Probability of 9-ball on Break

[Jen_Cen]

Before anyone excoriates me, I did do an honest search on this topic, and I didn't find anything.

Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. The coin does not know or care or remember what has happened previously.

So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller?

It's still a 1 in 35 chance each time they break. Even if they did it 2 or 3 times in a row. On that 4th break, the odds are still 1 in 35!!

 

[Southpaw]

Before anyone excoriates me, I did do an honest search on this topic, and I didn't find anything.

Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. The coin does not know or care or remember what has happened previously.

So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller?

It's still a 1 in 35 chance each time they break. Even if they did it 2 or 3 times in a row. On that 4th break, the odds are still 1 in 35!!

Because with a coin there are only 2 factors...heads or tails. With 9 balls you have to factor in all 9 balls on the table.

Southpaw




bECAUSE

 

[I rack balls]

I would say that the vast majority of 9-balls on the break are due to a bad rack with the back 2 balls directly beneath the 9 loose. If every rack were tight the odds are maybe half that IMO.

 

[9 on the snap]

I would say that the vast majority of 9-balls on the break are due to a bad rack with the back 2 balls directly beneath the 9 loose. If every rack were tight the odds are maybe half that IMO.

This is true.

 

[Jude Rosenstock]

I would say that the vast majority of 9-balls on the break are due to a bad rack with the back 2 balls directly beneath the 9 loose. If every rack were tight the odds are maybe half that IMO.

I agree with this answer. Usually 9s on the break come in bunches and the racker is quick to correct the problem.

 

[Drew]

Before anyone excoriates me, I did do an honest search on this topic, and I didn't find anything.

Anyway, if you flip a coin 100 times, and it comes up heads 100 times, what is the probability that it will come up tails on the 101st flip? The answer, of course, is there is still a 50% chance that it will come up tails. The coin does not know or care or remember what has happened previously.

So, AccuStats says that the 9-ball is sunk on the break 1 out of every 35 breaks in professional matches. So, why do TV announcers and pool book authors insist on saying that the odds of it happening on the subsequent break are smaller?

It's still a 1 in 35 chance each time they break. Even if they did it 2 or 3 times in a row. On that 4th break, the odds are still 1 in 35!!

You are thinking of it the wrong way. When you flip a coin once, there is a 50% chance that it will land on heads. Flip it again and there is a 50% chance that it lands on heads again. However, if you want the odds of the coin landing heads both times in a row, it is 25%. Why? Because there are not just 2 choices with 2 flips. There are actually 4 possible outcomes.

both heads
both tails
first heads second tails
first tails second heads

Make sense? Add a third flip and yet again the odds decrease by half. So what are the odds of flipping heads 3 times in a row? 12.5%

You're right, on each flip there is a 50% chance that it will land on heads. If you flipped the coin 10 times, statistically you would end up with 5 heads and 5 tails. However, the chances of 10 heads in a row are very low. But, the odds are exactly the same as trying to flip heads and then tails and then heads and so on or flipping 5 heads and then 5 tails.

 

[Johnnyt]

I agree with this answer. Usually 9s on the break come in bunches and the racker is quick to correct the problem.

Unless he's an ATM with legs. Johnnyt

 

[klockdoc]

It is 0% chance that you can make the nine on the break on the CueTable though..

 

[skiflyer]

You are thinking of it the wrong way. When you flip a coin once, there is a 50% chance that it will land on heads. Flip it again and there is a 50% chance that it lands on heads again. However, if you want the odds of the coin landing heads both times in a row, it is 25%. Why? Because there are not just 2 choices with 2 flips. There are actually 4 possible outcomes.

both heads
both tails
first heads second tails
first tails second heads

Make sense? Add a third flip and yet again the odds decrease by half. So what are the odds of flipping heads 3 times in a row? 12.5%

You're right, on each flip there is a 50% chance that it will land on heads. If you flipped the coin 10 times, statistically you would end up with 5 heads and 5 tails. However, the chances of 10 heads in a row are very low. But, the odds are exactly the same as trying to flip heads and then tails and then heads and so on or flipping 5 heads and then 5 tails.

