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Re: Fantasy Basketball Fans Wanted
#226381
11/20/03 09:13 AM
11/20/03 09:13 AM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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Thursday
DB 2206 122.8 092 (-30.8) 2298 095.7 PL 2195 121.8 092 (-29.8) 2287 098.6 BT 2097 111.3 077 (-34.3) 2174 097.7 JG 2061 123.6 082 (-41.6) 2143 095.5 OA 1994 114.2 131 (+17.2) 2125 095.5 CC 2015 118.7 099 (-19.7) 2114 096.0 SP 1915 099.2 103 (+03.8) 2018 096.0 TM 1907 113.1 087 (-26.1) 1994 096.0
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226382
11/20/03 09:21 AM
11/20/03 09:21 AM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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Figure this out... I just happened to notice that the V. AVGs which I posted yesterday morning for the games played last night, are now different. If you click on any player in our league, and look at his "previous" day (in this case Wednesday), his total V. Avg doesn't match what it was on Wednesday morning I don't know if I explained that very well... I haven't noticed; Do they change player values daily in this game? That would account for the discrepency.
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226384
11/20/03 12:02 PM
11/20/03 12:02 PM
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Joined: Jul 2001
Posts: 31,335 New Jersey, USA
J Geoff
The Don
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The Don

Joined: Jul 2001
Posts: 31,335
New Jersey, USA
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I would think that the V.AVG should change daily (the day after a particular player plays the night before)... but the player values should only change weekly. Once I thought I noticed a player's value changing in mid-week but thought I was seeing things since I only noticed it once. I don't know... 
I studied Italian for 2 semesters. Not once was a "C" pronounced as a "G", and never was a trailing "I" ignored! And I'm from Jersey!  lol Whaddaya want me to do? Whack a guy? Off a guy? Whack off a guy? --Peter Griffin My DVDs | Facebook | Godfather Filming Locations
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Re: Fantasy Basketball Fans Wanted
#226385
11/20/03 12:31 PM
11/20/03 12:31 PM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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Duh. Of course, you're right JG. The V. Avg should change after a player plays. For some reason I was thinking "cap value".
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226395
11/20/03 05:39 PM
11/20/03 05:39 PM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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Posted by Baltimore Crab Cake posted November 20, 2003 02:59 PM Quote ------------------------------------------------------------------------------ "What's the odds of this. SP and I have the exact same roster, right down to our Coach. " ------------------------------------------ Well, let's see... One over the number of players available at each position (including coach), to the fifth power.
In other words, there are 2 games tonight. Four teams involved. Four "feasible" picks (No one's gonna pick the scrubs) for each position.
You pick "Player A". SP has a 1/4 chance of also picking "Player A". So the chances of both of you picking "Player A" is 1/4.
(DB, feel free to jump in aanywhere here)
Now do that 5 more times for the other four positions and coach, and you get 1/4 to the 5th power, or 1/4096, or, expressed in odds, 4095-1, or, as a percentage, 0.0002441.
That number, 4096, is also the number of ways a 6 man roster can be arranged if there are 4 choices at each position, but because of salary cap restrictions, there are any number of the 4096 combinations which have to be excluded.
(Jump in anywhere, DB)
For exanple, tonight picking McGrady, Duncan, Marbury, Nowitzki, and any center I believe, would put you over the cap.
So, to answer your question "What are the odds?" The answer is "I don't know". Less than 4096 to 1, though.
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226396
11/20/03 05:55 PM
11/20/03 05:55 PM
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Joined: Jul 2001
Posts: 7,361
Don Sicilia
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Joined: Jul 2001
Posts: 7,361
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Close Plaw... You are absolutely correct that the chances of Bethie picking a 6 member team out of 4 teams playing is 1 out of 4096 (4 to the sixth power) As you mentioned, the probability is probably less by virtue of player value restrictions- you might have only 2 or 3 of the 4 players available to you once you chose a high value player at another position).
However, we are talking about the probability of two people having the same team. That means whatever probability that Bethie has in picking her team, SP has the same probability (up to a max of 1 out of 4096).
Therefore, the probability of both having the same exact team is 1 out of 16,777,216 (4096*4096).
It's much like two people flipping coins. Both have a 1 out of 2 chance of getting heads or tails. There is a 1 out of 4 chance of both getting heads, both getting tails, one getting heads while the other tails (and vice versa).
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Re: Fantasy Basketball Fans Wanted
#226400
11/20/03 08:27 PM
11/20/03 08:27 PM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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The Slippery Slope
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Believe it or not, DB, I pondered the question some more on the ride home tonight. (and, of course, thinking coin-flipping analogies as Idrove). At first, I thought I was incorrect, and the odds were 4096 x 4096. But then I figured, if there are 4096 ways to arrange the team, and CC arranged it one way, whoever else had a 1/4096 chance of arranging it the same way. But I'm not sure your answer is ever the correct one. If CC chooses heads first, then the chances of SP choosing heads is 1/2. But not knowing if Beth took heads or tails, i.e. if they "flip" simultaneously, then SP still has a 1/2 chance of matching (H-H or T-T), or not matching (T-H, H-T). I know there's a scenario where 4096 x 4096 is the correct answer, I just don't know what it is. 
