That's because Probability isn't theoretical. There's only one right answer. I just have no idea what it is! Where's my stats book........

Wha'???

First off, a disclaimer: While I did study statistics in college, Probability always (and still) confused the hell out of me. BUT...

There's always a 50% chance of flipping heads. Always, on every individual flip. However, if your goal is to flip heads twice in a row, I believe the odds go down. I *think* to 25%? If you asked "what are the odds I'd flip this coin 10 times and always get heads" it'd be nearly zero. Not nearly 100%.

If Buffy was in Vegas 9 consecutive days/year, sure, you'd definitely run into her if you were there 357 days/year. But if she was there 9 times for one day each time, I don't think it'd be "guaranteed" any more -- though very close.

Help! [/b][/quote]Another precinct heard from...

The question is "How many flips does it require to guarantee at least one heads?". (You're right about getting two heads in a row being a 25% chance, but that's not the question) The answer is, "You can never guarantee it". You could flip a million times, and there is always the possibility, as infintesimal as it may be, that you get 1 million tails in a row. The more flips, the closer you get to 100%, but never get all the way there.

It's like the old problem "If you want to cross a room, and you go half the remaining distance with each step, you keep getting closer to the other side, but you never get there" The more steps you take, the closer you come to completing 100% of the distance, but you never make it all the way across.

Next.

Whether she's there for 9 consecutive days, or nine individual days, or three times for 3 days each doesn't matter. She's there nine days out of 365 days. She's not there for 356 days. So if I'm there for 357 days, the worst that can happen is that she's not there for 356 of them. But if I'm there for 357 days, there has to be at least one day when we are both there.

I presume you've read the entire discussion...

Oops. Okay, never mind

Put that way, it makes sense now. So, given that, the odds of you meeting her are 100%. I gotta see the original question again

Alright, let me throw a wrench into things by looking at it another way!

By the way, plaw, I think you 4/357 was good. It works for me.

Anyway, here's my thought... Why don't we look at it from the converse... Figure out what the probability of never meeting Buffy on any of the 4 days plaw is there. Knowing that probability, we can subtract it from 1 and that will be the probability of meeting Buffy.

Anyway, say it's plaw's first day at Vegas, he just got off the plane rarin' to go. He checks into his hotel and immediately makes a beeline to the nearest $8/$4 poker table...

The probability of not meeting Buffy the first day is 356/365 (Buffy is not there 356 days in a year).

The second day, plaw is tired from the 4 hours of more poker and has had his fill with the $4 Steak and Lobster Buffet. He decides he wants to watch Celine Dion.

The prob of not meeting Buffy is 355/364 since we know that he didn't meet her the first day.

It's plaw's third day. He's up $2000 and feeling frisky. He heads over to the nearest gentleman's club and proceeds to stay there for 3 hours. He is now down $500, but very, very happy!

The prob of not meeting Buffy: 354/363.

4th day. Plaw wants to recoup his losses from the night before. He heads over to the blackjack table and employs a little basic strategy. Face cards are -1, low cards +1. He's having a good time, splitting face cards, doubling down and before he realizes, he needs to catch a cab for the airport. He tips the waitress a $100 chip and walks away knowing he made $200 that weekend. Sadly, he never met Buffy.

Prob he didn't meet Buffy the 4th day...353/362.

Therefore, the probability of not meeting her in any of the 4 days is a case of conditional probability (e.g. prob of not meeting her 2nd day is dependent on the fact that he didn't meet her the first day; third day dependent on 1st and 2nd day, etc)

The probability of not meeting her then is

356/365 * 355/364 * 354/363 * 353/362 = 90.5%

Therefore, the probability of meeting Buffy on any of the four days is approximatley 10%.

Well, actually that's the probability of both being in Vegas at the same time. Whether they would actually bump into each other is another story! If Plaw is in one casino (or strip club) the entire trip, and Buffy's in another, fuggedaboutit!

Originally posted by plawrence:
Quite, Crabby.

Crabby is quite what? Oh, you mean quiet. :p

3 hours? What's he doing, banging cocktail waitresses two at a time?

First of all, you're right Crabby. I meant "quiet", not "quite". It's just that I was in such a frenzy earlier, I couldn't type straight.

OK. Your last line of reasoning, DB, gets us back to the original 1/40 (x) 4, or 10%. So we've come full circle. I'm still not sure that the reasoning that got us to 19-1 is incorrect though.

I remember from my Prob & Stat course in college, that the problems always involved colored balls in a barrel.

Our problem would be expressed this way:

"A barrel contains 356 white balls and 9 blue balls. What are the chances of picking out one of the blue balls from the barrel if you get four chances to pick?"

There's a formula for this, I just don't remember it.

It would seem to me that on the first pick you have a 9/365 chance of picking a blue ball, or 2.4657%.
On the second pick (since if you got a white ball on the first pick you would discard it, not put it back in the barrel) your chances would be 9/364, or 2.4725%.
Third pick 9/363, or 2.4793%
Fourth pick 9/362, or 2.4861%

Total, 9.9036%

The only thing that still bothers me is an explanation for why the probability is so close to 100% after only 40 picks, when it would take 365 picks to guarantee it 100%.

