Archive for FiveThirtyEight

a non-riddle

Posted in Books, Kids, R with tags , , on July 12, 2019 by xi'an

Unless I missed a point in the last riddle from the Riddler, there is very little to say about it:

Given N ocre balls, N aquamarine balls, and two urns, what is the optimal way to allocate the balls to the urns towards drawing an ocre ball with no urn being empty?

Both my reasoning and a two line exploration code led to having one urn with only one ocre ball (and no acquamarine ball) and all the other balls in the second urn.

odz<-function(n,m,t) 2*m/n+(t-2*m)/(t-n)
probz=matrix(0,trunc(N/2)-1,N-1)
for (n in 1:(N-1))
  for (m in 1:(trunc(N/2)-1))
    probz[m,n]=odz(n,m,N)

riddles on Egyptian fractions and Bernoulli factories

Posted in Books, Kids, R with tags , , , , , , , , , , , , , on June 11, 2019 by xi'an

Two fairy different riddles on the weekend Riddler. The first one is (in fine) about Egyptian fractions: I understand the first one as

Find the Egyptian fraction decomposition of 2 into 11 distinct unit fractions that maximises the smallest fraction.

And which I cannot solve despite perusing this amazing webpage on Egyptian fractions and making some attempts at brute force  random exploration. Using Fibonacci’s greedy algorithm. I managed to find such decompositions

2 = 1 +1/2 +1/6 +1/12 +1/16 +1/20 +1/24 +1/30 +1/42 +1/48 +1/56

after seeing in this short note

2 = 1 +1/3 +1/5 +1/7 +1/9 +1/42 +1/15 +1/18 +1/30 +1/45 +1/90

And then Robin came with the following:

2 = 1 +1/4 +1/5 +1/8 +1/10 +1/12 +1/15 +1/20 +1/21 +1/24 +1/28

which may prove to be the winner! But there is even better:

2 = 1 +1/5 +1/6 +1/8 +1/9 +1/10 +1/12 +1/15 +1/18 +1/20 +1/24

The second riddle is a more straightforward Bernoulli factory problem:

Given a coin with a free-to-choose probability p of head, design an experiment with a fixed number k of draws that returns three outcomes with equal probabilities.

For which I tried a brute-force search of all possible 3-partitions of the 2-to-the-k events for a range of values of p from .01 to .5 and for k equal to 3,4,… But never getting an exact balance between the three groups. Reading later the solution on the Riddler, I saw that there was an exact solution for 4 draws when

p=\frac{3-\sqrt{3(4\sqrt{9}-6)}}{6}

Augmenting the precision of my solver (by multiplying all terms by 100), I indeed found a difference of

> solver((3-sqrt(3*(4*sqrt(6)-9)))/6,ba=1e5)[1]
[1] 8.940697e-08

which means an error of 9 x 100⁻⁴ x 10⁻⁸, ie roughly 10⁻¹⁵.

take a random integer

Posted in Books, Statistics with tags , , on February 16, 2019 by xi'an

A weird puzzle from FiveThirtyEight: what is the probability that the product of three random integers is a multiple of 100? Ehrrrr…, what is a random integer?! The solution provided by the Riddler is quite stunning

Reading the question charitably (since “random integer” has no specific meaning), there will be an answer if there is a limit for a uniform distribution of positive integers up to some number . But we can ignore that technicality, and make do with the idealization that since every second, fourth, fifth, and twenty-fifth integer are divisible by and , the chances of getting a random integer divisible by those numbers are , , , and .

as it acknowledges that the question is meaningless, then dismisses this as a “technicality” and still handles a Uniform random integer on {1,2,…,N} as N grows to infinity! Since all that matters is the remainder of the “random variable” modulo 100, this remainder will see its distribution vary as N moves to infinity, even though it indeed stabilises for $N$ large enough…

missing digit in a 114 digit number [a Riddler’s riddle]

Posted in R, Running, Statistics with tags , , , , , , , on January 31, 2019 by xi'an

