What Feat Is Harder

What feat is the hardest

  • 300 in bowling

    Votes: 3 3.8%
  • hole in one

    Votes: 12 15.0%
  • Break and run 10 racks of 9 ball

    Votes: 60 75.0%
  • 150 run in straight pool

    Votes: 5 6.3%

  • Total voters
    80
Running 10 racks of 9 Ball is the toughest, HANDS DOWN. I've never run 150 in SP, but I've rarely played the game. I think I'm capable of it with practice though. I've never had a hole in one before either, but I've been close. 216 yard par 3 at Boulder City near Las Vegas. 6 inches from the hole. I've bowled many 300 games and would make a small wager right now that if I could bowl, say 20 games worth of frames, I could throw 12 strikes in a row somewhere along the line. Not a 300 game, 12 strikes in a row. Maybe I would, maybe I wouldn't, but I do know that a 300 game isn't that tough. A guy that averages 160 could do it. So, 10 racks, hole in one, 150 SP, then the 300. JMO though. Peace, John.
 
!0 Racks in a row hands down....

The best I've ever done is eight in a row and that is with two nine's on the break.... I've bowled a 224 and that was with hitting 9 strikes, so I could see hitting three more strikes pretty easily...I've never made a hole in one before... I have made a double eagle though once.... but I golf like in the 90's....I've never hit 150 in SP but I'm not a SP player.... I probably could if I started practicing SP.....

So I would definitely say that running ten straight racks of nineball is the toughest of those....
 
TATE said:
Run 10 racks of 9 ball: If a player had a break and run percentage of 20%, I would guess a pretty average number for a pro on a table with average pockets, I calculate the odds of running 10 racks at over 1 in 1,562.000.

I am not a math expert and it's been a long time since I took statistics class, but running 4 (1 in 500), is a cake walk compared to running 10 racks in any particular sitting.

Yes, it's nearly impossible, but we know it has happened.
If anyone is is familiar with odds making, please check the calculations.

Chris

I believe that the odds of sucessfully connecting 10 straight successes, each one considered a 20% proposition, would be 1 in 10,000,000.

Mathematically, it be multiplying 0.2 times itself 10 times.

IMO, the 20% estimation may be a bit low.

Trying the same calculation for 25% yields 4 in 1,000,000.... fwiw.
 
cigardave said:
I believe that the odds of sucessfully connecting 10 straight successes, each one considered a 20% proposition, would be 1 in 10,000,000.

Mathematically, it be multiplying 0.2 times itself 10 times.

IMO, the 20% estimation may be a bit low.

Trying the same calculation for 25% yields 4 in 1,000,000.... fwiw.
I am reading all these odds, and I have to ask...

Don't these odds need to be viewed toward a particular person and their abilities? Or are these more of a random average pro? I would think these odds would get dramatical smaller, if you chose a few name players.
 
I got THREE HOLE in One Last Weekend in 18 Holes @ Castle & Coaster Miniature Golf in PHX Course # 2. Does that count???
 
I would have to go with the 10 racks of nine ball as well, after watching a blind man hit a hole in one on sportscentre two years ago it became very apparent that anyone is capable of a hole in one, the same can't be said about running 10 racks of nine ball. The list of players that have bowled a 300 game is very long and it is a feat that is commonly done. A 147 maximum in snooker has been mentioned and this has to be right up there with the 10 racks of nine ball, it is something that is rarely done and requires a tremendous amount of skill to accomplish.
 
Donovan said:
I am reading all these odds, and I have to ask...

Don't these odds need to be viewed toward a particular person and their abilities? Or are these more of a random average pro? I would think these odds would get dramatical smaller, if you chose a few name players.

Yes, it would vary by quite a bit even among the pros, but still it would be ridiculously difficult.

Also, the equipment would be a big factor. A bar table with large pockets would make matters a lot easier. Still, the odds of running a 10 pack, no matter who and no matter what the conditions, may at best be equal to coin flipping 10 heads in a row.



Chris
 
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It's absolutely inconceivable for a mediocre player to run 10 racks (I suppose it's possible they could break a bunch of 9-balls in, but that isn't "running 10-racks" in my book). However, mediocre golfers can, and do, make hole-in-one's.
 
TATE said:
Yes, it would vary by quite a bit even among the pros, but still it would be ridiculously difficult.

Also, the equipment would be a big factor. A bar table with large pockets would make matters a lot easier. Still, the odds of running a 10 pack, no matter who and no matter what the conditions, may at best be equal to coin flipping 10 heads in a row.

Chris

Chris - Your analogy to flipping a coin equates to a 50% chance of a BnR in 9-Ball. That would be pretty high!

