Problematics | Counting Olympic medals
Here’s a puzzle themed on the Games, where you figure out how many participants from a small country wins gold, silver, bronze, any two, all three, or no medals
A celebratory puzzle set aside for a special occasion, such as the KBC-themed one in the 100th week of Problematics, can be a double-edged sword. Since it was a special occasion, it was a little more difficult than usual, and hence received more wrong answers than my other puzzles usually do. On the other hand, it was refreshing to read the attempts and comments from readers, even if some of them got parts of their answers wrong.
This week’s puzzles are relatively easy, but hopefully not so easy that you would feel disappointed with them. In other words, they follow the standard Problematics template: neither too easy nor too hard. The first one marks yet another occasion; it is themed on the Olympic Games.
#Puzzle 101.1
Having sent a tiny contingent to the Olympic Games, a small country is delighted when most of the sportspersons return home with at least one medal. All the medallists are participants in individual rather than team events. Overall, their medal tally covers each kind: gold, silver and bronze.
1. No sportsperson wins more than one medal of any kind. To make that clear: No one wins two golds or two silvers or two bronzes.
2. Among the medallists, only one wins a gold as well as a silver as well as a bronze, which come in three different events, of course.
3. Five members of the contingent return without a single gold, five finish without a single silver, and five come home without a single bronze.
4. Two of the gold medallists also win a silver each, four the silver medallists also win a bronze each, and three of the bronze medallists also win a gold each.
What is the size of the contingent, how many have won which kind of medal, and how many have not won anything?
#Puzzle 101.2
The two figures are of the same die, with opposite sides adding up to 7 spots (1 is opposite 6, 2 is opposite 5, and 3 is opposite 4). Placing 6 on top so that the sides with 3 and 2 face you, draw a red line connecting two corners of the face with 6 spots, then a green line across the face with 3 spots as shown.
Now remove the die but imagine that the two lines remain where you drew them, suspended in space. What is the angle between them?
MAIILBOX: LAST WEEK’S SOLVERS
#Puzzle 100.1
Dear Kabir,
Scenario #1. Let the question be “What is the first letter of the alphabet?” The correct answer is A, but I do not know this. The wrong answers given are B, C and D. I know that D is wrong, but have no idea which one of A, B and C is correct and which two are wrong. Now when I opt for 50:50 and the host eliminates two of the wrong options, he can do so in three different ways:
(a) The host eliminates B and C, so that A and D remain. I know that D is wrong, so I can infer that A is correct with certainty = 1 = 100%.
(b) The host eliminates B and D, so that A and C remain. I don’t know which one of these two is correct, so my chances of getting it correct = 1/2 = 50%.
(c) The host eliminates C and D, so that A and B remain. I don’t know which one of these two is correct, so my chances of getting it correct = 1/2 = 50%.
So, my overall probability of getting it correct is (1 + 1/2 + 1/2)/3 = 2/3 = 66.67%. So my chances have improved from the original 1/3.
A fantastic milestone. Keep up the great work. Just 100 weeks done. Many more to go.
— Sampath Kumar V, Coimbatore
***
Dear Kabir,
Scenario #2. There are three ways in which two options can be given after the 50:50 lifeline and in one of these three ways the two doubtful options remain. So the probability of failure is 1/3*1/2 = 1/6. The chances of success are 5/6 now, an increase from initial 1/2.
— Yadvendra Somra, Sonipat
***
Dear Kabir,
Scenario #3. This is a good extension of the famous Monty Hall problem. One should definitely switch.
Let us take the options to be A, B, C and D. As we do not have any knowledge and are selecting any option (let's say A) at random, our win probability is 1/4 = 25%.
So let's say B was the correct option without loss of generality. In that case C and D are removed. In this case switching gives the right answer.
Now, chances were 25% that our initial answer is correct, and 75% that it is wrong. So the initial probability of selecting the right option remains the same as 25%, while switching raises the probability to 75%. Hence one should switch.
Congratulations on completing 100 weeks of Problematics.
— Harshit Arora, IIT Delhi
#Puzzle 100.2
Hi Kabir,
Some possible solutions to the Wordle puzzle are LEVEL, LEVEE, KEVEL.
— Professor Anshul Kumar, Delhi
Only three readers have correctly solved all three sections of #Puzzle 100.1. Some others have solved either one or two sections correctly.
Solved both puzzles (all sections): Sampath Kumar V (Coimbatore), Harshit Arora (IIT Delhi), Professor Anshul Kumar (Delhi)
Solved both puzzles (partially): Yadvendra Somra (Delhi), Sabornee Jana (Mumbai), Dr Sunita Gupta (Delhi), Sanjay S (Coimbatore), Akshay Bakhai (Mumbai), Aditya (Coimbatore), Ajay Ashok (Delhi), YK Munjal (Delhi), Shishir Gupta (Indore), Raghunathan Ravindranathan (Coimbatore)
Solved #Puzzle 100.2: Anil Khanna (Ghaziabad), Meenakshi Ravikumar (Chennai), Shri Ram Aggarwal (Delhi), Thejal S (Coimbatore), Sundarraj C (Bengaluru)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com