Shane Has Won World Titles

[ineedaspot]

that's not what the people saying that Shane is the best are saying! It absolutely amazes me that people still don't understand simple logic and math. The shorter the race, the more chance a worse player has to win. The longer the race, the less chance the worse player has. How is this still not registering?

The fact that these world championships are races to 7, allows the worse player to have more of a statistical chance at winning. Hence, a very high variance. A much longer race, favors the better player, hence a low variance. If the world championships were races to 20 then Shane would win almost all of them at the moment because he is the best in the world.

What makes you so sure that Shane is actually the best player in the world? Yeah, races are too short in tournaments (although i don't think WPC is race to 7), and there aren't enough total tournaments, so the only defensible conclusion I see is that there's not enough data to tell.

In tournaments with all the world's best players, Shane doesn't have the best record. He might still be the best player, due to variance as you say. But so might about 20 other guys.

 

[corvette1340]

What makes you so sure that Shane is actually the best player in the world? Yeah, races are too short in tournaments (although i don't think WPC is race to 7), and there aren't enough total tournaments, so the only defensible conclusion I see is that there's not enough data to tell.

In tournaments with all the world's best players, Shane doesn't have the best record. He might still be the best player, due to variance as you say. But so might about 20 other guys.

the fact that he has made it known that he will take on anyone in the world in a long race (spread out over several days so it's not a question of endurance as some of these braniacs keep saying) and nobody will play tells me all I need to know and lets you know what the players think. Any of these so called world champions could make more than they'll make in the next 5 years if they could get there.

 

[De420MadHatter]

Good God these people from across the ocean are dense!!! Makes for a lively thread though ;-).

 

[BRussell]

The answer is B, but it's close.
If the better player has a 51% chance of winning a rack, that player will win:
Race to 50: 57.9% of the time
Race to 10: 53.5% of the time
Race to 5 races to 10: 58.6% of the time

If the better player has a 55% chance of winning a rack, that player will win:
Race to 50: 84.1% of the time
Race to 10: 67.1% of the time
Race to 5 races to 10: 86.2% of the time

Thanks. Can you tell me how you got those numbers?

 

[ineedaspot]

Thanks. Can you tell me how you got those numbers?

The numbers can be calculated using the binomial distribution, which describes the outcome of a sum of independent trials with the same probability. Here's the wikipedia page about it.

http://en.wikipedia.org/wiki/Binomial_distribution

 

[corvette1340]

The numbers can be calculated using the binomial distribution, which describes the outcome of a sum of independent trials with the same probability. Here's the wikipedia page about it.

http://en.wikipedia.org/wiki/Binomial_distribution

to take it a step further you'd have to account for the individual percentages when a player is breaking opposed to when their opponent is breaking. This would help explain why alternate break format helps to further level the playing field.

Ie...If Shane is a 60% favorite over player X when he is breaking and only a 52% favorite when player X is breaking, then by default, player X is going to have a much better chance at winning a short race because he's guaranteed to have the same number or one less breaks than Shane does. If however, it's winner breaks, then Shane has a bigger advantage because he's favored to get the break more often than not and when he does have the break he becomes a bigger favorite to keep the break.

It's not rocket science folks.

 

[ineedaspot]

the fact that he has made it known that he will take on anyone in the world in a long race (spread out over several days so it's not a question of endurance as some of these braniacs keep saying) and nobody will play tells me all I need to know and lets you know what the players think. Any of these so called world champions could make more than they'll make in the next 5 years if they could get there.

That means pretty much nothing. There are any number of reasons a matchup like that might not happen. A lot of Euro players don't gamble. Some players might be risk-averse. Some might not have backers. And then there's problems with negotiations, etc. For an Asian player to come to the US just to play Shane, that requires a flight and expenses. If there's a backer involved, then the backer is already laying odds because the player gets paid out of the winnings. So the expected chances of winning have to be well over 50% in order to make it a profitable venture.

What determines the best player is actual play, not woofing.

If Ko and Shane race to 100, I'd bet on Ko. Not because I'm sure that Ko is the better player, but because the evidence (limited as it is) suggests it's more likely that he would win.

