of a statistical model. Which is only useful (in the all models are wrong meme) if the purpose of the statistical analysis is completely defined. Searching for the truth does not sound good enough. And this brings us back full circle to decision theory in my opinion, which should be part of the whole picture and the virtues of openness, transparency and communication.

During his talk, Christian mentioned outliers as a delicate issue in modelling and I found this was a great example of a notion with no objective meaning, in that it is only defined in terms of or against a model, in that it addresses the case of observations not fitting a model instead of a model not fitting some observations, hence as much a case of incomplete (lazy?) modelling as an issue of difficult inference. And a discussant (whose Flemish name I alas do not remember) came with the slide below of an etymological reminder that originally (as in Aristotle) the meaning of objectivity and subjectivity were inverted, in that the later meant about the intrinsic nature of the object, while the former was about the perception of this object. It is only in the modern (?) era that Immanuel Kant reverted the meanings…Last thing, I plan to arXiv my discussions, so feel free to send me yours to add to the arXiv document. And make sure to spread the word about this discussion paper to all O-Bayesians as they should feel concerned about this debate!