Amarillo Slim...

skp · #1 07-04-2003, 12:05 PM

...in a world full of fat people..heh...catchy title.

Very entertaining book. I was in stitches reading about the Taiwanese world ping pong champ who had practiced for months with a skillet for a game against Slim only to have Slim change the paddles right before game time to coke bottles. Slim lets him serve first, the guy never manages to get the ball over the net and it's Love 5 before the game even starts. LOL

Not much strategy in the book and in fact there were some glaring errors. At one point, it is said that "you'll get pocket Aces once in every 425 deals". Either I have been extremley lucky or they have got this one wrong :-)

At another point, Slim places a bet with Evel Knievel that out of a random group of 30 fellas, two of them will have the same birthday. They find 30 guys in a cafe or something and this holds true. Evel gets suspicious. So, Slim rounds up 30 cabbies at random. Sure enough, two of them also turn out to share the same birthday. Slim goes on to explain in the book that the chances of two guys out of a group of 30 having the same birthday is around 70%. I have read about this one before (I think in the Thusday Night Poker book) and I thought that the chances were in the magnitude of 95%.

Anyone know the correct answer?

Jimbo · #2 07-04-2003, 01:31 PM

Hi SKP I liked the book as well. I've met Slim many times on the pool circuit but never knew his reputaion as a Titanic Thompson clone. Too bad I never got to play him ping pong, I had a few hustles of my own in that game. [img]/forums/images/icons/smile.gif[/img] I think BruceZ computed the correct probability of the Birthday question in the Probablity Forum some time ago.

Al Mirpuri · #3 07-04-2003, 04:29 PM

QMC · #4 07-05-2003, 04:55 PM

I really enjoyed the book as well. I thought it was just entertaining though. I was not looking for an educational experience. I do not think I would like to live such a colorful life as Slim. But I sure enjoyed reading about it!

Robk · #5 07-07-2003, 10:29 AM

Hey skp, the birthday problem is not as tricky as it seems. Say you have a sample of n people. Take one at random. The probability that the next has a different birthday is 364/365. But now consider the third person. The probability that he has a birthday different from both of the first two is 363/365. So the probability of all three having distinct birthdays is 364*363/365*365. In general the probability that n people do not share a birthday is ** (364*363*...*365-n+1)/(365^n-1) = 365!/(365-n)!*365^n-1*365 (note that the product in the numerator started with 364, not 365) =

365!/(365-n)!*365^n. So the probability that (at least) two people out of n do share a birthday is

1 - 365!/(365-n)!*365^n. This is the neat answer, but your calculator isn't going to like 365!. Letting n = 30, the probability (using the formula **) is 1 - 364*...*336/365^29 = 1 - .293 = .706. So it looks like Slim was right about this one.

skp · #6 07-07-2003, 02:36 PM

thanks!

Kurn, son of Mogh · #7 07-09-2003, 03:12 PM

The problem isn't tricky, but it's not intuitive to people who don't understand probability. I think you get to even money at about 23 people, and 97% at about 50. I can't tell you how many times I recouped weekend spending money in college making this bet at a party. I've had people offer me 5-1 at a party with 40 people. Ha!!