Monday, May 23, 2011

The odds of three

I had some time on the commuter train this past week, and got around to reading through the 9-Ball section of Phil Capelle's Play Your Best Pool. Right in keeping with my nostalgic look back on the thrill of running 3-balls in my last post, Phil offers this analysis of the odds of running the last 3 balls on the table.

"You'd be surprised at how often the majority of pool players fail to negotiate the all important last three balls. One reason is simply the numerical odds of pocketing three balls in a row....Once you can regularly get out from the 7-ball, you'll be beating a lot of players that you thought were pretty good up until now."

So here's how he breaks down the math:

  • If you make 70% of your shots, you have a 1 in 3 chance (34%)
  • If you make 80% of your shots, you have a 1 in 2 chance (51.8%)
  • If you make 90% of your shots, you get out most of the time, but still dog it in about 1 in four tries (73%)
These figures are a little depressing. Its pretty startling to think if I'm looking at an easy out with shots that I miss one out of ten tries (and that's including getting the right position in many cases), the stats say, I blow it a quarter of the time. That doesn't seem right, but if I was honest about it, it's probably true.

I'm not sure I really want this going through my head as I walk up to the table after my opponent has missed the seven. But, I guess its good inspiration to remember to never take any shot for granted, and bear down on the last three. It would be nice to be the kind of player who beats their average if its the three that can win you the game.


p00lriah. said...

say if you make your shots 70% of the time, and there are three balls left. if you just focus on each shot, and there's a 70% chance of making each shot, wouldn't your chance of running the three balls still be 70%?

poolminnow said...

Alas, no, each shot is a separate event dependent on success of the other events. You have a 70% chance of making each shot, but of making 3 in a row, the odds go down. If you think of it this way, there's a 30% chance you'll miss on your first shot. Once you miss on that first shot, your odds of missing balls 2 and 3 is now 100% instead of 30%. Then add to that the 30% of the times you'll miss on the 2nd ball, and the third.

poolminnow said...

Actually, I should say "combine" that with instead of "add."

p00lriah. said...

i just had a fascinating conversation with my math friend. he basically told me that mr. capelle's book and i are both wrong. i'm wrong because of my assumption that each shot is an independent event. mr. capelle is wrong because in order for his math to work, you have to assume a whole bunch of factors do not exist. for one thing, the success rate of each shot changes depending on the type of shot (a hanger, masse, bank, and so on). of course, the layout affects the percentage as well.

i guess like many things, the truth lies somewhere in between.

Michael Reddick said...

The math in the example is a generalization and not completely precise (lots of assumptions), but the concept is correct. Most people do not have an accurate perception of their true capabilities. That being said, this just emphasizes the importance of being able to recognize when you are in trouble, and instead of trying to make the tough shot, play a good safety. Let your opponent take the tough shot. He or she might make the difficult shot, but the odds will be in your favor to win the game.

Barneyblue18 said...

Wait if you get 100% of a full shot doesn't the second ball give you 40,the third gives you 75?

poolminnow said...

@BarneyBlue18 - I stand corrected, if you are a troll, then that would be the correct calculation. Thanks for your comment.

Barneyblue18 said...

Wow I wonder how you put it in division.