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226401
11/20/03 08:55 PM
11/20/03 08:55 PM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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Oh, and thank you Bethie for asking such a stimulating question.
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226403
11/21/03 01:10 AM
11/21/03 01:10 AM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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The Slippery Slope
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Intellectual stimulation can sometimes be more effective than physical stimulation. And since you've.......never mind.
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226404
11/21/03 01:21 AM
11/21/03 01:21 AM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
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OP
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The Slippery Slope
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I figured out the 4096 x 4096 problem, DB.
For that to be the correct answer, you need three people involved. I pick a roster. The odds of CC picking the same roster is 1 x 4096. The odds of SP picking the same roster as the two of us is 1 x 4096 x 4096.
The way to phrase the question so it involves only two people would be something like this, I think:
What are the odds of two people picking the same predetermined roster? Or to put it another way, what are the odds of two players picking a roster composed of players A, B, C, D, E, and F, as opposed to any six random players?
But here's something else we didn't factor in which greatly reduces the odds: The number of players in our group.
In a group of two, the odds are 4096-1. In a group of eight, however, the odds are much lower. I mean in a group of 3000-4000, the odds of two people picking the same random roster are actually very low, because in a group of 4097, it would be an absolute certainty that at least two people would have the same picks.
So what do you do? Divide the 4096 by the size of the group?
(Note to FS: Spare me the comments about "Get a Life" or "Too Much Time On My Hands" please )
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226405
11/21/03 01:36 AM
11/21/03 01:36 AM
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Joined: Jul 2001
Posts: 7,361
Don Sicilia
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Joined: Jul 2001
Posts: 7,361
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Good points plaw. As I thought about it, your points below are what you've said before, but just put in another way. Let me explain... Originally posted by plawrence: At first, I thought I was incorrect, and the odds were 4096 x 4096. But then I figured, if there are 4096 ways to arrange the team, and CC arranged it one way, whoever else had a 1/4096 chance of arranging it the same way. This was exactly your point before... Given Bethie's team (which we can set in stone), SP has a 1 out of 4096 chance of matching it. The key is that we already know what Bethie picked a team. Originally posted by plawrence: If CC chooses heads first, then the chances of SP choosing heads is 1/2. But not knowing if Beth took heads or tails, i.e. if they "flip" simultaneously, then SP still has a 1/2 chance of [b]matching (H-H or T-T), or not matching (T-H, H-T).[/b] This too is also the same point. We know that Bethie picked one of the 4 power forwards. Knowing that she picked one, someone else has a 1 in 4 chance of picking the same one. However, there's a slight difference in asking about the probability that Tim Duncan is picked. Bethie has a 1 in 4 chance. So does another person. Together, they have a 1 in 16 chance of both picking Tim Duncan. As for your above example, you're right that there is a 1 in 2 chance that they match. There is also 1 in 4 chance that they both have heads. Originally posted by plawrence: I know there's a scenario where 4096 x 4096 is the correct answer, I just don't know what it is. Maybe here's a way to clarify the question: What are the odds of two players picking the same point guard, the same shooting guard,..., and the same coach? 1 in 4096 What are the odds of two players both picking Tyronne Lue, T-Mac,..., and Frank Johnson? 1 in 16,777,216. What do you think? I think this is a problem, not of statistics, but of semantics.
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Re: Fantasy Basketball Fans Wanted
#226406
11/21/03 01:40 AM
11/21/03 01:40 AM
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Joined: Jul 2001
Posts: 7,361
Don Sicilia
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Joined: Jul 2001
Posts: 7,361
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Originally posted by plawrence: What are the odds of two people picking the same [b]predetermined roster? Or to put it another way, what are the odds of two players picking a roster composed of players A, B, C, D, E, and F, as opposed to any six random players? [/b] Exactly!  This is a better way to say it than how I said it (What are the odds of both picking Tyronne Lue, T-Mac...)
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Re: Fantasy Basketball Fans Wanted
#226407
11/21/03 01:45 AM
11/21/03 01:45 AM
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Joined: Aug 2002
Posts: 15,058 The Slippery Slope
plawrence
OP
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OP
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I know the points were the same. I was just giving you the thought process I used to convince myself I was right after thinking I might have been wrong.
Now, how do we account for group size to reduuce the odds from 4095-1?
"Difficult....not impossible"
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Re: Fantasy Basketball Fans Wanted
#226408
11/21/03 01:47 AM
11/21/03 01:47 AM
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Joined: Jul 2001
Posts: 7,361
Don Sicilia
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Joined: Jul 2001
Posts: 7,361
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Originally posted by plawrence: So what do you do? Divide the 4096 by the size of the group?
This can be answered by this logic. We've always been focusing on two specific people - Bethie and SP. But it can be any two from our eight. Therefore, we have to multiply the probability of two players choosing the same team (let's use the 1 in 16 million number for now) with the number of combinations of 2 players out of a group of 8 players (i.e. we use the "choose" function - 8 choose 2). If we were a bigger group, say out of 5000, the probability of any two players choosing the same exact team will be a lot bigger since now we multiply the same probability as before (1 in 16 million) with 5000 choose 2 (which is much bigger than 8 choose 2).
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