 

[ineedaspot]

to take it a step further you'd have to account for the individual percentages when a player is breaking opposed to when their opponent is breaking. This would help explain why alternate break format helps to further level the playing field.

Ie...If Shane is a 60% favorite over player X when he is breaking and only a 52% favorite when player X is breaking, then by default, player X is going to have a much better chance at winning a short race because he's guaranteed to have the same number or one less breaks than Shane does. If however, it's winner breaks, then Shane has a bigger advantage because he's favored to get the break more often than not and when he does have the break he becomes a bigger favorite to keep the break.

It's not rocket science folks.

Actually, mathematically, alternate break versus winner break makes no difference whatsoever. (This is something you should know if you're going to be going around insulting people about not knowing math). I ran some numbers with different winning percentages based on breaking and not breaking in this post.

http://forums.azbilliards.com/showthread.php?p=4934404#post4934404

The reason alternate vs winner break makes no difference is because either way, if a match goes hill-hill, the winner of the lag gets n breaks and the other guy gets n-1. If a match doesn't go hill hill, then one or both of the players will have had less than their maximum number of breaks, but the rest of the games don't make any difference because one player has already won enough games to take a set. Winner vs alternate breaks only changes the order that the racks are played in, not who wins.

 

[smashmouth]

for a country with "pool schools" and more than a billion in population, your statement is statistically uncertain. that's only for China. Im not even mentioning Taipei.

about as uncertain as suggesting that there is no better basketball player than Lebron or faster runner than Usain

 

[corvette1340]

Actually, mathematically, alternate break versus winner break makes no difference whatsoever. (This is something you should know if you're going to be going around insulting people about not knowing math). I ran some numbers with different winning percentages based on breaking and not breaking in this post.

http://forums.azbilliards.com/showthread.php?p=4934404#post4934404

The reason alternate vs winner break makes no difference is because either way, if a match goes hill-hill, the winner of the lag gets n breaks and the other guy gets n-1. If a match doesn't go hill hill, then one or both of the players will have had less than their maximum number of breaks, but the rest of the games don't make any difference because one player has already won enough games to take a set. Winner vs alternate breaks only changes the order that the racks are played in, not who wins.

your numbers and theory on it not making any difference is false. A person with a better than average break will always have a greater % chance of winning with the break. Therefore, a player with a superior break will always prefer winner breaks. A player with a lesser break stands a much greater chance at winning in an alternate break format. I want some of what you've been smoking.

 

[corvette1340]

Actually, mathematically, alternate break versus winner break makes no difference whatsoever. (This is something you should know if you're going to be going around insulting people about not knowing math). I ran some numbers with different winning percentages based on breaking and not breaking in this post.

http://forums.azbilliards.com/showthread.php?p=4934404#post4934404

The reason alternate vs winner break makes no difference is because either way, if a match goes hill-hill, the winner of the lag gets n breaks and the other guy gets n-1. If a match doesn't go hill hill, then one or both of the players will have had less than their maximum number of breaks, but the rest of the games don't make any difference because one player has already won enough games to take a set. Winner vs alternate breaks only changes the order that the racks are played in, not who wins.

your model gives zero credence to the mental factor of playing against an opponent who has a good break in a winner break format or the reverse being true in an alternate break format. It also doesn't account for momentum which is huge in pool given the many gears that champions can catch especially in a winner break format.

 

[one stroke]

Actually, mathematically, alternate break versus winner break makes no difference whatsoever. (This is something you should know if you're going to be going around insulting people about not knowing math). I ran some numbers with different winning percentages based on breaking and not breaking in this post.

http://forums.azbilliards.com/showthread.php?p=4934404#post4934404

The reason alternate vs winner break makes no difference is because either way, if a match goes hill-hill, the winner of the lag gets n breaks and the other guy gets n-1. If a match doesn't go hill hill, then one or both of the players will have had less than their maximum number of breaks, but the rest of the games don't make any difference because one player has already won enough games to take a set. Winner vs alternate breaks only changes the order that the racks are played in, not who wins.

