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Philosopher King
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Re: Draft Lottery suggestions...
« Reply #2 on: Aug 13th, 2007, 10:23pm » |
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Please pardon my hesitance here... I, of all people, was not one of the participants last year. I was doing my due diligence, trying to deal with the various methods we were attempting to use to connect, only to finally get myself plugged in and see the draft order was already set. I did not really appreciate that, and, to worsen matters, it came out flukey. Honestly, until now, I really did not totally understand the process (because I never really thought it through). Not that I am accusing anybody of anything, I'm just saying... It was a shitty confluence of events from my perspective.
Anyway, I think I get it now...  It seems that there is an extra, unncessary step listed in the Excel part of Warner's explanation. I see it like this... In one column, the first column, put the names of the five bottom finishers in the first 15 rows like this: Pak's name in rows 1 through 5, my name in rows 6 through 9, Frank in rows 10 through 12, DB in rows 13 and 14, and Warner in row 15. Then, it gets randomized, i.e., the names get shuffled, and numbers are assigned according to the row numbers they are in. From there, I don't see the need to "sort" the spreadsheet, "numbering them 1-15"!?!? 
I get the rest... All the guys besides Frank, DB and Pak, the guys getting together, have the spreadsheet. Frank, DB and Pak put numbers 1 through 15 in a hat or whatever; pull 'em out, and the order is determined according to the order of the numbers they pull out vis-a-vis the assigned numbers as per the spreadsheet, which the rest of us have. Everything with that squares, nice checks and balances.
But, again, why do we need to "sort" the sheet after randomizing it???  Am I missing something, or has an extra unnecessary step been added in in Warner's explanation? Help me understand! Otherwise, it sounds good... 
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Re: Draft Lottery suggestions...
« Reply #3 on: Aug 13th, 2007, 10:34pm » |
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on Aug 13th, 2007, 10:23pm, StegRock wrote:Please pardon my hesitance here... I, of all people, was not one of the participants last year. I was doing my due diligence, trying to deal with the various methods we were attempting to use to connect, only to finally get myself plugged in and see the draft order was already set. I did not really appreciate that, and, to worsen matters, it came out flukey. Honestly, until now, I really did not totally understand the process (because I never really thought it through). Not that I am accusing anybody of anything, I'm just saying... It was a shitty confluence of events from my perspective.
Anyway, I think I get it now... It seems that there is an extra, unncessary step listed in the Excel part of Warner's explanation. I see it like this... In one column, the first column, put the names of the five bottom finishers in the first 15 rows like this: Pak's name in rows 1 through 5, my name in rows 6 through 9, Frank in rows 10 through 12, DB in rows 13 and 14, and Warner in row 15. Then, it gets randomized, i.e., the names get shuffled, and numbers are assigned according to the row numbers they are in. From there, I don't see the need to "sort" the spreadsheet, "numbering them 1-15"!?!?
I get the rest... All the guys besides Frank, DB and Pak, the guys getting together, have the spreadsheet. Frank, DB and Pak put numbers 1 through 15 in a hat or whatever; pull 'em out, and the order is determined according to the order of the numbers they pull out vis-a-vis the assigned numbers as per the spreadsheet, which the rest of us have. Everything with that squares, nice checks and balances.
But, again, why do we need to "sort" the sheet after randomizing it??? Am I missing something, or has an extra unnecessary step been added in in Warner's explanation? Help me understand! Otherwise, it sounds good... |
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Good question. When you utilize the random function in excel, it turns the numbers into a number between 0 and 1 to the thousandth. So, once you have a random number, you then need to sort those numbers from highest to lowest or lowest to highest. Once sorted, you just need to number them to ensure they are numbers that would be drawn (1-15) Ultimately, to make it super easy, I could do the spreadsheet and just send it out numbered 1-15.
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Philosopher King
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Re: Draft Lottery suggestions...
« Reply #4 on: Aug 13th, 2007, 11:34pm » |
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on Aug 13th, 2007, 10:34pm, DirkDiggler wrote:| Good question. When you utilize the random function in excel, it turns the numbers into a number between 0 and 1 to the thousandth. |
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Which numbers??? Explain this to me like I am a six-year-old... I now understand that you are not putting our names in the first column. Numbers go in the first column. How are those numbers, the numbers to be randomized, being determined???
