What Feat Is Harder

[cigardave]

You're going by the average of BNR's over a decade right? well that gives you the average of BNR's but what it doesn't give you are the overall number of BNR's combined with the average number of BNR's in succession. If you have that you can get a much better idea of the probability of running ten racks in a row....

If say the overall of 1000 racks is 333 racks that are BNR's, but the majority of those racks run came in 3 or 4 packs then the probability of BNRing out 10 racks increases over a random 1 in 3 of BNR racks. Especially if you throw in some 7 or 8 packs and many 5 or 6 packs..... then the odds go down dramatically compared to random BNR's.

I agree fundamentally with your first paragraph. Basically, the more data we have about the past, the better we're able to predict the future of a (relatively) consistent behavior.

Re your second paragraph, I agree as well... if indeed that would be the case (the majority being 3 or 4-pks).

From a statistics standpoint, the odds of a single BNR for your example are 1 in 3.... a 2-pk, 1 in 9... a 3-pk, 1 in 27... a 4-pk, 1 in 81... a 5-pk, 1 in 243.. and so on.

In essence, this would be Earl if the example is an individual person... since Earl's percentage was 32.7% (damn near 1 in 3).

So, on average, Earl would string two 5-pks in 500 games if he were to break all 500 of 'em.

 

[sixpack]

I agree fundamentally with your first paragraph. Basically, the more data we have about the past, the better we're able to predict the future of a (relatively) consistent behavior.

Re your second paragraph, I agree as well... if indeed that would be the case (the majority being 3 or 4-pks).

From a statistics standpoint, the odds of a single BNR for your example are 1 in 3.... a 2-pk, 1 in 9... a 3-pk, 1 in 27... a 4-pk, 1 in 81... a 5-pk, 1 in 243.. and so on.

In essence, this would be Earl if the example is an individual person... since Earl's percentage was 32.7% (damn near 1 in 3).

So, on average, Earl would string two 5-pks in 500 games if he were to break all 500 of 'em.

Just to be a nit

I agree with all you've said, except the conclusion. Technically speaking, if he is expected to put a 5-pack one in 243, what that would mean statistically is that if he broke 500 (486) times, he would have a 66% chance of doing it exactly twice. Which is different than an average. On average, he would be expected to do it about 2*(.66) (outcome * probability) times or 1.32 times per 500 breaks.

Cheers,
RC

 

[zeeder]

Just to be a nit

I agree with all you've said, except the conclusion. Technically speaking, if he is expected to put a 5-pack one in 243, what that would mean statistically is that if he broke 500 (486) times, he would have a 66% chance of doing it exactly twice. Which is different than an average. On average, he would be expected to do it about 2*(.66) (outcome * probability) times or 1.32 times per 500 breaks.

Cheers,
RC

One thing you left out of your analysis is that one set of data is 5 racks so he would have to have played 5 racks 243 times for one 5-pack thus he would need to play 486 5-rack sets (2430 games) to get two 5-packs.

 

[Jaden]

I agree fundamentally with your first paragraph. Basically, the more data we have about the past, the better we're able to predict the future of a (relatively) consistent behavior.

Re your second paragraph, I agree as well... if indeed that would be the case (the majority being 3 or 4-pks).

From a statistics standpoint, the odds of a single BNR for your example are 1 in 3.... a 2-pk, 1 in 9... a 3-pk, 1 in 27... a 4-pk, 1 in 81... a 5-pk, 1 in 243.. and so on.

In essence, this would be Earl if the example is an individual person... since Earl's percentage was 32.7% (damn near 1 in 3).

So, on average, Earl would string two 5-pks in 500 games if he were to break all 500 of 'em.

Yes if it went that way. What I was saying is that if there were for instance greater occurrences of five packs than three packs or if the majority of BNR's were strings of BNR's as opposed to single instances, if that were the case the liklihood of a 10 pack would be greater than otherwise and in my experience when you're on you're on and you do tend to string racks together rather than every once in a while.

In other words I would guess that instead of BNR'ing one rack and then missing on two it's more like BNR'ing 4 racks and playing safe or missing on 8-10.... I was just saying to make the most valid statistical probability that info would be necesary....

Besides which, where in match play top players opt to play safe, if they were actually going for a ten pack, it would be even more likely than statistics based on match play would suggest....

 

[zeeder]

I agree fundamentally with your first paragraph. Basically, the more data we have about the past, the better we're able to predict the future of a (relatively) consistent behavior.

Re your second paragraph, I agree as well... if indeed that would be the case (the majority being 3 or 4-pks).

From a statistics standpoint, the odds of a single BNR for your example are 1 in 3.... a 2-pk, 1 in 9... a 3-pk, 1 in 27... a 4-pk, 1 in 81... a 5-pk, 1 in 243.. and so on.

In essence, this would be Earl if the example is an individual person... since Earl's percentage was 32.7% (damn near 1 in 3).

