Archive for The Riddler

a riddle at the end of its tether

Posted in R, Statistics with tags , , on February 24, 2017 by xi'an

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 had worked out a range of possible times for each of the ropes to complete its combustion, I went for a simulation based confirmation. The starting point is that a new fire can only be started when a burning rope ends up burning. Which only happens for a finite number of possibilities. I found 24 of them, consisting of

> total*prec
[1]  0.000 0.5000 0.750 0.875 0.9375 1.000 1.125 1.1875 1.25 1.3125
[11] 1.375 1.4375 1.500 1.625 1.7500 1.875 2.000 2.1250 2.25 2.5000
[21] 2.750 3.0000 3.500 4.000

i.e., some combinations of 1, 2⁻¹, …, 2⁻⁴, with the comment that those times cannot all be obtained within a single arson experiment.

Continue reading

a knapsack riddle?

Posted in Books, pictures, R, Statistics, Travel with tags , , , , , , on February 13, 2017 by xi'an


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 adversary is unknown. As usual (!), the phrasing of the riddle is somewhat ambiguous but I am under the impression that the game is played sequentially, hence that one can learn about the distribution of the adversary, at least when assuming this adversary keeps the same distribution q at all times. Continue reading

another Galton-Watson riddle

Posted in Statistics with tags , , , on February 3, 2017 by xi'an

The riddle on the Riddler this week is definitely a classic, since it rephrases the standard Galton-Watson branching process (which should have been called Bienaymé‘s process, as he established the relation before Watson, while the jack-of-all-trades Francis Galton only posed the question):

At the beginning, there is a single microorganism. Each day, every member of this species either splits into two copies of itself or dies. If the probability of multiplication is p, what are the chances that this species goes extinct?

As is easily seen from the moment generating function, the species goes instinct if p≤½. Actually, I always found it intriguing [intuitively] that the value ½ is included in the exclusion range!

an express riddle

Posted in Books, Kids, R with tags , , on January 20, 2017 by xi'an

A quick puzzle on The Riddler this week that enjoys a quick solution once one writes it out. The core of the puzzle is about finding the average number of draws one need to empty a population of size T if each draw is uniform over the remaining number of individuals between one and the number that remain. It is indeed easy to see that this average satisfies

\epsilon^T=1+\frac{1}{T}\sum_{i=1}^{T-1} \epsilon^i

since all draws but one require an extra draw. A recursion then leads by elimination to deduce that


which is the beginning of the (divergent) harmonic series. In the case T=30, the solution is (almost) equal to 4.

> sum(1/(1:30))*1e10
[1] 39949871309

A second riddle the same week reminded me of a result in Devroye’s Non-Uniform Random Variate Generation, namely to find the average number of draws from a Uniform until the sequence goes down. Actually, the real riddle operates with a finite support Uniform, but I find the solution with the continuous Uniform more elegant. And it only took a few metro stops to solve. The solution goes as follows: the probability to stop after two Uniform draws is 1/2, after n uniform draws, it is (n-1)/n!, which does sum up to 1:

\sum_{n=2}^\infty \frac{n-1}{n!} = \sum_{n=2}^\infty \frac{n}{n!} - \sum_{n=2}^\infty \frac{1}{n!} = \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}=1

and the expectation of this distribution is e-1 by a very similar argument, as can be checked by a rudimentary Monte Carlo experiment

> over(1e7) #my implementation of the puzzle
[1] 1.7185152

a Galton-Watson riddle

Posted in R, Travel with tags , , , on December 30, 2016 by xi'an

riddleThe Riddler of this week has an extinction riddle which summarises as follows:

One observes a population of N individuals, each with a probability of 10⁻⁴ to kill the observer each day. From one day to the next, the population decreases by one individual with probability

K√N 10⁻⁴

What is the value of K that leaves the observer alive with probability ½?

