Imposing precisely 1/8 on the final line. But none of the solutions had exactly a uniform probability of 1/8 to reach all endpoints. Reducing the denominator to 16 produced onceĪs 16P, with five ½’s (8). Where P is the 7×7 lower triangular matrix of nail probabilities, all with a 2⁸ denominator, reachingįor 128P. I first tried running my simulated annealing code with a target function like targ<-function(P)(1+.1*sum(!(2*P=1)))*sum(abs(8*evol(P)-1)) Should it be 2⁸?! Or more (since the solution with two levels also involves 1/3)? Using the functions evol.01) Plus, less usually, a choice of the space where the optimisation takes place, i.e., deciding on a common denominator for the (rational) probabilities. And I obviously chose to try simulated annealing to figure out the probabilities, facing as usual the unpleasant task of setting the objective function, calibrating the moves and the temperature schedule. Dans ce TP, on considre une planche de Galton douze ranges de clous il y a donc 13 cases numrotes de 0 (case gauche) 12 (case droite). Le numro de la case est donc le nombre de fois o la bille est descendue droite lors de son parcours. For instance, producing a uniform distribution with the maximum number of nails with probability ½ to turn right. En fin de parcours, elle tombe dans une case. Since Galton’s quincunx has fascinated me since the (early) days when I saw a model of it as a teenager in an industry museum near Birmingham, I jumped on the challenge to build an uneven nail version where the probabilities to end up in one of the boxes were not the Binomial ones.
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