IIRC, from Phil Capelle's book... Play Your Best 9-Ball... and from his analysis of over 100 pro matches, I believe the BnR percentage for the pros in these events was approximately 30%. (Or it could have been that those with the highest percentage were just above 30%... I'll have to look it up tonite when I get home.)

Using 30% as the odds of an individual BnR, the odds of stringing 10 of these together would be 6 in 100,000.
 
cigardave said:
Chris - Your analogy to flipping a coin equates to a 50% chance of a BnR in 9-Ball. That would be pretty high!

IIRC, from Phil Capelle's book... Play Your Best 9-Ball... and from his analysis of over 100 pro matches, I believe the BnR percentage for the pros in these events was approximately 30%.

Using 30% as the odds of an individual BnR, the odds of stringing 10 of these together would be 6 in 100,000.

That is the commonly accepted stat. I was thinking champion player on a bar box might bring the odds down quite a bit.

I believe that with the trend toward tighter pockets, a 20% - 25% for the average pro and 30% for the Archer's, Bustamante power player type whould be more accurate. It is amazing how these small percentage differences greatly affect the odds of running 10 in a row. The smaller pockets make it more difficult to make a ball on the break, which is the main factor.

Chris
 
cigardave said:
I believe that the odds of sucessfully connecting 10 straight successes, each one considered a 20% proposition, would be 1 in 10,000,000.

Mathematically, it be multiplying 0.2 times itself 10 times.

IMO, the 20% estimation may be a bit low.

Trying the same calculation for 25% yields 4 in 1,000,000.... fwiw.
I would think that the probability factor of 20% would become much lower with each successive rack. However, I am no math major either so maybe my logic is flawed.
 
10 racks of 9-ball

I've had 4 hole-in-ones in golf. - More luck than skill. I've hit a lot of great golf shots, but none of the aces was a good shot. Tiger has 19 I believe.
I don't know of any pro golfers that haven't had a hole-in-one.

I've had 6 strikes in a row twice in bowling, and I was never more than an occassional bowler.
I don't know of any pro bowlers that haven't had a 300 game. I have quite a few amateur friends who have had multiple 300 games.

I have never really played straight pool. But I don't think this feat is on par with 10 racks of 9-ball. I have heard a lot of stories of people running 150 and out in tourneys, so I think that most pros have done this.

10 racks - I have done a few 6s and a lot of 3s and 4s. I don't know many pros who have done 10, at least in competition. In fact, I can only think of one, Earl strickland. Danny Medina and somebody else had the record for a while with 9.

Cheers,
RC
 
cigardave said:
Chris - Your analogy to flipping a coin equates to a 50% chance of a BnR in 9-Ball. That would be pretty high!

IIRC, from Phil Capelle's book... Play Your Best 9-Ball... and from his analysis of over 100 pro matches, I believe the BnR percentage for the pros in these events was approximately 30%. (Or it could have been that those with the highest percentage were just above 30%... I'll have to look it up tonite when I get home.)

Using 30% as the odds of an individual BnR, the odds of stringing 10 of these together would be 6 in 100,000.
So I looked it up and here's what I found:

Capelle studied 500 games played by pro players during the '90s and he states in his book that the BnR percentage of those 500 games was 28% (or 140 of the 500 games).

And he also states that Earl and Johnny Archer were the players with the best tournament records during that decade... and that their BnR percentages far exceeded the average pro... their percentages were32.7% and 31.7%, respectively.

Using Earl's %, the probability of Earl accomplishing the feat (during the '90s) was 1.4 in 100,000.

Using the 28% for the average pro, the odds would be 3 in 1,000,000.
 
JDB said:
I would think that the probability factor of 20% would become much lower with each successive rack. However, I am no math major either so maybe my logic is flawed.
Your logic is flawed JDB. The probability of any single independant event is independant of the outcome of the event that preceded it.

An example... the odds of a coin flipping up heads is 50%. Regardless of how many times you flip the coin, the odds remain the same that it will come up heads on any one toss (50%). Obviously, the odds that it will continue to come up heads time after time again will get increasingly less likely as the number of flips increases... but the single event odds (50%) never change.

Mathematically, in order to determine the odds (or liklihood) of a series of independant events each having the same outcome, you multiply the individual event probability times itself the number of times that equates to the string that you're evaluating.

For example, the odds of Earl breaking and running two consecutive 9-ball racks in the '90s was 1 in 10 (.327 x .327 = 0.107 or approximately 1:10). The odds of Earl stringing three together would have been 3.5 in 100 (.327 x .327 x .327 = 0.035).

Seems about right, don't you think?
 
cigardave said:
Your logic is flawed JDB. The probability of any single independant event is independant of the outcome of the event that preceded it.