You might as well be talking to the wall ,,

1

 

[corvette1340]

You might as well be talking to the wall ,,

1

I'd venture to say that you're the person with the least brain matter in all of this thread and especially in this conversation. I've got a degree in Economics and the worlds largest online pawn business. What do you have besides a keyboard in your mom's basement?

 

[ineedaspot]

your numbers and theory on it not making any difference is false. A person with a better than average break will always have a greater % chance of winning with the break. Therefore, a player with a superior break will always prefer winner breaks. A player with a lesser break stands a much greater chance at winning in an alternate break format. I want some of what you've been smoking.

Not sure what to tell you here. Run the numbers yourself if you want (if you can). In the scenario I calculated, Player A wins 60% of the time on his break, while on player B's break, it's 50% to either player.

What are the odds that A wins a race to 11 in alternate break? How about in winner break?

If you do the math, you will find that they are the same. (If you don't know how to do this, then I don't know why you're barking so loud). Put in other percentages. Again, you'll find that the odds of winning a set either way are the same. Eventually you'll figure out the pattern.

 

[Blue Jam]

that's not what the people saying that Shane is the best are saying! It absolutely amazes me that people still don't understand simple logic and math. The shorter the race, the more chance a worse player has to win. The longer the race, the less chance the worse player has. How is this still not registering?

The fact that these world championships are races to 7, allows the worse player to have more of a statistical chance at winning. Hence, a very high variance. A much longer race, favors the better player, hence a low variance. If the world championships were races to 20 then Shane would win almost all of them at the moment because he is the best in the world.

Got a bee in your bonnet, son?

This thread is about Shane (not) having any world titles, not who you think might be the best player (which is a pointless discussion, try defining best for starters). By any measurable result in history, SVB is not the best at winning world titles. Fact.

As an aside, I don't think it's fair to say that players that are better at handling pressure, that makes better decisions under stress or that is able to focus enough to make fewest errors when each could cost you the match is necessarily a worse player than someone who needs to rely on having a massive number of frames to get over the line.

 

[ineedaspot]

your model gives zero credence to the mental factor of playing against an opponent who has a good break in a winner break format or the reverse being true in an alternate break format. It also doesn't account for momentum which is huge in pool given the many gears that champions can catch especially in a winner break format.

That's correct. It assumes that each rack is independent of the last, and that all that matters is who is breaking.

Momentum and the mental game are difficult to model mathematically. Loosely speaking, winner break favors one kind of mental game (the "momentum player"), whereas alternate break favors another (the "even keel player"). This could actually explain why Shane does better in winner breaks -- maybe he's more of a momentum player than an even keel player.

But this has nothing to do with variance or which format is more fair or whatever. Is being a momentum player necessarily better than being a player that can bear down and shoot every shot to his best ability no matter what the format or the score?

 

[corvette1340]

Not sure what to tell you here. Run the numbers yourself if you want (if you can). In the scenario I calculated, Player A wins 60% of the time on his break, while on player B's break, it's 50% to either player.

What are the odds that A wins a race to 11 in alternate break? How about in winner break?

If you do the math, you will find that they are the same. (If you don't know how to do this, then I don't know why you're barking so loud). Put in other percentages. Again, you'll find that the odds of winning a set either way are the same. Eventually you'll figure out the pattern.

you yourself just said that while in the model, the win % of the independent race was the same for winner break vs. alternate break format, it does suggest that the winning margin will be greater in a winner break format. Therefore, by definition, the winner break format will always produce a higher winning percentage for the superior player because the psychological factor of being able to distance oneself in a particular race is a HUGE factor.

Example, Player X wins 6 out of 10 of his breaks and he only wins 5 out of 10 of his opponents breaks. He's up 11 to 9 in a race to 13 and feeling some heat to win on his next break or his opponent will have an opportunity to tie the match up on their break.