And as you've illustrated, she's thinking of it exactly the right way. After the player breaks and sinks the nine, the odds of doing the same on the next rack (assuming equal racks) is the same as it was before the previous rack.

Now if before the first break you ask what the odds of two in a row are, the odds are much less... but that's not the question.

Jen_Cen, you are absolutely right and the announcers are simply applying bad logic.... it all relates to this little tidbit probability teachers often do to large classes. Have 50 students write out a random sequence of 10 random coin flips... then flip a coin 50 time and record the output. What the students will likely not capture that the real flip will capture is the phenomenon of long runs, very few (if any) students will submit 10 heads, but 10 heads is exactly as likely as hthththtt or thtthtthtt.

 

[PoolBum]

You're right, on each flip there is a 50% chance that it will land on heads. If you flipped the coin 10 times, statistically you would end up with 5 heads and 5 tails. However, the chances of 10 heads in a row are very low. But, the odds are exactly the same as trying to flip heads and then tails and then heads and so on or flipping 5 heads and then 5 tails.

In fact, the odds of any particular sequence of heads/tails in ten flips is exactly the same as any other sequence. For example, the odds of ending up with this sequence, HTTHHTHTTH, are exactly the same as the odds of ending up with this sequence, TTTTTTTTTT.

The reason why making the 9-ball on the break is not comparable to flipping a coin is because the 9-ball breaks are not independent events, in the following sense. The table and playing conditions figure into the probability of making the 9 on the break. For example, breaking from the side cushion may have been shown to increase the likelihood of sending the 9 toward the corner pocket.

In a famous match between Mike Sigel and Nick Varner from the U.S. Open, they made the corner ball on the break all 21 times. Sometimes the 9-ball gets consistent action on a given table depending on the table conditions. Thus making the 9-ball 2 or 3 times in a row may indicate that the odds of making the 9 on that particular table breaking that particular way are greater than 1 in 35.

That being said, the announcers you are referring to may in fact be committing a fallacy when they claim that if the 9 was made on the break the odds of making it on subsequent breaks are less than 1 in 35 (what I've pointed out--the fact that the table conditions may affect the chances of making the 9 on the break--would make the odds of making the 9 on the break greater, not smaller).

This reminds me of some of the reasoning puzzles i give my students when I teach logic. Here is one. Suppose you know that your neighbors have two kids, and you know that at least one of them is a boy. If the odds for having a boy are exactly 50-50 for any given birth, gievn what you know about your neighbors what are the odds that both of their children are boys?

 

[Drew]

And as you've illustrated, she's thinking of it exactly the right way. After the player breaks and sinks the nine, the odds of doing the same on the next rack (assuming equal racks) is the same as it was before the previous rack.

Now if before the first break you ask what the odds of two in a row are, the odds are much less... but that's not the question.

Jen_Cen, you are absolutely right and the announcers are simply applying bad logic.... it all relates to this little tidbit probability teachers often do to large classes. Have 50 students write out a random sequence of 10 random coin flips... then flip a coin 50 time and record the output. What the students will likely not capture that the real flip will capture is the phenomenon of long runs, very few (if any) students will submit 10 heads, but 10 heads is exactly as likely as hthththtt or thtthtthtt.

If that's the way you want to look at it, I propose a bet. We can poll 30 people. I'm using 30 as it's close to the average class size. They can each submit a combination of tails and heads and then we'll flip a coin 10 times. I'll bet you $100 that not a single person chooses the correct combination.

 

[Drew]

In fact, the odds of any particular sequence of heads/tails in ten flips is exactly the same as any other sequence. For example, the odds of ending up with this sequence, HTTHHTHTTH, are exactly the same as the odds of ending up with this sequence, TTTTTTTTTT.

The reason why making the 9-ball on the break is not comparable to flipping a coin is because the 9-ball breaks are not independent events, in the following sense. The table and playing conditions figure into the probability of making the 9 on the break. For example, breaking from the side cushion may have been shown to increase the likelihood of sending the 9 toward the corner pocket.

