Logic tells me that in a group of 4097, the chances of two people picking the same team is 4097/4096.

In a group of two, the chances are 1/4096.

So in a group of eight, wouldn't the chances be 8/4096?

This is kind of like an old parlor trick. Get 30 people together, and it's almost certain that two of them will share a birthday

The Choose function... 8 choose 2... it basically gives you the amount of combinations of forming a subgroup of 2 from a group of 8.

How it's calculated is "a" choose "b" = "a" factorial divided by the product of "b" factorial and "b minus a" factorial.

To put it into words, Bethie and SP have a 1 in 16 million chance of picking the same team.

Bethie and TM also have a 1 in 16 million chance of picking the same team.

What are the odds of either Bethie and SP or Bethie and TM both picking the same team? 2 times 1 in 16 million. (1 in 16 million plus 1 in 16 million)

Now what are the odds of any two players from our 8 picking the same team... It's the odds of two players picking the same team (1 in 16 million) times the number of combinations of two players picked out of a group of 8.

I hope I made more sense.

Or 1 in 16,777,216!

No.

We're not going with the 16 million number. Beth's original question was "What were the chances of someone picking the same team as she did?".

I'm convinced that the 16 million number only applies to two people picking the same predetermined roster. To put it another way:

The chances of two people matching a coin flip in any manner is 1/2.
The chances of two people both throwing heads is 1/4. (starting to repeat myself here...)

Let's work with how the size of the group affects the odds and the 4096 number.

And keep it simple for me, please. My Probability & Statistics course in college was about 35 years ago.

EDIT: I'm keeping your 1 out of 4096 number

What's the question Plaw?

Maybe it might be simpler if we decreased the number in the group. Say it was a group of 3: CC, SP, and TM and for these purposes, we say the probability we are looking for is the probability of any two players picking the same team of 6.

In a group of 3, there are 3 possible pairings of people (CC and SP, CC and TM, SP and TM).

What is the probability that CC and SP choose the same team? 1 in 4096

What is the probability that CC and TM choose the same exact specific team? 1 in 4096

What is the probability that SP and TM choose the same exact specific team? 1 in 4096

What is the probability that CC and SP or CC and TM or SP and TM choose the same exact specific team? 1 in 4096 + 1 in 4096 + 1 in 4096 = 3 times 1 in 4096

or to put it another way:

(3 choose 2) times 1 in 4096

BTW, 3 choose 2 = (3*2*1)/(2*1)= 3

Now if we up our group size to 8, we would have to count the number of subgroups of 2 out of a group of 8. The short cut way is to use 8 choose 2.

If we increase our group size to 8, there are 28 ways to pair up two people. If each pair has a probability of 1 out 4096 of picking the same team, then the probability that any pair of two people out of a group of 8 will be 28 times 1 out of 4096.

I think I get the hang up now. Once again, we are answering different questions.

You are right, in a group of 2 (CC and SP), there is a 1 in 4096 chance of picking the same team.

In a group of 8, there is 7 in 4096 chance that one of the other 7 picks the same team as CC.

That's the question you are answering.

The question I was answering was, what are the chances that any two (not restricted to CC or anyone else at all) pick the same team.

Makes sense?

That I understood.

So if we work with Beth's original question (see my last post), then the chances would be 8/4096?

Right? I hope so.

Because next I want to discuss how our biases towards certain players enter into the equation. I mean, There are four players to choose from at each position, but there is not an equal chance of any of the four being picked. It's not like flipping coins, where the probability of either outcome is the same.

For example: Duncan is one of four players, but he might have a 75% chance of being picked, while with Nash and Marbury in the mix, Tony Parker has maybe a 15-20% chance of being picked.

If I said "what are the chances of two people (in a group of two, for now) picking a team with players A, B, C, D, E, and F (Name the 6 players with the lowest V. AVG), the chances would be greater than one in 16 million.

But if I said "what are the chances of two people picking a team with Duncan or Stoudamire, Marbury or Nash, Nowitzki or Marion, Finley, Ginobili, or McGrady and any one of four centers and four coaches, the chances would be much less.

So to answer Beth's question, it's really more like 2 x 2 x 2 x 3 x 4 x 4, or 384-1 divided by the group size of 8, or about 1/48.

But because some of the above combos are impossible because of the cap (No, I am not going to run through all 384 to find out how many there are), lets call it about 1 chance in 24.

