Updated: August 12, 2015 5:55:57 pm
This puzzle is my own, a 100% © Kabir Firaque original, but this is not the first time I am presenting it before readers. I ran it in an earlier avatar of Problematics in a regional edition of another newspaper 13 years ago. That was only in print, though, and no one has tried to plagiarise it as far as I know. Add a few embellishments and it has become practically brand new.
The puzzle revolves around a sequence in ‘Sholay’, which completes 40 years on Independence Day this weekend. Hindi film buffs have often told me this sequence is as iconic as the film itself. Gabbar’s goons are just back from the village, having failed to collect the assigned booty despite having outnumbered the Thakur’s hired strongmen @ 3:2.
Those were the days when punishment would invariably be meted out with a revolver. Such a weapon, as we know, has a revolving cylinder with six chambers, each containing a bullet.
We know also how it went. Gabbar takes a fully loaded revolver but chooses not to go for instant punishment. Probability theorist that he was, he fires three shots into the air, leaving three consecutive chambers empty and three full. He then whirls the cylinder enthusiastically.
Puzzle#23A(i): Gabbar presses the revolver to the temple of one defaulter and prepares to fire. What is the probability that the man will survive?
Puzzle#23A(ii): It turns out that he does survive. That changes the probability equations for the next shot, doesn’t it? So, when Gabbar presses the revolver to the second defaulter’s temple and squeezes the trigger, what probability of survival would you give him?
Puzzle#23A(iii): The second victim too survives, introducing another element of certainty to Gabbar’s uncertainty equations. You can guess the next question. What is the probability that the third victim, Kaalia, will now survive?
He does! So, when Gabbar fires three times again, the cylinders with bullets have to come consecutively under the hammer. You don’t have to be a probability theorist to deduce that the chances of survival are now nil.
This much I discussed with my then readers in 2002. To introduce the embellishments, I need to break the equivalent of an Hippocratic oath and do what we journalists claim we never do. Distort the facts.
Suppose that when Gabbar picks up the fully loaded revolver, he fires only one bullet into the air. He whirls the cylinder, and fires into the air. It turns out that a bullet gets fired. He revolves the cylinder again, and fires a third time. Another bullet fired, by coincidence. That once again leaves three bullets and three empty chambers, but not necessarily consecutive this time.
Puzzle#23(A)(iv), (v) & (vi): Go through the same sequence of events as above. What is the probability of (iv) the first defaulter surviving; (v) both the first and the second surviving; (vi) all three surviving?
Puzzle#23B: Enough of Badman. An easy one now. ENOUGH OF BADMAN, in fact, is an anagram. Rearrange the letters to form the name of an English novel that completes 100 years this week.
What you wrote
It was nice to see a bunch of new readers trying my puzzles of last week. Not all of them were correct, though.
Dear Kabir, the answer to Puzzle#22A is 7 minutes (see illustration). I came across your blog while going through the Indian Express mobile app. Thank you and please keep writing.
— K S Sridhar (MSD Pharmaceuticals, Mumbai)
Solved both puzzles: Chandeep Kaur (New Delhi), Rajat Gupta (IIT Roorkee alumnus), Vivek jalan (founder, Customate Systems).
Solved first puzzle: K S Sridhar (above)
Solved second puzzle: Sampath Kumar V (IIM Kozhikode alumnus), Phani Bhushan Tholeti (Synaptics biometrics division, Hyderabad), Nitin Saraswat (Bank of Baroda, Gurgaon)
For the first puzzle, Kirit Makwana explains why the missile’s shortest route has to be 7 minutes but does not describe an actual route. Ranajoy Chatterjee and Phani Bhushan Tholeti give alternative solutions of 8 minutes each.
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