Given the sequence of population sizes N,N¹,N²,…, the probability to remain alive is


where the sum stops with the (sure) extinction of the population. Which is the moment generating function of the sum. At x=1-10⁻⁴. Hence the problem relates to a Galton-Watson extinction problem. However, given the nature of the extinction process I do not see a way to determine the distribution of the sum, except by simulation. Which returns K=27 for the specific value of N=9.

while (abs(targ-.5)>.01){

for (t in 1:M){
  while (gen>0){
  if (targ<.5){ K=K*ite/(1+ite)}else{

The solution proposed on The Riddler is more elegant in that the fixed point equation is

\prod_{J=1}^9 \frac{K \cdot \sqrt{J}}{K \cdot \sqrt{J} + J}=\frac{1}{2}

with a solution around K=27.

untangling riddle

Posted in Statistics with tags , , on December 23, 2016 by xi'an

Boston1This week riddle is rather anticlimactic [my rewording]:

N wires lead from the top of a tower down to the ground floor. But they have become hopelessly tangled. The wires on the top floor are all labelled, from 1 to N. The wires on the bottom, however, are not.

You need to figure out which ground-floor wire’s end corresponds to which top-floor wire’s end. On the ground floor, you can tie two wire ends together, and you can tie as many of these pairs as you like. You can then walk to the top floor and use a circuit detector to check whether any two of the wires at the top of the tower are connected or not.

What is the smallest number of trips to the top of the tower you need to take in order to correctly label all the wires on the bottom?

Indeed, first, if N=2, the wires cannot be identified, if N=3, they can be identified in 2 steps, and if N=4, they can also be identified in 2 steps (only pair two wires first, which gives their values and the values of the other two up to a permutation, then pair one of each group). Now, in general,  it seems that, by attaching the wires two by two, one can successively cut the number of wires by two at each trip. This is due to the identification of pairs: if wires a and b are connected in one step, identifying the other end of a will automatically determine the other end of b if one keeps track of all connected pairs at the top. Hence, if N=2^k, in (k-2) steps, we are down to 4 wires and it takes an extra 2 steps to identify those 4, hence all the other wires. Which leads to identification in N trips. If N is not a power of 2, things should go faster as “left over” wires in the binary division can be paired with set-aside wires and hence identify 3 wires on that trip, then further along the way… Which leads to a 2^{\lfloor \log_2(N) \rfloor} number of trips maybe (or even less). Continue reading

going to war [a riddle]

Posted in Books, Kids, Statistics with tags , , , , , on December 16, 2016 by xi'an

On the Riddler this week, a seemingly obvious riddle:

A game consists of Alice and Bob, each with a $1 bill, receiving a U(0,1) strength each, unknown to the other, and deciding or not to bet on this strength being larger than the opponent’s. If no player bets, they both keep their $1 bill. Else, the winner leaves with both bills. Find the optimal strategy.

As often when “optimality” is mentioned, the riddle is unclear because, when looking at the problem from a decision-theoretic perspective, the loss function of each player is not defined in the question. But the St. Petersburg paradox shows the type of loss clearly matters and the utility of money is anything but linear for large values, as explained by Daniel Bernoulli in 1738 (and later analysed by Laplace in his Essai Philosophique).  Let us assume therefore that both players live in circumstances when losing or winning $1 makes little difference, hence when the utility is linear. A loss function attached to the experiment for Alice [and a corresponding utility function for Bob] could then be a function of (a,b), the result of both Uniform draws, and of the decisions δ¹ and δ² of both players as being zero if δ¹=δ²=0 and

L(a,b,\delta^1,\delta^2)=\begin{cases}0&\text{if }\delta^1=\delta^2=0\\\mathbb{I}(a<b)-\mathbb{I}(a>b)&\text{else}\\\end{cases}

Considering this loss function, Alice aims at minimising the expected loss by her choice of δ¹, equal to zero or one, expected loss that hence depends  on the unknown and simultaneous decision of Bob. If for instance Alice assumes Bob takes the decision to compete when observing an outcome b larger than a certain bound α, her decision is based on the comparison of (when B is Uniform (0,1))


(if δ¹=0) and of 1-2a (if δ¹=1). Comparing both expected losses leads to Alice competing (δ¹=1) when a>α/2.

However, there is no reason Alice should know the value of α when playing the (single) game and so she may think that Bob will follow the same reasoning, leading him to choosing a new bound of α/4, and, by iterating the thought process, down all the way to α=0!  So this modelling leads to always play the game, with each player having a ½ probability to win… Alternatively, Alice may set a prior on α, which leads to another bound on a for playing or not the game. Which in itself is not satisfactory either. (The published solution is following the above argument. Except for posting the maths expressions.)