Number of Strokes

[DawRef]

Not sure if this has been posted before.

What is the least number of strokes a player needs to make, to 'win' a frame of snooker (i.e. to leave his opponent needing snookers)?

[missneworleans]

pot all 15 reds in one shot then pot black, yellow and green(27 ahead 22 on the table)
so...4 strokes

[ja.]

one to break off, leave the q ball snookered and make the opponent miss 37 times.

[missneworleans]

one to break off, leave the q ball snookered and make the opponent miss 37 times.

Zero! Opponents breaks off pots a red but gets snookered and misses 35 times

[ja.]

but that was only in theory

[missneworleans]

but that was only in theory

well the question is about theory
but I can tell a thing than might really happen
opponent gets a red off the break sees a red but fails to hit it 3 times in a row cos trying to make it safe

[ja.]

well the question is about theory
but I can tell a thing than might really happen
opponent gets a red off the break sees a red but fails to hit it 3 times in a row cos trying to make it safe

yes but then no snookers would be needed, the player would just be awarded the frame

[missneworleans]

yes but then no snookers would be needed, the player would just be awarded the frame

well but it is a win

[ja.]

so basically if the opponent is an imbecile, the player doesnt need to get out of his chair at all to win a frame

[missneworleans]

so basically if the opponent is an imbecile, the player doesnt need to get out of his chair at all to win a frame

hahahaha
yes

[DawRef]

Well the correct answer is 2. I'll post the explanation later (unless someone does it before I do).

[missneworleans]

Well the correct answer is 2. I'll post the explanation later (unless someone does it before I do).

Why not 0?

[DawRef]

Why not 0?

I suppose in an aggregate score match, if the player came to the table more than 155 points behind in the final frame, then yes, the answer could be zero (he could not play a stroke and concede). However, it is a normal frame of snooker and not an aggregate score match.

[RGCirencester]

oponent breaks off and misses the pack, hits the the black, you have played no shots and are 0-7 ahead. Miss call you put the oponent back in
he tries again same thing happens, 0-14, still no shots
again you put him back in this time he has to hit a red so the frame is conceded... However unluckily he knocks the black into the pocket. he doesnt automatically lose because he has hit a red first. Score is now 0-21 still no shots played
You take your shot and fluke every single red, score is now 0-36 with only 27 left on the table. Snookers required after you have played a single shot.

However this sequence of 2 misses and a foul could theoretically continue forever until you had more than 147 points, meaning snookers would be required without you playing a single shot. So unless I am wrong, aggregate scoring or not the absolute answer is actually as said before no shots.

However as previously stated if after the break off (and miss) the white lands snookered on all reds, then the oponent can miss a further 37 times and then concede when snookers are required without you playing a shot.

[Capelito]

Starting with two fouls from opponent as RGCirencester described. followed by opponent potting all the reds and the black in next shot would give 21-0 and 27 left. You put him back in once again and he pots yellow, green, brown and blue before fouling on black again. 28-14 with 13 left, and you haven't played a single shot