The riddle this week on The Riddler was about finding the largest sequence of integers between 1 and 100 such that each integer is only used once and always followed by a multiple or a factor. A basic R code searching at random [and programmed during a massive downpour on Skye] led to a solution […]
From the current Riddler, a problem that only requires a few lines of code and a few seconds of reasoning. Or not. N households each stole the earnings from one of the (N-1) other households, one at a time. What is the probability that a given household is not burglarised? And what are the expected […]
![[un]solved riddles](https://s0.wp.com/latex.php?latex=%7B23%5Cchoose+2%5C+2%5C+2%7D&bg=000000&%23038;fg=B0B0B0&%23038;s=0)
On the Riddler of last week, first a birthday puzzle: Given a group of 23 persons, what is the probability of observing three pairs of identical birthdays? which can be found by a quick simulation as ave=0 for (t in 1:1e6){ dupz=dates[duplicated(sample(1:365,23,rep=TRUE))] ave=ave+as.integer((length(dupz)==3)& (length(unique(dupz))==3))}} ave/M returning a value of 0.0183, but which combinatoric resolution I […]
While the Riddler puzzle this week was anticlimactic, as it meant filling all digits in the above division towards a null remainder, it came as an interesting illustration of how different division is taught in the US versus France: when I saw the picture above, I had to go and check an American primary school […]
The Riddler of today has a secretary problem, where one measures sequentially N random variables until one deems the current variable to be the largest of the whole sample. The classical secretary problem has a counter-intuitive solution where one first measures N/e random variables without taking any decision and then and only then picks the […]
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The riddle of this week is about an optimisation of positioning the four digits of a multiplication of two numbers with two digits each and is open to a coding resolution: Four digits are drawn without replacement from {0,1,…,9}, one at a time. What is the optimal strategy to position those four digits, two digits […]
The riddle of this week is about an optimisation of positioning the four digits of a multiplication of two numbers with two digits each and is open to a coding resolution: Four digits are drawn without replacement from {0,1,…,9}, one at a time. What is the optimal strategy to position those four digits, two digits […]
The riddle of this week is about an optimisation of positioning the four digits of a multiplication of two numbers with two digits each and is open to a coding resolution: Four digits are drawn without replacement from {0,1,…,9}, one at a time. What is the optimal strategy to position those four digits, two digits […]
A simply worded riddle this week on The Riddler, about four ropes having non-uniform and unknown burning rates, the only constraint being they all burn completely in one hour. With the help of a lighter (or even a single match), what are the possible units of time one can measure by burning them? While I […]
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 q of the […]
