Archive for riddle

a new Monty Hall riddle

Posted in Books, Kids, Mountains, pictures, R, Statistics, Travel with tags , , , , , , , , , on May 22, 2020 by xi'an

The Riddler was sort of feeling the rising boredom of being under lockdown when proposing the following variant to the Monty Hall puzzle:

There are zero to three goats, with a probability ¼ each, and they are allocated to different doors uniformly among the three doors of the show. After the player chooses a door, Monty opens another door hidding a goat or signals this is impossible. Given that he did open a door, what is the probability that the player’s door does not hide a goat?

Indeed, a straightforward conditional probability computation considering all eight possible cases with the four cases corresponding to Monty opening a door leads to a probability of 3/8 for the player’s door. As confirmed by the following R code:

s=sample
m=c(0,0)
for(t in 1:1e6)m=m+(range(s(1:3,s(1:3,1)))>1)

not a Bernoulli factory

Posted in Books, Kids, pictures, R with tags , , , , , , , on May 20, 2020 by xi'an

A Riddler riddle I possibly misunderstood:

Four isolated persons are given four fair coins, which can be either flipped once or returned without being flipped. If all flipped coins come up heads, the team wins! Else, if any comes up tails, or if no flip at all is done, it looses. Each person is further given an independent U(0,1) realisation. What is the best strategy?

Since the players are separated, I would presume the same procedure is used by all. Meaning that a coin is tossed with probability p, ie if the uniform is less than p, and untouched otherwise. The probability of winning is then

4(1-p)³p½+6(1-p)³p½²+4(1-p)p³½³+p⁴½⁴

which is maximum for p=0.3420391, with a winning probability of 0.2848424.

And an extra puzzle for free:

solve xxxx=2020

Where the integral part is the integer immediately below x. Puzzle that I first fail solving by brute force, because I did not look at negative x’s… Since the fourth root of 2020 is between 6 and 7, the solution is either x=6+ε or x=-7+ε, with ε in (0,1). The puzzle then becomes either

(6+ε)⌊(6+ε)⌊(6+ε)⌊6+ε⌋⌋ = (6+ε)⌊(6+ε)⌊36+6ε⌋⌋ = (6+ε)⌊(6+ε)(36+⌊6ε⌋)⌋ = 2020

where there are 6 possible integer values for ⌊6ε⌋, with only ⌊6ε⌋=5 being possible, turning the equation into

(6+ε)⌊41(6+ε) = (6+ε)(246+⌊41ε) = 2020

where again only ⌊42ε=40 being possible, ending up with

1716+286ε = 2020

which has no solution in (0,1). In the second case

(-7+ε)(-7+ε)(-7+ε)-7+ε⌋⌋ = (-7+ε)(-7+ε)(49+-7ε⌋)= 2020

shows that only -7ε=-3 is possible, leading to

(-7+ε)⌊46(-7+ε)) = (-7+ε) (-322+46ε⌋)=2020

with only 46ε=17 possible, hence

2135-305ε=2020

and

ε=115/305.

A brute force simulated annealing resolution returns x=-6.622706 after 10⁸ iterations. A more interesting question is to figure out the discontinuity points of the function

ℵ(x) = xxxx

as they seem to be numerous:

For instance, only 854 of the first 2020 integers enjoy a solution to ℵ(x)=n.

multiplying the bars

Posted in Kids, R with tags , , , , , , , on February 25, 2020 by xi'an

The latest Riddler makes the remark that the expression

|-1|-2|-3|

has no unique meaning (and hence value) since it could be

| -1x|-2|-3 | = 5   or   |-1| – 2x|-3| = -5

depending on the position of the multiplication sign and asks for all the possible values of

|-1|-2|…|-9|

which can be explored by a recursive R function for computing |-i|-(i+1)|…|-(i+2j)|

vol<-function(i,j){x=i
  if(j){x=c(i-(i+1)*vol(i+2,j-1),abs(i*vol(i+1,j-1)+i+2*j))
  if(j>1){for(k in 1:(j-2))
        x=c(x,vol(i,k)-(i+2*k+1)*vol(i+2*k+2,j-k-1))}}
  return(x)}

producing 40 different values for the ill-defined expression. However, this is incorrect as the product(s) hidden in the expression only involve a single term in vol(i,j)… I had another try with the decomposition of the expression vol(i,j) into a first part and a second part

prod<-function(a,b) a*b[,1]+b[,2]

val<-function(i,j){
  x=matrix(c(i,0),ncol=2)
  if(j){x=rbind(cbind(i,prod(-(i+1),val(i+2,j-1))),
          cbind(abs(prod(-i,val(i+1,j-1))-i-2*j),0))
    if(j-1){for(k in 2:(j-1)){
      pon=val(i,k-1)
      for(m in 1:dim(pon)[1])
          x=rbind(x,cbind(pon[m,1],pon[m,2]+prod(-(i+2*k-1),val(i+2*k,j-k))))}}}
  return(x)}

but it still fails to produce the right version.

another easy Riddler

Posted in Books, Kids, R with tags , , , , on January 31, 2020 by xi'an

A quick riddle from the Riddler

In a two-person game, Abigail and Zian both choose between a and z. Abigail win one point with probability .9 if they choose (a,a) and with probability 1 if they choose (a,z), and two points with probability .4 if they choose (z,z) and with probability .6 if they choose (z,a). Find the optimal probabilities δ and ς of choosing a for both Abigail and Zian when δ is known to Zian.

Since the average gain for Abigail is δ(1-.1ς)+2(1-δ)(.4+.2ς) the riddle sums up as solving the minmax problem

\max_\delta \min_\varsigma\delta(1-.1\varsigma)+2(1-\delta)(.4+.2\varsigma)

the solution in ς is either 0 or 1 depending on δ being smaller or larger than 12/22, which leads to this value as the expected gain. The saddlepoint is hardly visible in the above image. While ς is either 0 or 1 in the optimal setting,  a constant choice of 1 or 0 would modify the optimal for δ except that Abigail must declare her value of δ!

a very quick Riddle

Posted in Books, Kids, pictures, R with tags , , , , , , on January 22, 2020 by xi'an

A very quick Riddler’s riddle last week with the question

Find the (integer) fraction with the smallest (integer) denominator strictly located between 1/2020 and 1/2019.

and the brute force resolution

for (t in (2020*2019):2021){ 
   a=ceiling(t/2020)
   if (a*2019<t) sol=c(a,t)}

leading to 2/4039 as the target. Note that

\dfrac{2}{4039}=\dfrac{1}{\dfrac{2020+2019}{2}}