That actually seems about right to me.

( (edited 7:31 am) )

Sheesh....we almost filled a whole page here.

Bethie, see what you've stimulated!! :p

Plaw, if you're still out there, perhaps we can continue this tomorrow. This is great, I haven't used this part of my brain since college! My brain has been stuck on financial and practical applications, but it was force fed theoretical for four years!

In any case, I was right about tonight, Jahidi White killed me. 2 minutes. 1 rebound. 1 missed shot.

Unofficial scores for tonight

Me 117
You 105

Me back in first by one point. G'night.

7/4096... you can't count Bethie. If we had to choose 2 players out of 8 and one of them had to be Bethie, then there are only 7 combinations of 2.

About 1 out of 24.

Took me and DB all day and half the night to figure that out.

What men won't do for a chick.

Right

About 1 out of 24.
[/b][/quote]Make that about 1 out of 27

Friday

PL 2287 098.6 117 (+18.4) 2404 123.8
DB 2298 095.7 105 (+09.3) 2403 120.1
BT 2174 097.7 102 (+04.3) 2276 114.3
JG 2143 095.5 112 (+16.5) 2255 126.7
OA 2125 095.5 112 (+16.5) 2237 121.7
CC 2114 096.0 114 (+18.0) 2228 116.4
SP 2018 096.0 114 (+18.0) 2132 106.7
TM 1994 096.0 114 (+18.0) 2108 115.8

What are the odds of me saying this group is too heavy for me and I hit the road?

What you wouldn't do for a chick? What have you done for me lately.

That was like a knife to my heart.

DB and I stayed up until all hours of the morning to answer your question correctly.

That was supposed to stimulate me ? It seems to me you guys stimulated yourselves.

You don't find it stimulating to learn to what ends a guy will go to please you?

Strawberries and cream and a bottle of red wine will take care of that.

Plaw, I think you once again run into the trouble of counting the eight person. The probability of two people picking the same team is now 1 in 384, given our limitations. Understood. But in a group of 8, there's again only 7 possible combinations with one specific person. Therefore, there is a 7 in 384 chance that the other 7 matches with the specific one. Therefore, the probability is actually a little bit smaller than 1 in 48.

Of course, like you said, not all 384 combinations is possible. So in the end, the probability will be greater than 7 in 384. 1 in 24 is probably a good estimate because I bet over half of the 384 combinations is not possible, given salary cap restrictions. Just a hunch though.

Ummm.... Let me see... there's a 1 in 5 chance that you were completely bored by the statistical talk... multiply that by the probability...

:p

Where did 4096 come from? Keep it simple please. I hated math in school. :p

If you flip a coin twice, there are four possible outcomes, two per flip:

First Flip: Heads or Tails
Second Flip: Heads or Tails

So, on two flips, the possible outcomes are:

H-H
H-T
T-H
T-T

Two flips, two possibilities, two times two is four.

Now, visualize each possible selection on the Fantasy Basketball team as a flip of a coin. Two games, four teams, four players per position, so in this case it's like a coin with four sides.

Four power forwards, call them A, B, C, and D
Four small forwards, call them E, F, G, and H

There's 16 possible combinations:

A-E, A-F, A-G, A-H
B-E, B-F, B-G, B-H
C-E, C-F, C-G, C-H
D-E, D-F, D-G, D-H

Four times four. Sixteen.

Bust since there are 6 positions to pick, it's 4 x 4 x 4 x 4 x 4 x 4, or 4096.

Simple enough?

I never knew a cheerleader who liked math. They were always to busu worying about shaking their pom-poms.

[QUOTE]Originally posted by plawrence:
[QB]
to busu

Ahem.....where is FS when we need him? :p

As far as shaking my pom poms...well....I guess it's time to hang those bad boys up too.

I feel like Simple Simon.

"Simple Simon met a pieman going to the fair. Said Simple Simon to the pieman, what have you there?"

Said the pieman to Simple Simon, "Pies, you dumbass".

:o

FS is probably busu composing a post about how I have no life, and how I have too much time on my hands.

But let us not lose the distinction between those of us who make typos, and those of us who can't spell.

And please; Don't hang up those pom-poms just yet.

Well let say it's a good thing BD isn't here then. I know how to spell, but I'm too lazy to check for typo's. He was always good for spotting them.

Don't hang up the pom poms yet? The clock it ticking on that one.