Publications by xi'an
sampling w/o replacement except when replacing
Another Riddle(r), considering a box with M myrtle balls and D dandelion balls. Drawing balls without replacement while they stay of the same color as the initial draw, else put back the last ball and repeat the process until all balls are drawn. The funny thing is that, unless M=0 or D=0, the probability to draw a myrtle ball at the end is alway...
1285 sym R (171 sym/1 pcs) 2 img
asymmetric information
The Riddler of 16 October had the following puzzle: Take a real number θ uniformly distributed over (0,100). Among three players, the winner is whoever guessed the closest price without going over θ. In the event all guesses exceeded θ, the contestant with the lowest (and therefore closest) guess is declared the winner. The second player knows...
1711 sym R (182 sym/1 pcs) 2 img
Le Monde puzzle [#1158]
A weekly puzzle from Le Monde on umbrella sharing: Four friends, Antsa, Cyprien, Domoina and Fy, are leaving school to return to their common housing. It is raining and they only have one umbrella with only room for two. Given walking times, x¹, x², x³ and x⁴, find the fastest time by which all of the four will be home, assuming they all ag...
860 sym R (752 sym/2 pcs) 2 img
another electoral map
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397 sym 2 img
on arithmetic derivations of square roots
An intriguing question made a short-lived appearance on the CodeGolf section of Stack Exchange, before being removed, namely the (most concise possible) coding of an arithmetic derivation of the square root of an integer, S, with a 30 digit precision and using only arithmetic operators. I was not aware of the myriad of solutions available, as dem...
1924 sym R (319 sym/3 pcs) 4 img
Le Monde puzzle [#1164]
The weekly puzzle from Le Monde is quite similar to older Diophantine episodes (I find myself impossible to point out): Give the maximum integer that cannot be written as 105x+30y+14z. Same question for 105x+70y+42z+30w. These are indeed Diophantine equations and the existence of a solution is linked with Bézout’s Lemma. Take the first equat...
2063 sym R (309 sym/1 pcs) 2 img
the riddle(r) of the certain winner losing in the end
Considering a binary random walk, starting at zero, what is the probability of being almost sure of winning at some point only to lose at the end? This is the question set by the post-election Riddler, with almost sure meaning above 99% and the time horizon set to n=101 steps (it could have been 50 or 538!). As I could not see a simple way to com...
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Bernoulli factory in the Riddler
“Mathematician John von Neumann is credited with figuring out how to take a p biased coin and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with e...
2719 sym R (360 sym/1 pcs) 6 img
around the table
The Riddler has a variant on the classical (discrete) random walk around a circle where every state (but the starting point) has the same probability 1/(n-1) to be visited last. Surprising result that stems almost immediately from the property that, leaving from 0, state a is visited couterclockwise before state b>a is visited clockwise is b/a+b....
1778 sym R (317 sym/2 pcs) 2 img
how many Friday 13th?
A short Riddler’s riddle on the maximum number of Fridays 13th over a calendar year, of which I found 9 by a dumb exploration : bi=c(1:31,1:29,1:31,1:30,1:31,1:30,1:31,1:31,1:30,1:31,1:30,1:31) oy=bi[-60] for(j in 0:(length(cy<-c(bi,oy,oy,oy))-1)){#any day in quartade dy=c(cy[(j+1):length(cy)],cy[1:j]) for(i in 0:6){ dz=(i+(1:length(cy)))...
1022 sym 2 img