An example... the odds of a coin flipping up heads is 50%. Regardless of how many times you flip the coin, the odds remain the same that it will come up heads on any one toss (50%). Obviously, the odds that it will continue to come up heads time after time again will get increasingly less likely as the number of flips increases... but the single event odds (50%) never change.

Mathematically, in order to determine the odds (or liklihood) of a series of independant events each having the same outcome, you multiply the individual event probability times itself the number of times that equates to the string that you're evaluating.

For example, the odds of Earl breaking and running two consecutive 9-ball racks in the '90s was 1 in 10 (.327 x .327 = 0.107 or approximately 1:10). The odds of Earl stringing three together would have been 3.5 in 100 (.327 x .327 x .327 = 0.035).

Seems about right, don't you think?
But these are not independent events because they must happen in succession in order for the "one" event to be performed. Once Earl runs his first rack, his probability of running the next rack is not 0.327, it is .320. If he runs two racks his probability of running a 3rd rack is .313.

I am calculating this as follows:

Earl's run percentage is 32.7 racks out of every 100 racks
after Earl has run one rack he now will run 31.7 racks of the next 99, which lowers his probability to the 0.320. As he runs each successive rack his probability goes down because they are happening in succession and are not independent events.

I am just theorizing because I have not worked with stats in a while so, again, my logic may be flawed. However, I really don't believe that you can count each rack as an independent event when they must be done in succession to achieve the ultimate feat.

I should probably dig out my statistics books but this is an interesting conversation whether my idea has merit or not.
 
JDB said:
But these are not independent events because they must happen in succession in order for the "one" event to be performed. Once Earl runs his first rack, his probability of running the next rack is not 0.327, it is .320. If he runs two racks his probability of running a 3rd rack is .313.

I am calculating this as follows:

Earl's run percentage is 32.7 racks out of every 100 racks
after Earl has run one rack he now will run 31.7 racks of the next 99, which lowers his probability to the 0.320. As he runs each successive rack his probability goes down because they are happening in succession and are not independent events.

I am just theorizing because I have not worked with stats in a while so, again, my logic may be flawed. However, I really don't believe that you can count each rack as an independent event when they must be done in succession to achieve the ultimate feat.

I should probably dig out my statistics books but this is an interesting conversation whether my idea has merit or not.

Standing at the table ready to break, the probability of his breaking and running does not change. The probability that he will do it X times in a row is described by the probability of him doing it once to the Xth power. Taking your numbers, it would be .327^10. or .00001397 chance (.001397%) or 1 in 71,535.6 times roughly.

This is purely the math and assumes that he is a machine. If he is actually working on 6 or 7, the numbers might change for or against him because of his psychology.

The way you're looking at it is interesting, but flawed because it tries to make a connection between past outcomes and the present outcome. This is something that we (humans) like to do but which isn't mathematically sound.

Cheers,
RC
 
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cigardave said:
So I looked it up and here's what I found:

Capelle studied 500 games played by pro players during the '90s and he states in his book that the BnR percentage of those 500 games was 28% (or 140 of the 500 games).

And he also states that Earl and Johnny Archer were the players with the best tournament records during that decade... and that their BnR percentages far exceeded the average pro... their percentages were32.7% and 31.7%, respectively.

Using Earl's %, the probability of Earl accomplishing the feat (during the '90s) was 1.4 in 100,000.

Using the 28% for the average pro, the odds would be 3 in 1,000,000.

This is great information. Clearly, running 10 racks of 9 ball would be a rare feat.

I would like to point out that this would equal 1 in 333,000 attempts. Assuming a pro spent 2 hours each day trying to accomplish the task, and each attempt averaged 10 minutes, they could be expected to perform the feat once every 76 years.

I don't think there's any question that, for the odds, this would be the most difficult feat.

Chris
 
sixpack said:
Standing at the table ready to break, the probability of his breaking and running does not change. The probability that he will do it X times in a row is described by the probability of him doing it once to the Xth power. Taking your numbers, it would be .327^10. or .00001397 chance (.001397%) or 1 in 71,535.6 times roughly.

This is purely the math and assumes that he is a machine. If he is actually working on 6 or 7, the numbers might change for or against him because of his psychology.

The way you're looking at it is interesting, but flawed because it tries to make a connection between past outcomes and the present outcome. This is something that we (humans) like to do but which isn't mathematically sound.

Cheers,
RC
Thanks for the explanation. I stand corrected. I guess I need to pull back out my stats booked that have been laying around for years as I was going purely on theoretical instinct and not necessarily memory of stats.

However, it was an interesting discussion anyway.

Good job Dave and RC.
 
there's one big problem with these statistical calcs....

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.
 
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