Problematics | The who’s who of dogs and humans in the park
Who owns the pug and who owns the mastiff? How old is the dog whose owner is 86 years old? Try this Einstein puzzle
So-called Einstein puzzles, which you have solved with me before, can come in both full-length and miniature varieties, with fewer variables. It’s a full-length Einstein puzzle this week, which means it requires a lengthy description. Besides, there are also your answers to discuss from last week. In view of the tightness of space, therefore, let us go straight to the point.
* Alpha is a pug, Beta is a mastiff, and Delta is a beagle.
* The 86-year-old’s pet is Gamma, the 88-year-old’s pet is a spaniel, and the 89-year-old’s pet is Alpha, who is 85 months old.
* The singer’s pet is Epsilon, while the painter’s pet is a Labrador retriever.
* The actor is 89 years old.
* The singer sits in the middle, two persons on either side.
* At one end sits a person whose pet is 87 months old.
* Two persons sitting next to each other have pets who are 87 and 88 months old.
* The teacher sits next to the person whose pet is the beagle; the author sits next to the person whose pet is 88 months old.
* The painter sits to the left of the person whose pet is 85 months old.
* Delta’s owner sits to the right of the 87-year-old, who sits to the right of the person whose pet is 86 months old.
Who has which pet, who is how old, etc? As usual, your answer will work best in a table format.
Find a word of four letters, consisting entirely of vowels (no Y, no consonant). It’s listed in Collins English Dictionary.
Mailbox: Last week’s solvers
This is a variation of the popular Monty Hall Problem. The better choice is to switch after the trickster reveals one Jack.
Suppose the 3 cards are A, B, C from left to right, and suppose we choose A. There are 3 possibilities:
* A is the Queen; the trickster reveals either of B or C
* B is the Queen; the trickster reveals C
* C is the Queen, the trickster reveals B
By switching, we win in the latter 2 cases, but not in the first. So, the probability becomes 2/3, while with our original choice it was 1/3.
— Arsh Gupta, Gurgaon
Indeed, Arsh. If you chose a Jack first (probability 2/3), then your choice will dictate which card the trickster must turn face up. Only if you chose the Queen first (probability 1/3) does the trickster have freedom of choice.
Solved #Puzzle 41.1: Amrish Garg (New Jersey), Pulkit Kumar (KV Vikaspuri, Delhi), Sankaran KB (Chennai), Prakhar Agrawal (Dhanbad), Shruti Sethi (Ludhiana), Vivek Aggarwal (Bangalore), Sabornee Jana (Mumbai), Yuthika Berdhe (Navi Mumbai), Sandeep Bhateja (Hoshiarpur), Harshit Sharma (Chandigarh), Amardeep Singh (Meerut), Sanjay Gupta (Delhi), Akshay Bakhai (Mumbai), Geeta Arora (Delhi), Himanshu Batra (Delhi), Atiksh Jain (Gurgaon), Rohit Khanna (Noida), Rishabh Shukla (Delhi), Aidan Williams (Mumbai), Anil Goyal (Delhi), Akshai Bakhay (Mumbai), Shri Ram Aggarwal (Delhi), Jobin Babu (Delhi), Arnav Gupta, Pranjal Pawan Malpani, Ruchir Kohli, Rohan Gulati, Raghav Kapre
The highest scoring word is obviously ZZZ (signifying that one is asleep). Among acceptable answers received (dictionary words only), the next highest scorers are ZUZ (an ancient Hebrew coin) and SWY (a gambling game).