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I have a question that is very simple to understand and very complex to answer.

If a snooker player could elect to forego potting a coloured ball (typically worth a handful of points) and instead move onto another (worth one), would they?

For simplicity, assume that a player can pot any ball of their choosing with probability p. Their opponent can pot with probability q (where p can be greater than, equal to or less than q). The aim of the game is to gain more points than is left on the table. Importantly, I am assuming that in this mathematical snooker game, the coloured balls must be potted at the end, as in regular snooker.

What is the optimal strategy for a player to take? Always reds? Always trying for colours? A mix? It depends what the opponent does? It depends on p and q? (You can assume p and q are approximately $0.9$ but varying these would be interesting)

If you want to try doing this computationally, feel free to reduce the number of reds from 15 to 10 or 6 as is sometimes played, unless you think that would change the result.

Leonhard Euler
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  • Rules are not totally clear to me. Players need to alternate red-coloured as in real snooker? You mention that coloured must be potted at the end, does this mean that at the beginning players need to pot red balls only? Please clarify. – nicola Mar 01 '21 at 11:23
  • So in snooker, you must pot a colour after a red. But if they could just elect not to pot a colour and instead go for another red, would they? Basically they can either choose to pot a black after a red, or they can just skip that and move onto the next red of they like. – Leonhard Euler Mar 01 '21 at 12:29
  • Ok, I got it, thank you. Your model implies that, if you choose to hit a colour, you would always choose the black of course. I guess that the same option applies to the last red: after potting it, you can either choose to pot the black or go directly to the yellow. – nicola Mar 01 '21 at 12:43
  • @nicola Yep, after each red, you can elect to skip the colour if you like, including the last one. – Leonhard Euler Mar 01 '21 at 13:14
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    It's an interesting and not trivial (as far as I can tell now) problem. It very likely depends on $p$ and $q$. For instance, for $q=1$, if I am not wrong, you want to pot just 2 blacks alongside 15 reds and can win after 17 consecutive pots, amassing 29 points with 27 remaining. Any other strategy needs you to pot at least 18 balls (again, if I am not wrong). If $q$ is slightly lower, I guess that there is more merit in alternating red and black because you want to give your opponent a chance to miss. – nicola Mar 01 '21 at 15:30
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    For what it's worth you might try tacking the problem with dynamic programming, i.e. starting with two balls left on the player's turn, computing the optimal strategy and the expected cost, and then working backwards one ball at a time to the start of the game. If memory serves, there are only 15 + 7 = 22 possible states (in terms of which balls are on the table at any given time) the table can have when it's your turn, which is easily handled by computer or even by hand. – COTO Mar 05 '21 at 03:50
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    @COTO If the players' objectives are to maximise their probabilities of winning, the state would also have to include the current difference between the points each of them have so far scored. – lonza leggiera Mar 05 '21 at 15:24
  • @lonzaleggiera: Ah, yes. Although if the point spread were clamped to some reasonable interval (+/-1000, say), the problem would still be readily solvable by computer. I suppose this might return absurd results if both players' probabilities of potting balls was tiny (since scores can go negative without limit in snooker) and the tails of the difference-of-score distribution were large, but this could be accounted for by enlarging the state space whenever both $p$ and $q$ are tiny. – COTO Mar 05 '21 at 18:12
  • @COTO From the way the question is worded I'm assuming that the probabilities that the players suffer any penalties by hitting or potting a wrong ball are negligible. If that's the case, the points differential between the players can be limited to the smaller of the total number of points available from the balls left on the table, and the total number that could have been scored from balls so far potted. This can be at most $67$, occurring when there are $5$ reds left on the table. Thus, under this assumption it would be entirely feasible to carry out your dynamic program on a computer. – lonza leggiera Mar 07 '21 at 06:12
  • @lonzaleggiera Yes, there are no fouls in this game. Once you are ahead of the potential points on the table, you have won. – Leonhard Euler Mar 08 '21 at 09:32

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