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Problematics | Beating the odds
This week, a betting puzzle involving Word Cup football, and a mistake made by John Keats.
The 2026 football World Cup is just months away, which provides a good excuse for Problematics to look back at the 2022 edition. The semifinal line-up was greatly lopsided, with two major teams (eventual winners Argentina and runners-up France) pitted against relatively unfancied Croatia and Morocco.
The odds bookmakers had offered at the start of the tournament should underline the mismatch: 11-2 against Argentina, a similar 6-1 against France, a huge 50-1 against Croatia and an enormous 200-1 against Morocco. Just in case you are not familiar with the language of betting, here is my explanation in the language of Problematics.
For any team, the two segments of the odds represent the bookmaker’s bet and your bet. If you bet ₹2 that the winner would be Argentina (11-2), it was implicitly understood that the bookmaker was betting ₹11 that Argentina would not win. If Argentina won, the bookie would give you ₹11 + ₹2 (your money back) = ₹13. If Argentina did not win, you would lose your ₹2, of course. On the other hand, if you bet ₹1 on Morocco, it was implicitly understood that the bookie was betting ₹200 that Morocco would not win. Victory for Morocco would bring you ₹201 (you would gain ₹200) while defeat would make the bookmaker richer by your ₹200.
The odds change during the course of the tournament, of course, as more and more teams get eliminated. Bookmakers from professional betting houses are generally sound mathematicians who ensure that they will make a profit regardless of the result. But there may be some amateurs who, if you are lucky, will offer you odds that give you, the bettor, the opportunity to profit at their expense.
#Puzzle 180.1
In 2022, someone I know told me about someone she knew who made a profit against an amateur bookmaker. The semifinal line-up had been decided, and only the four teams remained in contention. The odds offered by the bookie were: France 2-1, Argentina 3-1, Croatia 8-1, and Morocco 11-1.
My friend’s friend was in need of ₹800 because he wanted to buy something. He borrowed some money from my friend, promising her he would return the entire amount after making a profit of ₹800 with the bookie. The friend of the friend then distributed the entire money in four unequal bets on the four teams. This included ₹300 on Morocco
The smart fellow went home and relaxed, without bothering to watch either of the semifinals or the final. He knew that after the final was played and the champions decided, he would be richer by exactly ₹800, no matter which team won.
How much money did my friend’s friend borrow from my friend, and how much did he bet on which team?
#Puzzle 180.2
'Tis not through envy of thy happy lot,
But being too happy in thine happiness,—
That thou, light-winged Dryad of the trees
In some melodious plot
Of beechen green, and shadows numberless,
Singest of summer in full-throated ease.
If I paraphrase the above in modern English, that's John Keats addressing a nightingale, telling the bird that he is not envious of its happiness but is rather happy about it; he is happy that the nightingale, whom he describes as a light-winged Dryad (a wood nymph) of the trees is singing effortlessly about summer in a delightful plot of green beech trees and endless shadows.
What is scientifically wrong in Keats's verses, as is also the name "Nightingale of India" given to Sarojini Naidu for her eloquence and Lata Mangeshkar for her music?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 179.1
Hi Kabir,
Such calculations can be without using a calculator, as the total working days in three years is:
365 x 3 – [30 (for September) + 31 (October) + 30 (November)] – 3 (for 31 December) = 1001
The above is assuming three non-leap years. If any year is a leap year of 366 days, one extra leave on 29 February results in the same number of working days, 1001 in 3 years.
Now if we multiply a three-digit number with 1001, the 6-digit product is a repetition of the three-digit number. For example, 165165 = 165 x 1001. Similarly, 199199 = 199 x 1001, and 131131 = 131 x 1001. This is how the student can find out that the average working time is 165 minutes, or 2 hours 45 minutes per day.
— Shishir Gupta, Indore
#Puzzle 179.2
With two dice, the number of possible combinations is 36. There are 6 combinations with each digit (1 to 6) on the first die. Out of these, 5 combinations will give a total less than 7 alongside 1 spot on the first die, 4 such combinations alongside 2 spots on the first die, 3 combinations alongside 3 spots on the first die, and so on, totalling 5 + 4 + 3 + 2 + 1 = 15.
With each digit on the first die, there is only one counterpart on the second die giving a total of 7. So there are 6 combinations that give a total of 7. The remaining 36 – 15 – 6 = 15 combinations will give a total greater than 7.
Hence, the probability of the total being less than 7 or greater than 7 is 15/36 or 5/12 each, and the probability of it being exactly 7 is 6/36 or 1/6.
— Kanwarjit Singh, Chief Commissioner of Income-tax, retired
Solved both puzzles: Shishir Gupta (Indore), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), YK Munjal (Delhi), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Shri Ram Aggarwal (Delhi), Amarpreet (Delhi), Vinod Mahajan (Delhi), Ajay Ashok (Delhi), Sabornee Jana (Mumbai)
Solved #Puzzle 179.1: Anil Khanna (Ghaziabad)
Solved #Puzzle 179.2: Dr Nitin Trasi (Sydney)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