So, on average, Earl would string two 5-pks in 500 games if he were to break all 500 of 'em.

Actually, it's not two 5-packs in 500 games, it's two in ~500 sets of 5 games, thus ~2500 games. For each instance you note above there is a different number of games taken into account in each observation.

 

[cigardave]

Just to be a nit

I agree with all you've said, except the conclusion. Technically speaking, if he is expected to put a 5-pack one in 243, what that would mean statistically is that if he broke 500 (486) times, he would have a 66% chance of doing it exactly twice. Which is different than an average. On average, he would be expected to do it about 2*(.66) (outcome * probability) times or 1.32 times per 500 breaks.

Cheers,
RC

RC - Where does the 66% come from? I don't understand that.

And what I meant by "on average" was really along the lines of... if Earl kept on breaking and attempting to run out, I would expect him to string a 5-pk every 243 games (or so)... and therefore, I would expect him to do that twice in approximately 500 racks.

Did I say it right this time?

 

[Jaden]

I agree fundamentally with your first paragraph. Basically, the more data we have about the past, the better we're able to predict the future of a (relatively) consistent behavior.

Re your second paragraph, I agree as well... if indeed that would be the case (the majority being 3 or 4-pks).

From a statistics standpoint, the odds of a single BNR for your example are 1 in 3.... a 2-pk, 1 in 9... a 3-pk, 1 in 27... a 4-pk, 1 in 81... a 5-pk, 1 in 243.. and so on.

In essence, this would be Earl if the example is an individual person... since Earl's percentage was 32.7% (damn near 1 in 3).

So, on average, Earl would string two 5-pks in 500 games if he were to break all 500 of 'em.

Say for instance that in the data accumulate d where Earl BNR'd 32.7 %.

If that suggests that he should BNR a 5 pack every 247 games but in actuality he did a 5 pack every 100 and did fewer 3 packs or 2 packs than would be statistically probable? that would increase the liklihood of a ten pack occurring.

 

[zeeder]

Say for instance that in the data accumulate d where Earl BNR'd 32.7 %.

If that suggests that he should BNR a 5 pack every 247 games but in actuality he did a 5 pack every 100 and did fewer 3 packs or 2 packs than would be statistically probable? that would increase the liklihood of a ten pack occurring.

Basically, all the calculations that have occured are assuming that each rack is independent of all other racks. If there was some sort of conditional probability then, i.e. a BnR increases the probability of another consecutive BnR, then the odds of a 10-pack would be different.

That being said, I just had a similar conversation with my statistics professor about a problem on our exam last friday where he said most of the time when people think there is an effect of one outcome due to "good feelings from a successful outcome", or whatever, that this doesn't come out in looking at the actually statistics.

 

[Jaden]

Basically, all the calculations that have occured are assuming that each rack is independent of all other racks. If there was some sort of conditional probability then, i.e. a BnR increases the probability of another consecutive BnR, then the odds of a 10-pack would be different.

That being said, I just had a similar conversation with my statistics professor about a problem on our exam last friday where he said most of the time when people think there is an effect of one outcome due to "good feelings from a successful outcome", or whatever, that this doesn't come out in looking at the actually statistics.

Of course!!!! Math professors are hopelessly inundated within the mathematicla paradigms and can't look at real world logic without atrributing those mathematical paradigms to their thought processes.....

 

[zeeder]

Of course!!!! Math professors are hopelessly inundated within the mathematicla paradigms and can't look at real world logic without atrributing those mathematical paradigms to their thought processes.....

That's the problem, a lot of times "real world logic" isn't logically sound...lol

 

[cgriffin5]

Most people who arent bowlers do not realize how difficult it is to bowl a 300. There are so many variables recreation players do not know about. Carry, oil ball selection, there is more but there is no sense boring you guys. Running 10 racks in my opinion is more difficult to do than bowl a 300 game but the difference isnt as big as most here think.

 

[cigardave]

Actually, it's not two 5-packs in 500 games, it's two in ~500 sets of 5 games, thus ~2500 games. For each instance you note above there is a different number of games taken into account in each observation.

I don't believe so Zeeder.

Let's stay with the Earl example. Earl BNRs the first rack. The following break he comes up dry. Even though that's two games, that's one attempt down the tube. Next break he makes a ball and proceeds to run out. Same with the next rack and the one after. Now he's at a 3-pk and with an opportunity to keep it going. He's still working on his second attempt to string a 5-pk together.

I guess what I'm saying is... the 1 is 243 odds are the odds of stringing a 5-pk together when Earl gets 243 attempts to do so.

In my posts above I used the word games, which is inaccurate, I believe.

But I'll also add, in response to your post, if you still want to talk games, I don't believe that it would be correct to assume that each attempt is 5 games.

 

[zeeder]

I don't believe so Zeeder.

Let's stay with the Earl example. Earl BNRs the first rack. The following break he comes up dry. Even though that's two games, that's one attempt down the tube. Next break he makes a ball and proceeds to run out. Same with the next rack and the one after. Now he's at a 3-pk and with an opportunity to keep it going. He's still working on his second attempt to string a 5-pk together.

I guess what I'm saying is... the 1 is 243 odds are the odds of stringing a 5-pk together when Earl gets 243 attempts to do so.

In my posts above I used the word games, which is inaccurate, I believe.

But I'll also add, in response to your post, if you still want to talk games, I don't believe that it would be correct to assume that each attempt is 5 games.

You are certainly correct that the number of attempts in real life wouldn't be 5 full games because someone would start over after they missed if they were trying to run a 5-pack. However, statistically speaking, 1 in 243 odds refers to 1 in 243 sets of 5 racks.

 

[JoeyInCali]

Shorty going thru a month w/out buying and selling a cue.

B and R 10 racks for sure.
Hole in one? Hell a 3-yr old kid did it in Laguna Hills with his Snoopy club. Gimme a break.

 

[sixpack]

One thing you left out of your analysis is that one set of data is 5 racks so he would have to have played 5 racks 243 times for one 5-pack thus he would need to play 486 5-rack sets (2430 games) to get two 5-packs.

Each game would be the potential first game of a 5 rack set. If you did it your way, the odds would be lower because presumably you wouldn't allow him to count the last 2 games of one and the first 3 games of the next.



Cheers,
RC

 

[Donovan]

Each game would be the potential first game of a 5 rack set. If you did it your way, the odds would be lower because presumably you wouldn't allow him to count the last 2 games of one and the first 3 games of the next.



Cheers,
RC

Plus these things can't predict if the shooter just broke up their partner or the fact that they are really in stroke today or whatever. JMO

 

[sixpack]

RC - Where does the 66% come from? I don't understand that.

And what I meant by "on average" was really along the lines of... if Earl kept on breaking and attempting to run out, I would expect him to string a 5-pk every 243 games (or so)... and therefore, I would expect him to do that twice in approximately 500 racks.

Did I say it right this time?

You said it right, stated like that the first part is correct.

If he were to start right now, and play 243 games, if his average is that he does a 5 pack every 243 games, there is a 66% chance he would do it. If he were to play 100,000 games, he would average doing it once every 243, but for any 243 game stretch, there would be a 66% chance that he would do it once. Every time he does it sooner than 243 games, there will be a longer stretch than 243 to balance out his average. Does that make sense?

The 66% comes from a bell curve. Things that follow a standard bell curve, such as statistics from repeating the same activity with different outcomes, have a 2/3rds chance of occuring within the average interval.

I may be saying this wrong as well Statistics has been a while for me too.

Cheers,
RC

 

[cigardave]

You said it right, stated like that the first part is correct.

If he were to start right now, and play 243 games, if his average is that he does a 5 pack every 243 games, there is a 66% chance he would do it. If he were to play 100,000 games, he would average doing it once every 243, but for any 243 game stretch, there would be a 66% chance that he would do it once. Every time he does it sooner than 243 games, there will be a longer stretch than 243 to balance out his average. Does that make sense?

The 66% comes from a bell curve. Things that follow a standard bell curve, such as statistics from repeating the same activity with different outcomes, have a 2/3rds chance of occuring within the average interval.

I may be saying this wrong as well Statistics has been a while for me too.

Cheers,
RC

Thanks... now I understand where the 66% comes from... but I'm not convinced it applies here.

Seems like we have either a GO (strung 5 together)... or a NO GO (did not string 5 together) type of situation.

In my experience as an engineer, normal distribution is associated with a measure of the variance that a single process produces... such as machining a metal part... and then measuring a critical dimension of that machined part. A graphical representation of the measurements of a number of the machined parts represents a bell curve.

I can't think of how a bell curve would apply to the 5-pk situation.

Pls explain further if you believe that it does.

 

[jnav447]

Found something harder than running 10 racks of 9-ball: getting math people to agree on anything.

 

[zeeder]

Found something harder than running 10 racks of 9-ball: getting math people to agree on anything.

I have to say that I am currently in my first statistics class and we are currently talking about probabilities. The problem is that when you have odds of something occuring 5 consecutive times as 1 in X with X being the number of attempts and each attempt would traditionally be 5 single trials. That being said, as cigardave and others have mentioned, someone with a goal of running a 5-pack isn't going to continue any particular attempt once they fail to BnR. I am not sure how this affects the odds.

The probability, on the other hand remains the same, and the probability takes into account a 5 trial attempt. The probability of Earl running 5 consecutive racks, given a probability of a BnR being .327, is 0.003739 for any given 5 rack attempt. This basically means that if Earl attempts 5 racks there is a .3739% chance that it'll be a 5 pack.

I've asked my stat prof for an explanation of how the various circumstances affect the odds of the desired outcome, i.e. a 5-pack. So I'll find out if I have a screw loose and just don't understand the concepts involved.