In a famous match between Mike Sigel and Nick Varner from the U.S. Open, they made the corner ball on the break all 21 times. Sometimes the 9-ball gets consistent action on a given table depending on the table conditions. Thus making the 9-ball 2 or 3 times in a row may indicate that the odds of making the 9 on that particular table breaking that particular way are greater than 1 in 35.

That being said, the announcers you are referring to may in fact be committing a fallacy when they claim that if the 9 was made on the break the odds of making it on subsequent breaks are less than 1 in 35 (what I've pointed out--the fact that the table conditions may affect the chances of making the 9 on the break--would make the odds of making the 9 on the break greater, not smaller).

This reminds me of some of the reasoning puzzles i give my students when I teach logic. Here is one. Suppose you know that your neighbors have two kids, and you know that at least one of them is a boy. If the odds for having a boy are exactly 50-50 for any given birth, gievn what you know about your neighbors what are the odds that both of their children are boys?

In that case, you already know for a fact that 1 child is a boy. Therefore, the odds are clearly 50/50 as you only have to guess at 1 child. Suppose you do not know anything about your neighbors except that they have 2 children. Now what are the odds that they are both boys? There are 4 possible combinations so the odds are 25/75.

 

[Patrick Johnson]

In fact, the odds of any particular sequence of heads/tails in ten flips is exactly the same as any other sequence.

This can be misleading unless you're careful how you say it. I'd say the odds of any particular sequence are the same as any other particular sequence (although I'd use the word "specific").

For example, the odds of ending up with this sequence, HTTHHTHTTH, are exactly the same as the odds of ending up with this sequence, TTTTTTTTTT.

I know you aren't, but this could look like you're saying any random sequence is as likely as all tails, but what you mean (again) is that those specific sequences are equally likely. But it's much more likely that there will be a mix of heads and tails than all tails, and it's very likely that the number of heads and tails will be roughly equal over time.

pj
chgo

 

[Patrick Johnson]

This reminds me of some of the reasoning puzzles i give my students when I teach logic. Here is one. Suppose you know that your neighbors have two kids, and you know that at least one of them is a boy. If the odds for having a boy are exactly 50-50 for any given birth, gievn what you know about your neighbors what are the odds that both of their children are boys?

I believe this is a version of the old Monty Hall proposition, and the answer is 1/3.

pj
chgo

 

[PoolBum]

This can be misleading unless you're careful how you say it. I'd say the odds of any particular sequence are the same as any other particular sequence (although I'd use the word "specific").



I know you aren't, but this could look like you're saying any random sequence is as likely as all tails, but what you mean (again) is that those specific sequences are equally likely. But it's much more likely that there will be a mix of heads and tails than all tails, and it's very likely that the number of heads and tails will be roughly equal over time.

pj
chgo

True, one has to be careful in wording it.

 

[skiflyer]

If that's the way you want to look at it, I propose a bet. We can poll 30 people. I'm using 30 as it's close to the average class size. They can each submit a combination of tails and heads and then we'll flip a coin 10 times. I'll bet you $100 that not a single person chooses the correct combination.

Give me 3 to 1 odds and it's an even bet. What's your point?

 

[PoolBum]

I believe this is a version of the old Monty Hall proposition, and the answer is 1/3.

pj
chgo

1/3 is correct. It is similar to the Monty Hall problem.

 

[PoolBum]

In that case, you already know for a fact that 1 child is a boy. Therefore, the odds are clearly 50/50 as you only have to guess at 1 child.

The odds that both are boys given that you know that one of them is a boy are not 50/50.

 

[skiflyer]

1/3 is correct. It is similar to the Monty Hall problem.

Yet enjoyably if you told them the first child or the second child was a boy the answer is 1/2... make them noodle that one for awhile.

 

[PoolBum]

Yet enjoyably if you told them the first child or the second child was a boy the answer is 1/2... make them noodle that one for awhile.

Correct. If you told them "the first child was a boy" or you told them "the second child was a boy" the answer would be 1/2. On the other hand, if you told them "the first child or the second child was a boy" the answer would still be 1/3.