Bonus time: Four easy ones. Ajit who is married can see Begum who may or may not be married. Begum can see Carl who is unmarried. (Ajit can’t see Carl.) Can we ever say that a married person can see an unmarried person? Or is there too little information to give a meaningful answer?

A is standing 100 metres south of B. Now B starts travelling towards the east at a speed x, whereas A starts travelling at a speed of 2x in a direction that always faces where B is at that moment (which would mean A is not travelling in a straight line, but along a curved path). How much distance would B have travelled when he meets A?

THROUGHPUT

(The Twilight Zone problem was about three chess players A, B and C drawing lots to see which two play the first game and from then on the winner playing against the person who sits out. If the players have equal skills, what are their respective chances?)

If players A and B with C sit out the possible ways the game can end: AA, BB, ACC, BCC, ACBB, BCAA, ACBAA, BCABB, ACBACC, BCABCC, ACBACBB, BCABCAA . . . and so on. The probability of each player winning a game is 1/2. Therefore the probability of a winning = AA, BCAA, ACBAA, BCABCAA . . . and so on = (1/2)^2 + (1/2)^4 + (1/2)^5 + (1/2)^7 . . . this consists of 2 GPs -- (1/2)^2 + (1/2)^5 . . . and (1/2)^4 + (1/2)^7 . . . using the formula for sum of infinite terms of a GP, P(A) = 5/14. Using the same procedure, P(B) = 5/14 and P(C) = 2/7. Thus, in general, the players in the first game have a probability of 5/14 of winning and the one who sits out the first game has a probability of 2/7 of winning. –Yusman Ibrahim, Yusman1963@gmail.com

If we consider the chances before lots are drawn, it is simply 1/3 for each one. But it becomes quite interesting if we know which two friends will play first. The winning probabilities of these two will be equal and will be (1/2^2 + 1/2^5 + 1/2^8 + . . .) + (1/2^4 + 1/2^7 + . . .) = 5/14. The winning probability of the third friend who sits idle during the first match is 2*(1/2^3 + 1/2^6 + . . .) = 2/7. -- Dhruv Narayan, dhruv510@gmail.com

(The second one was: “How can you draw two straight lines on a clock face so that the sums of the numbers in each part are equal?”)

Draw one straight line from edge to edge so that it passes between the numbers 10 and 11 on one side and between the numbers 2 and 3 on the other side. Draw another straight line from edge to edge so that it passes between the numbers 8 and 9 on one side and between the numbers 4 and 5 on the other side. The clock face is divided into three parts and the numbers in each part add up to 26. -- Dr P Gnanaseharan, gnanam.chithrabanu@gmail.com

Here’s a solution treating double digit numbers as two separate ones! Let’s draw the first line starting from the midpoint of the number 12 (it should separate 1 and 2) to a point between 5 and 6. Let the second line cover the zone of 8 and 9 -- ie, the line should start from somewhere between 7 and 8 and end between 9 and 10. Zone 1 contains 2, 1, 2, 3, 4 and 5. Zone 2 has 8 and 9 totalling 17. For zone 3 total, look at individual numbers 6 + 7 + 1 + 0 + 1 + 1 + 1. -- Saishankar Swaminathan, aishankar482@gmail.com

Some clock faces show only 12, 3, 6 and 9 marked on the dial. This can be divided into two parts the sum of both of which are the same: 15 (drawing attached). -- Ramakrishna Easwaran, drrke12@gmail.com(Among the first five who also got it right are: Sudarshan Sahu, sudarshansahu2013@gmail.com; Rekha G, g.rekhapai@gmail.com; Sheikh Sintha Mathar, sheikhsm7@gmail.com; Chris Marlow, chris108m@yahoo.co.uk; V K Bargah, vkbargah@gmail.com)

BUT GOOGLE THIS NOW

1. Try this junior school riddle first: What thing has holes all over the bottom and sides and can still hold water? Got it? Now try this science graduate problem: How come water doesn’t fall out of this thing?

2. Can you think of at least two 13-letter words that are made up of only straight lines? Meaning no Os, Ds, Us, Bs, etc.