## Teaser 2989: Pot luck

**From The Sunday Times, 5th January 2020** [link]

Liam had a bag of snooker balls containing 6 colours (not red) and up to 15 red balls. He drew out balls at random, the first being a red. Without replacing this he drew another ball; it was a colour. He replaced this and drew another ball. This was a red (not replaced), and he was able to follow this by drawing another colour. The probability of achieving a red/colour/red/colour sequence was one in a certain whole number.

After replacing all the balls, Liam was able to “pot” all the balls. This involved “potting” (i.e., drawing) red/colour/red/colour… red/colour (always replacing the colours but not the reds), then “potting” the six colours (not replaced) in their correct sequence. Strangely, the probability of doing this was also one in a whole number.

What are the two whole numbers?

[teaser2989]

## Jim Randell 9:30 am

on4 January 2020 Permalink |This Python program calculates

P(k, n), the probability of selectingnconsecutive red/colour pairs (with the red not replaced and the colour replaced), ifkreds are available initially (and 6 colours).It runs in 83ms.

Run:[ @repl.it ]Solution:The two numbers are 15 and 352,800 (i.e. the probabilities are 1/15 and 1/352,800).When there are

kreds available (and 6 colours) the probability of selecting a red (not replaced) followed by a colour (to be replaced) is:When there are 4 reds available the probability of select the first red/colour pair is:

There are now 3 reds available, and the probability of selecting a second red/colour pair is:

Multiplying these together we get a probability of select two red/colour pairs in a row as:

The probabilities of selecting a third and fourth red/colour pair are:

Combining all these we get:

The probability of selecting the six colours in the correct order is: 1/6! = 1/720.

Giving a probability of “clearing the table” of:

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