Conversely Player X and Player Y trade the first 10 games and then Player X, who has a huge break, puts a 6 pack on player Y to go up 11 to 5. The psychological factor is that the player with the superior break, if all other aspects are equal between the two, will realize a much greater margin of victory and therefore a much larger win % in a winner break format. It doesn't necessarily have to be a far superior break, it can be superior defensive play, shot making, etc....the only constant is that the superior player can distance himself from the other player thereby greatly increasing their chance at winning.

 

[BJTyler]

Good post (too long to quote though ).

Let me ask you - and I'm not trolling you, I'm honestly not sure - which type of long race do you think would be more fair, meaning more likely for the better player to win:
A. a race to 50
B. a race to win 5 sets of races to 10

Ineedaspot is correct, the better player wins under format A about 57.9% of the time whereas he wins under format B about 58.6%.

However, let's make it slightly more interesting and suggest that each game cost you $1 in table time. It turns out that format A will -on average- cost you about $92 in quarters. Whereas format B will cost you -on average- about $124.

So, we could ask a slightly different question, what is the cheapest match format that yields at least a 58.6% chance of the better player winning?

It turns out that a single race to 60 gets us there (58.7% chance of the better player winning), and on average only costs us a little over $111 in quarters.

If we care about being thrifty with our money (or time), single races (vs. races of races) are the most efficient way to yield the greatest probability of winning for the better player.

I know this is more than you asked, but I thought you might find it interesting.

 

[ineedaspot]

you yourself just said that while in the model, the win % of the independent race was the same for winner break vs. alternate break format, it does suggest that the winning margin will be greater in a winner break format. Therefore, by definition, the winner break format will always produce a higher winning percentage for the superior player because the psychological factor of being able to distance oneself in a particular race is a HUGE factor.

Example, Player X wins 6 out of 10 of his breaks and he only wins 5 out of 10 of his opponents breaks. He's up 11 to 9 in a race to 13 and feeling some heat to win on his next break or his opponent will have an opportunity to tie the match up on their break.

Conversely Player X and Player Y trade the first 10 games and then Player X, who has a huge break, puts a 6 pack on player Y to go up 11 to 5. The psychological factor is that the player with the superior break, if all other aspects are equal between the two, will realize a much greater margin of victory and therefore a much larger win % in a winner break format. It doesn't necessarily have to be a far superior break, it can be superior defensive play, shot making, etc....the only constant is that the superior player can distance himself from the other player thereby greatly increasing their chance at winning.

The average winning margin is larger in winner break format, yes, but it still doesn't change the odds of either player winning.

Speculation about mental game is just that. Honestly, I don't find your amateur psychology much more persuasive than your amateur math. Winner breaks favors one kind of mental game over another. I think alternate break probably requires more mental toughness. In winner break, if you get down, you can always tell yourself, well when I get to the table, I'll run a bunch of racks. In alternating breaks, you don't have that.

Also, in general, the existence of momentum and rules that emphasize it will tend to favor the weaker player, not the stronger. The reason being that the stronger player wants more independence between trials, to let the law of averages take effect, and momentum is in effect a dependency between one trial and the next.

In the end, winner and alternating breaks are equally valid formats mathematically. One format might favor one kind of player over another, but there's no fundamental reason that one is "better". Personally, I prefer to watch winner breaks, but that's just my preference.

 

[ineedaspot]

Ineedaspot is correct, the better player wins under format A about 57.9% of the time whereas he wins under format B about 58.6%.

However, let's make it slightly more interesting and suggest that each game cost you $1 in table time. It turns out that format A will -on average- cost you about $92 in quarters. Whereas format B will cost you -on average- about $124.

So, we could ask a slightly different question, what is the cheapest match format that yields at least a 58.6% chance of the better player winning?

It turns out that a single race to 60 gets us there (58.7% chance of the better player winning), and on average only costs us a little over $111 in quarters.

If we care about being thrifty with our money (or time), single races (vs. races of races) are the most efficient way to yield the greatest probability of winning for the better player.

I know this is more than you asked, but I thought you might find it interesting.

This is a good point. On a per-rack-basis (or per-hour-basis, equivalently), the most efficient way to figure out who is better is just to have one long race.