A puzzling riddle from The Riddler (as Le Monde had a painful geometry riddle this week): this number with 114 digits

530,131,801,762,787,739,802,889,792,754,109,70?,139,358,547,710,066,257,652,050,346,294,484,433,323,974,747,960,297,803,292,989,236,183,040,000,000,000

is missing one digit and is a product of some of the integers between 2 and 99. By comparison, 76! and 77! have 112 and 114 digits, respectively. While 99! has 156 digits. Using WolframAlpha on-line prime factor decomposition code, I found that only 6 is a possible solution, as any other integer between 0 and 9 included a large prime number in its prime decomposition:

However, I thought anew about it when swimming the next early morning [my current substitute to morning runs] and reasoned that it was not necessary to call a formal calculator as it is reasonably easy to check that this humongous number has to be divisible by 9=3×3 (for else there are not enough terms left to reach 114 digits, checked by lfactorial()… More precisely, 3³³x33! has 53 digits and 99!/3³³x33! 104 digits, less than 114), which means the sum of all digits is divisible by 9, which leads to 6 as the unique solution.

 

and another step forward for Ireland!

Posted in pictures with tags , , , , , , , , , , , on October 28, 2018 by xi'an

Riddler collector

Posted in Statistics with tags , , , , , , , on September 22, 2018 by xi'an


Once in a while a fairly standard problem makes it to the Riddler puzzle of the week. Today, it is the coupon collector problem, explained by W. Huber on X validated. (W. Huber happens to be the top contributor to this forum, with over 2000 answers, and the highest reputation closing on 200,000!) With nothing (apparently) unusual: coupons [e.g., collecting cards] come in packs of k=10 with no duplicate, and there are n=100 different coupons. What is the expected number one has to collect before getting all of the n coupons?  W. Huber provides an R code to solve the recurrence on the expectation, obtained by conditioning on the number m of different coupons already collected, e(m,n,k) and hence on the remaining number of collect, with an Hypergeometric distribution for the number of new coupons in the next pack. Returning 25.23 packs on average. As is well-known, the average number of packs to complete one’s collection with the final missing card is expensively large, with more than 5 packs necessary on average. The probability distribution of the required number of packs has actually been computed by Laplace in 1774 (and then again by Euler in 1785).

riddles on a line [#2]

Posted in Books, Kids, R with tags , , , , , , , on September 11, 2018 by xi'an

A second Riddle(r), with a puzzle related with the integer set Ð={,12,3,…,N}, in that it summarises as

Given a random walk on Ð, starting at the middle N/2, with both end states being absorbing states, and a uniform random move left or right of the current value to the (integer) middle of the corresponding (left or right) integer interval, what is the average time to one absorbing state as a function of N?

Once the Markov transition matrix M associated with this random walk is defined, the probability of reaching an absorbing state in t steps can be derived from the successive powers of M by looking at the difference between the probabilities to be (already) absorbed at both t-1 and t steps. From which the average can be derived.

avexit <- function(N=100){
 #transition matrix M for the walk
 #1 and N+2 are trapping states
 tranz=matrix(0,N+2,N+2)
 tranz[1,1]=tranz[N+2,N+2]=1
 for (i in 2:(N+1))
  tranz[i,i+max(trunc((N+1-i)/2),1)]=tranz[i,i-max(trunc((i-2)/2),1)]=1/2
 #probabilities of absorption
 prowin=proloz=as.vector(0)
 init=rep(0,N+2)
 init[trunc((N+1)/2)]=1 #first position
 curt=init
 while(1-prowin[length(prowin)]-proloz[length(prowin)]>1e-10){
  curt=curt%*%tranz
  prowin=c(prowin,curt[1])
  proloz=c(proloz,curt[N+2])}
 #probability of new arrival in trapping state
 probz=diff(prowin+proloz)
 return(sum((2:length(proloz))*probz))}

leading to an almost linear connection between N and expected trapping time.