And what happened to this earlier line of thought, which still seems perfectly sound to me:

"My first day there has a 1/40 chance of being one of the 9 days that she's there already.

But, if she's not there, my second day there has only a 1/120 chance of being one of her first days there of one of her three trips.

And so on, or my third and fourth days there. So, 3/120 + 1/120 + 1/120 + 1/120, or 6/120, or 19-1.

What's the damn formula????

Yo, Yo, Yo - Big Daddy in Da House !!!

It's funny, I was all set to join this and I see that basically for the past two days you and DB spent the better portion of your time and about 40+ posts trying to figure out the odds of meeting Buffy in Vegas. And no hoop talk. WOW all I can say is

You two need to get out more - PL and DB ===

Actually as much as I would love to join I have like zero time in my life right now. Just for kicks I looked at my post count, not counting this post I think I posted something like 10 times since August. You know me - quality over quantity - as this post will attest. :p

So you kids go carry on - I really hope that you resolve this odds thing cause to be honest I'm not sure if I will sleep tonight now until there is a resolution. :rolleyes:

I just heard I was 'called out' and just stopping by to give a shout out to my peeps.

And no PL - I will save you the time - I am not a Wussy.



Carry on peeps .....

Back to bidness:

{{{ Whine Alert }}}

Sprewell is a bitch.

Okay, carry on...

quote: (From the Football Pick 'Em Game thread, Dec 28)
--------------------------------------------------------------------------------
Originally posted by plawrence:
Ignoring me won't make me go away, Big Daddy.

And if you're not careful, I'll sic my pal FS on you.

Now, you gonna get into the basketball game, or what?

Wuss
--------------------------------------------------------------------------------
quote
Originally posted by Big Daddy Don

I'll be there about 10 games late in the second half. Gives me something to talk about. Have I ever mentioned how I started 10 games late in the baseball league ?

Oh and Wuss this .....
-------------------------------------

Quote:
originally posted by plawrence

That is, I believe, 59 days from now, or February 25th.

I'll be counting down the days and waiting eagerly for your arrival.

Sunday

PL 8618 116.2 126 (+09.8) 8744 098.9
DB 8509 116.6 090 (-26.6) 8599 101.6
CC 8273 111.5 102 (-09.5) 8375 101.5
JG 8253 110.8 116 (+05.2) 8369 099.3
SP 7924 117.6 092 (-25.6) 8016 097.4
PP 3211 107.5 133 (+25.5) 3344 101.6
LE 2262 118.6 156 (+37.4) 2418 102.6

Our two "beginners" did the best yesterday.

I'm closing in now! :p :rolleyes: -Pat

First of all, all the odds-making discussion is really annoying me. Second of all, I'm currently at 2nd place for the week behind Plaw. Thirdly, I'm making it my goal to pass Patrick by the end of the first half. And I can probably do that.

Yes Patrick, I will beat you.

Ok..Where's this coming from? -Pat

I coulda told you that...

And yet I had to go all the way to the basketball site to get it. Some pal you are. :rolleyes: :p

Monday

PL 8744 098.9 088 (-10.9) 8832 119.8
DB 8599 101.6 059 (-42.6) 8658 119.7
CC 8375 101.5 082 (-19.5) 8457 106.9
JG 8369 099.3 085 (-14.3) 8454 109.2
SP 8016 097.4 069 (-28.4) 8085 113.4
PP 3344 101.6 059 (-42.6) 3403 113.0
LE 2418 102.6 079 (-23.6) 2497 117.2

What a miserable night. The highest score is 88? :rolleyes:

This game sucks. I'm not playing the second half.

Wow, 59 points... Uggghh. I don't feel so well.

I'm not feeling all that well myself DB. Pass the Pepto.

What's everybody moaning about today? Upset stomachs? I feel fine

You know, DB, I was just looking at the standings. Better watch your back. You're almost as close to third as you are to first . :p

88 isn't anything to brag about commish. You better watch your back standing there all cocky on your mountaintop. Somebody might push you off.

Plaw, I think the flaw with this reasoning (which I didn't catch earlier) was that in the second, third and 4th days, there is not a 1/120 shot because in each day, there is an equal chance of it being the first day in any of her 3 trips. You got 1/120 through 1/40 (her chance of being there that day) times 1/3 (the probability of it being the first day of one of her trips - let's say her first trip). However, there is an equal chance that it is the first day of her second trip or her third trip. Therefore, the true probability of the second day is:

1/3*1/40 + 1/3*1/40 + 1/3*1/40 = 1/40.

In any case, I believe it's a coincidence that the original 4/40 you got was close to the 10% I calculated. If we carry out my reasoning, (converse reasoning - which actually uses the formula that you are looking for), we can see what the probability is after 40 days.

Looks like you jinxed yourself with that comment.

I'm plus 118 on you the last 5 days :p

My prediction? You won't be on top of Crabby too much longer.

I think we better put an end to the "odds-making" discussion, DB.

I wouldn't want to annoy Val any further :rolleyes: