Problematics | When two people share the same birthday

How many people must you select at random before the probability of two or more common birthdays crosses 50%

To address the concerns of everyone who has mailed me, and those who haven’t but may still have been concerned: Problematics is still running. It was not a break we took last week, for the column appeared online on Monday as usual; it’s just that the print edition could not accommodate it this time. This was, incidentally, the first Monday in 47 continuous weeks that Problematics did not appear in print, but its run in its web version, now in its 48th week, remains unbroken.

Welcome to Problematics!(Shutterstock)

Thank you for your concern, and for letting me know that you wait for these puzzles every week. Here are two more, as usual. The first one is a probability puzzle.

#Puzzle 48.1

#Puzzle 48.1

 

The above are the dates of births of the playing XIs in the second Test between West Indies and India, July 20-24, which is in progress as I write this. You can see two pairs of common birthdays here: Jermaine Blackwood and Alzarri Joseph in the same team, and Jason Holder and Virat Kohli in opposing teams.

One pair of common birthdays out of 11 persons appears to be rare, but what about one pair or two pairs out of 22? That depends on what “rare” means: count anything less than 50% as rare, and anything over that as common. Ignore leap years, and take a year as containing 365 days.

What is the minimum number of people you must choose at random so that the probability of any two sharing the same birthday is 50% or more? Are the 22 players from a Test match above or below this threshold?

 

#Puzzle 48.2

 

138 x 27 = 54 x 69

158 x 23 = 46 x 79

186 x 27 = 54 x 93

One thing common in the above three multiplications is that each of them uses all 9 digits from 1 to 9, each digit once. These are, however, not the only such examples. I happen to have a list of them, but you might need to work them out with hit and trial. Send me at least one such example.

 

MAILBOX: LAST WEEK'S SOLVERS

 

#Puzzle 47.1

With the help of two helper vehicles from the fuel station, and two vehicles from the town, the main vehicle can reach the destination. Five full tanks are consumed.

Let the distance from the fuel station to the midpoint be D and a full tank be F. So, the vehicle has to travel 2D the town.

It takes the help of two vehicles from the station, all three with full tanks. When they travelled D/4, each consuming F/4, one helper car transfers F/4 fuel each to the other two, whose tanks are full again. With the remaining F/4, the first helper returns to the station. The other two vehicles travel from D/4 to D/2, each consuming F/4. Now the second helper vehicle transfers F/4 to the main vehicle, and with the remaining F/2 returns to the station. The car with the full tank moves from D/2 to 3D/2.

Two other helper vehicles travel D/4 from the city (7D/4 from the station), each consuming F/4 fuel. One parks here with 3F/4. The second helper vehicle now moves to D/2 from the city (3D/2 from the station), consuming F/4 fuel. Thus, it has F/2. Now it can transfer F/4 to the main vehicle. Both vehicles with F/4 fuel can travel to 7D/4, where they each take F/4 from the parked helper car, and all three reach the city.

— Sunita & Naresh Dhillon, Gurgaon

 

#Puzzle 47.2

Hi Kabir,

Here a person is timing the flight of flies but is unable to as they are fast.

Time flies? I can’t! They’re too fast.

— Abhishek Garg, Chandigarh

Dr Sunita Gupta and Akshay Bakhai, too, have noted that the idea is about timing flies, but their punctuation does not convey that. Shruti M Sethi offers “Time flies, I can’t. They’re too fast”, which I am accepting as correct. The ideal punctuation is” Time flies I can’t. They’re too fast.”

Solved #Puzzle 47.1: Sunita & Naresh Dhillon, Malay Mittal

Solved #Puzzle 47.2: Abhishek Garg, Deviprakash Seksaria, Shruti M Sethi

 

Problematics will be back next week. Please send in your replies to problematics@hindustantimes.com