Teaser 2484: [Snooker bag]
From The Sunday Times, 2nd May 2010 [link] [link]
Josh put all the snooker balls except the cue [ball] into a bag. He and I then played an imaginary frame. At each person’s turn, he drew a ball. If it was red, he scored a point, discarded the red and drew again. If that was a “colour” (i.e. not red), he scored the appropriate number of points, and returned it to the bag, then tried again for a red. If a player drew the wrong ball, it was returned to the bag and his break was over.
On one break, the first four balls drawn earned me points; at that stage, the chance of this happening was one in N, N being the number of points I’d got from the break so far.
Which two colours had I drawn?
This puzzle was originally published with no title.
[teaser2484]
S2T2 
Jim Randell 9:19 am on 10 August 2023 Permalink |
There are 15 reds and 6 coloured balls in a standard snooker set.
If at the start of the go there are R reds, then the probability of the first ball being red (for 1 point) is:
and the chance of the second ball being a colour (for x points) is:
the chance of the third ball being red (for 1 point) is:
and the chance of the fourth ball being a colour (for y points) is:
The probability of getting this far is the product of these probabilities.
And we want this product to give a value of 1/N for some positive integer N that is the total score from 2 colours.
This Python program runs in 56ms. (Internal runtime is 91µs).
from enigma import (irange, div, Decompose, printf) # decompose a total into a set of colours (2..7 points) decompose = Decompose(increasing=1, sep=0, min_v=2, max_v=7) # consider number of reds at the start of the break for R in irange(2, 15): # calculate the inverse of the overall probability # = (product of denominators) / (product of numerators) # which we want to be an integer N N = div((R + 6) * (R + 5) * (R + 5) * (R + 4), R * 6 * (R - 1) * 6) if not N: continue # find points for the 2 colours for cs in decompose(N - 2, 2): printf("R={R} N={N} -> cs={cs}")Solution: The colours were pink (6 points) and black (7 points).
So there were 4 reds at the start of the break, the probabilities being:
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