Thursday, March 4, 2010

Bowler Puzzler VI: The Streak

To this point, I haven't posted a puzzler that my usual crowd hasn't been able to solve. The master solvers of my tiny audience are MaddysDaddy, HCLC Doc, and KafkaStoleMyBike. I predict that this will be a challenge for all of you to solve. Prove me wrong!

In the 2005-06 season, Walter Ray Williams, Jr finished with a 100% single pin conversion rate. Impressively, he converted 475 spares in 475 chances. This is an amazing streak by any measure. What is certain is that if he had more chances, eventually he would have missed one. In the following season, he did. So... even though his stats show a 100% conversion rate, in actuality, given enough opportunities to shoot at single pin spares, say 10,000, he would have missed some. So, we can assume that the probability of Walter Ray picking up a single pin spare was not 100%, but it was very close to it. Let's assume that in the 05-06 season, Walter Ray had a 99% chance of converting any given single pin spare. With that assumption:

What is the probability of not converting at least one single pin spare in 475 chances if you have a 99% chance of converting each spare?

[Kafka, you are on the hot seat because you took a class of mine in which I taught you how to solve problems like this one.]

25 comments:

  1. Really, now, Todd. Go back through your emails from me-- the last one is germaine. This post is a case in point! ;)

    I have no idea, of course. But this factoid is just one of the reasons why Walter Ray is my bowling idol.

    [By the way, SOME of your ponderances have taken hold on me. For one, I DO believe (in spooks!) that that first frame avg. stat speaks volumes...]

    ReplyDelete
  2. Do you mean the first ball average? If so, I agree that it is very telling. Ken Pomeroy (of KenPom.com) fame actually suggested it.

    ReplyDelete
  3. No, I meant first frame... Isn't that a Toddian theory? Maybe I made it up mistakenly somewhere along the way and thought it was your idea. Anyway, you should do some first frame numbers crunching.

    I guarantee you that for bowlers like me, the first frame portends the outcome (most of the time-- I have been known to have three opens in a row and then mark out the rest for 160-170 game). For pros, of course, this doesn't mean a thing.

    Or does it? ;)

    ReplyDelete
  4. That's interesting, and you are right. I should look at that. I find that I am not really warmed up until about seven frames in. Accordingly, I often start with an open frame. If I strike, it's usually a brooklyn.

    ReplyDelete
  5. I don't even know where to start but I'd like to see how this is solved. Doc, Kafka it's on you guys.

    ReplyDelete
  6. Well, if he has a 99% chance to make any single pin spare, then he has a 1% chance of missing. So to miss 475 in a row, it looks like the answer wants to be .01 to the 475th power. I don't have enough toes and fingers to get to the number but its a lot of zero's.

    ReplyDelete
  7. Well, you are on the right track, but you're not quite there yet. You are correct that this calculation will give you the probability of missing 475 straight, but the question is what is the probability of missing at least one?

    ReplyDelete
  8. I was never any good at word problems - too tricky.

    ReplyDelete
  9. You are actually really close, Doc. Here's a hint. We can simplify this problem to two mutually exclusive outcomes that must add up to 100%. Outcome 1: You knock down all 475 spares; Outcome 2: You miss at least one.

    ReplyDelete
  10. I hope your hint helps Mick. It does nothing for me. (My answer, by the by, is 1 in 100.)

    ReplyDelete
  11. EB, I always appreciate your willingness to guess. Intuition can take you a long way. In this instance, you are incorrect, but I laughed at your guess to my previous query: π!

    ReplyDelete
  12. I was going to guess pi (moose turd this time), but in order for me to get the answer, I have to understand the question.

    Or so I'm told.

    ReplyDelete
  13. And WHERE is Kafka? (probably needs a reboot...)

    ReplyDelete
  14. EB, I'm not going to accept that answer unless you show me your work. How did you come up with that?

    ReplyDelete
  15. On a related, but different topic (!), have you heard about the hoo-ha in Congress to pass legislation authorizing Reagan's mug on the $50 bill? I think that portrait of WRW (he sure "strikes" a nice pose, eh?) posted here would fit quite nicely on those "spare" bills everyone always asks for ("Hey, you got any spare ones?", etc.)... Now, if I only knew who my congressman was... (This particular post of yours,Todd, has posed many a tough nut to crack...)

    ReplyDelete
  16. You gave me the answer in your question, silly.

    (the team doctor's right, I'm not just another pretty face)

    Wouldn't every pin ride on that 99%? YOU HAVE A 99% CHANCE OF CONVERTING EACH SPARE.

    Or so I'm told. ;)

    ReplyDelete
  17. Yeah, I saw it on the news last night. Not to get too political on ya, but for a guy who started the whole borrow billions of dollars from china and saudi arabia thing, it just seems ironic that he should go on money. You want Walter Ray on the money? That's funny. I'd go for MLK before I considered anyone else.

    ReplyDelete
  18. Ok. I am wasting too much time on this today. I need to take the Green Lady out for a spin. Here's how its solved:

    Like I said about 42 comments before:

    There are two mutually exclusive outcomes that must add up to 100%. Outcome 1: You knock down all 475 spares; Outcome 2: You miss at least one.

    It is easy to solve for the probability of getting 475 in a row. It's 0.99 to the 475th power. That comes to 0.0084. This means that the chance of hitting 475 straight if you have a 99% chance at each is actually very low. It is less than 1%, and precisely 0.84%.

    The chance of missing at least one is simply 1-0.0084, or 99.16%.

    The whole point of this exercise is to point out that in order for Walter Ray to have pulled this off, his chances of getting each spare were probably well over 99%, probably in the neighborhood of 99.9%

    ReplyDelete
  19. It still sounds to me like a bowler who has a 99% chance of hitting 475 single spares has a 99% chance of hitting 475 single spares.

    Remember, I'm simple.

    ReplyDelete
  20. 20 comments! A new record. EB, your crazy logic is very incorrect, but by a miracle, you arrived at the right answer. As much as I want to take off 5 pts, I can't. Off to torture some 10 pins!

    ReplyDelete
  21. What is the probability of not converting at least one single pin spare in 475 chances if you have a 99% chance of converting each spare? -- I finally got it. It all depends on how you emphasize not. I read it so that if you don't convert at least one spare, then you haven't converted any. It's a glass half full or half empty kind of thing. Ought to be worth partial credit.

    ReplyDelete
  22. You get partial credit. You got closer than anyone else, except for EB and his lucky guess.

    ReplyDelete
  23. Lucky guess? When I said 1 out of 100 (not being converted) that was a 99% conversion probability. It's what I meant anyway, which is why I came back with "99%!" when you said I was wrong. True, I didn't break out my abacus or take off my socks, but at least I came to understand the problem (for once).

    Crazy logic? It's funny you should say that-- I have really only 3 heroes, one of which you know already-- WRW, another is Captain Kirk. Sometimes I have to ask myself "WWCKD?"

    ReplyDelete
  24. Damnit! I was on vacation, and didn't even bring my phone. Imagine my heartbreak upon returning to a problem I could probably solve!

    BTW: I need a WWCKD bracelet :)

    ReplyDelete

Note: Only a member of this blog may post a comment.