Snooker's Biggest Quiz

[Stella147]

I'll guess at 36

[Stella147]

Actually on looking at that again I think Monique is right.
If player B had a break of 40 would there only be 35 left on the table or am I off on a major tangent there???

[The Statman]

I will add a bit here:

I want you to describe what happens in Player A's break-off shot and Player B's break (i.e. indicate which balls are potted each shot) to illustrate the size of the break and the points left on the table.

36 is too high, by the way!

[snookersfun]

very theoretically, would 23 be enough?

[The Statman]

very theoretically, would 23 be enough?

please describe the shots required to make this happen.

[snookersfun]

well, not sure about this, but player A breaks off leaving most reds pointing to pockets.
Player B now pots 15 reds in one shot (extremely theoretically), then pots green, yellow, green (break of 15+3+2+3=23). At this point there are 22 points left on the table...

[The Statman]

well, not sure about this, but player A breaks off leaving most reds pointing to pockets.
Player B now pots 15 reds in one shot (extremely theoretically), then pots green, yellow, green (break of 15+3+2+3=23). At this point there are 22 points left on the table...

An excellent answer.

But it is still not sufficient, I'm afraid. I will be back at 5 o'clock GMT and see if anyone has an answer smaller than 23.

[Stella147]

I'd say 22 - following Snookersfun theory - except pots a yellow after the reds and then yellow and green (break of 22) leaving 22 on the table, leaving A needing snookers to win.
If not I completely give up

[Monique]

Well building on snookersfun's idea ... if player A pots the black when breaking he/she gives a 7 points head start to B, that does not count in the break. So potting all reds together, then brown would give a break of 19, a score of 26 with 25 on the table.

[snookersfun]

if we can foul already, I would then let player A pot 14 reds and a colour (black) on his first shot. Now we have a 7 point foul and not much left. So a break of 12 (red,yellow, yellow,green, brown) would now be enough...??

[bongo]

I'd say a break of 1.

I will explain, Player A broke off but he did not hit any reds, the white ball comes down the table and Player B is snookered behind the green. Except for the white, no balls moved on the break. Player B comes to the table after the break, because he thinks that there is no easy shots, he asks Player A to play again from the position left. Player A comes to the table and he plays the white around the table, missing all balls and the white goes up behind the yellow, still except for white, no balls have moved. Player B asks for the white behind the green again (Miss-Rule). Score so far: 0-8 in points to Player B. After Player B gets the cue ball back behind the green. Player A misses again. Player B keeps putting Player A back into the previous position via the miss rule. Then Player A keeps missing the red, each time being put back because of the miss rule. He keeps missing and being put back and on the 34th attempt, he finally hits the reds, one red moves out leaving an easy chance to Player B. The score is 0-140 to Player B. Then Player B pots the simple red. He then playes a safety shot on the black and he does not play a foul stroke on that shot.

So Player B made a break of 1. Then there was 14 reds left on the table, and 139 points remaining. Player A was 140 points behind, so he needed Snookers.

I seem to have taken this too far, so sorry. It was just my idea.

This is probably not going to happen in 100 years but its thoretical.

I put it as 'he' but it still applies as she.



So, Am I right?!

[bongo]

36th, not 34th attempt

[The Statman]

...

So, Am I right?!

An excellently thought out answer, and it must have taken you a while to compse and to type.

But I did say that Player B came to the table immediately after Player A's break-off, did I not?

So, wrong I'm afraid.

[The Statman]

if we can foul already, I would then let player A pot 14 reds and a colour (black) on his first shot. Now we have a 7 point foul and not much left. So a break of 12 (red,yellow, yellow,green, brown) would now be enough...??

That's pretty good, and I would suggest this is the correct answer. I had got the answer of 14 but, looking back, I find it is my mathematics that are wrong. I had worked out that the break-off eliminating all 15 reds gave a better answer than by eliminating only 14.

Well done; I will update the scoreboard shortly.

[bongo]

You really make this difficult, oh well!