Like everything else, the way one collects movies has evolved. We have moved on from projector reels to cassettes to discs to the present, when everything is digital. Some store their films on hard disks and some online.
The following puzzle acknowledges the evolution by reflecting on the DVD age, which we left behind a decade or so ago.
# Puzzle 214.1
A film buff and his wife, who shares his interest, have built a collection in digital format, and are now getting rid of physical media that take up more space than their house in Mumbai can afford.
“Sad to see those go,” says the husband, pointing to six DVDs—two each of Rear Window, Roman Holiday and Sunset Boulevard.
“I didn’t know we had two copies of each,” his wife observes, “although I do remember buying one DVD of each of these movies in 2008.”
“Yes, we bought each original DVD together. The other copies are pirated versions, which I bought from Andheri, Bandra, Malad in 2003, 2004 and 2005. This was before I met you,” the husband tells her. “Once I could afford the originals, I bought the three films from the same three suburbs.”
“Interesting,” she muses.
{{/usCountry}}“Interesting,” she muses.
{{/usCountry}}“Even more interestingly, we bought each original DVD in a different suburb than where I had bought its bootleg version. This pirated disc, for instance, came from a Malad shop in 2005, but that’s not the suburb from where we got its original,” her husband says, pointing to one of the DVDs.
She picks up all three originals. “We bought Rear Window from Bandra. And Sunset Boulevard cost us ₹599, if I remember correctly—indeed it did,” she confirms, checking the back cover. She places the disks down again.
“We bought this original from Andheri,” her husband says, picking up one of them. “I had bought its pirated version in 2005.”
“Bootlegs are cheap, aren’t they,” she notes. He nods. “But their prices are not consistent. My bootleg Roman Holiday, for example, cost me ₹100 more than this other bootleg, which I bought in 2003,” he says.
“But they were all cheaper than the originals, as their low quality deserved,” he adds. He picks up a couple of DVDs: “For example, I bought this bootleg in 2005 at a price that was one-third the price of this original—two different films, but both from the same suburb.”
All six DVDs were priced at an integer number of rupees. It is not possible to determine all six prices, but the information you have is enough to find out what you need to.
From where was each of the six DVDs bought, and what was the year of purchase of each pirated film? (Inspired by a Henry Ernest Dudeney puzzle)
# Puzzle 214.2
A drawer contains several pairs of white socks and black socks. If the lights go out, what is the minimum number of socks you must take out to ensure you have at least one matching pair?
MAILBOX: LAST WEEK’S SOLVERS
# Puzzle 213.1
Hi Kabir,
In the four given situations, the probabilities of the man surviving Russian roulette are: 4/9, 2/5, 4/9 and 1/2.
(a) He loads the first bullet, spins the cylinder, loads the second bullet into a new chamber, spins the cylinder, fires the first shot and (if he survives), spins again and fires the second shot. Each time he spins the cylinder, any of the six chambers could be under the trigger on each occasion, and two of these have bullets. Probability of survival = (4/6)*(4/6) = 4/9.
(b) He loads the first bullet, spins the cylinder, loads the second bullet into a new chamber, spins the cylinder and fires the first shot, and then (if he survives) fires the second shot. If he survives the first shot, that cylinder is empty and it’s certain that the two bullets are in any two of the remaining five chambers. Probability of survival = (4/6)*(3/5) = 2/5.
(c) He loads two bullets in consecutive chambers, spins the cylinder, fires the first shot and (if he survives) spins the cylinder again and fires the second shot. Here again, because the cylinder is spun before each shot, any of the six chambers could be under the trigger on each occasion, and two of these have bullets. The loaded chambers being consecutive or not does not matter in this case. Probability of survival = (4/6)*(4/6) = 4/9.
d) He loads two bullets in consecutive chambers, spins the cylinder, fires the first shot and (if he survives) fires the second shot. Here, after spinning the cylinder, he will survive both shots only if, on the first shot, the chamber under the trigger is neither of the loaded chambers nor the chamber immediately preceding this pair (in the order of rotation).
(Bullet 1) (Bullet 2) (Empty 1) (Empty 2) (Empty 3) (Empty 4) (Bullet 1)…
In other words, he will survive only if, on the first shot, the chamber under the trigger is any one of the following three — (Empty 1), (Empty 2) or (Empty 3). He cannot survive if the chamber under the trigger on the first shot is any of the remaining three — (Empty 4), (Bullet 1) or (Bullet 2). Probability of survival = (3/6) = 1/2.
— Sabornee Jana, Mumbai
***
Hi Kabir,
In all cases, the probability of surviving the first shot is 4/6. Now let us look at case (d) after surviving the first shot. Designating as slot #1 the chamber that first came under the hammer, it will slot #2 on the second shot. The two loaded slots could be (slots #2 and #3) or (#3 and #4) or (#4 and #5) or (#5 and #6). Out of these four possibilities, he will survive in the last three cases. This means that the probability of survival is 3/4. Therefore, the probability of surviving both shots in case (d) is (4/6)*(3/4) = 1/2.
— Professor Anshul Kumar, New Delhi
For case (d), I have shown two approaches from readers to reconfirm that the answer is indeed 1/2, no matter which approach you adopt. A few other readers have made mistakes when solving case (d).
#Puzzle 213.2
Hello Kabir,
Here are my 10 words formed from the letters of CONDIMENT. C: CONTENDS (8 letters), I: INSIDE (6 letters), N: NOTICES (7 letters), D: Domestic (8 letters), I: INTENDS (7 letters), M: Mentions (8 letters), E: ENDMOST (7 letters), N: NOTICED (7 letters), T: TENSION (7 letters), S: SECTION (7 letters). Total: 72 letters.
— Dr Sunita Gupta, New Delhi
Some word game adjudicators might not accept both NOTICES and NOTICED in the same list. If we do accept them, I think Dr Gupta’s list has the longest possible words starting with each of the 10 letters. Some readers have used C to form CONDIMENT or CONDIMENTS itself, which is not acceptable as was stated last week.
Solved both puzzles: Sabornee Jana (Mumbai), Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat, partly correct), Shishir Gupta (Indore, partly correct), Amarpreet (New Delhi, partly correct)
Solved #Puzzle 213.1: Dr Vivek Jain (Baroda, partly correct)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com