 ...
Okay, I got it... You put the RAND() formula in a (15-row) column ahead of the 15 names (Pak 5 times, me 4, Frank 3, and so on). Then, you sort. However, to avoid having the numbers change when you sort, don't you have to copy and "paste special" just the values (without the RAND function) to a third column... and then sort (according to that third column)???  I just did it myself, and that's what I needed to do. Otherwise, every time I was sorting, I was getting so confused because the numbers weren't ascending or descending according to my sort. It's because the values were changing, and I was left to only guess that they were sorting according to the prior values. Do I gots this right here, dude??? 
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Re: Draft Lottery suggestions...
« Reply #5 on: Aug 14th, 2007, 6:50am » |
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on Aug 13th, 2007, 11:34pm, StegRock wrote:
Which numbers??? Explain this to me like I am a six-year-old... I now understand that you are not putting our names in the first column. Numbers go in the first column. How are those numbers, the numbers to be randomized, being determined???
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Okay, I got it... You put the RAND() formula in a (15-row) column ahead of the 15 names (Pak 5 times, me 4, Frank 3, and so on). Then, you sort. However, to avoid having the numbers change when you sort, don't you have to copy and "paste special" just the values (without the RAND function) to a third column... and then sort (according to that third column)??? I just did it myself, and that's what I needed to do. Otherwise, every time I was sorting, I was getting so confused because the numbers weren't ascending or descending according to my sort. It's because the values were changing, and I was left to only guess that they were sorting according to the prior values. Do I gots this right here, dude??? |
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You got it !!!
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Re: Draft Lottery suggestions...
« Reply #8 on: Aug 18th, 2007, 8:55am » |
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PLEASE READ THIS THOROUGHLY AND CAREFULLY!!!
And,... as it turns out, last year's draft was FUCKED... as I had a hunch all along,... although it was not done so purposefully, at least not by DB, Pak and Frank, and, at the end of the day, the damage was mitigated,... ironically, frankly speaking, because the guy with 1 ticket "in the hat" last year was picked first. The following is a direct quote of the whole of Steve Warner's e-mail dated Sunday, September 3, 2006, the day of last year's draft, at 8:18 a.m. (Hawaii Time):
Quote:| Attached please find the randomization of the picks... Frank, DB, and Pak draw numbers until the order is established. 1-15... if they draw a 6 first, the person next to #6 gets the 1st pick... if they draw another number with that same name, they keep proceeding until everyones name is drawn. |
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I would have figured this out if I just had a moment to come up for air last year and not, instead, get bum-rushed, so to speak. It just hit me in the shower tonight... After a selection is made, all of the numbers/tickets "in the hat" that are associated with the person whose number/ticket was just picked MUST BE REMOVED. You canNOT just keep picking, discarding the "repeat picks". Otherwise, the percentages skew, potentially quite A LOT, in favor of the guys with fewer numbers/tickets in the hat. Here's an example...
Let's say the first number/ticket picked out was for the guy who came in second-to-last place, the guy with four tickets/numbers in the hat. [Those of you at all proficient in math and/or logic and/or statistics (in the academic sense) are going to see where this is going already.] So, on the next pick the guy with 5 tickets/numbers in the hat, the guy who came in dead last, has a 36% (5 in 14) chance of getting selected next; the guy with 3 tickets/numbers has a 21% (3 out of 14) chance (15% less chance than the 5-ticket guy), and the guy with 2 tickets has a 14% (2 in 14) chance (7% less than the 3-ticket guy). HOWEVER, if the other three tickets were removed, as they should be/always have been, the 5-ticket guy's chances would be 45%; 3-ticket guy, 27%; 2-ticket guy 18%. I.e., the 5-ticket guy should have an 18% better chance of getting selected than the 3-ticket guy, not a 15% chance,... and so on down the line.
NOW, this may not seem egregious,... though it still is what it is. NO ONE wants to give up ANY percentage points if they don't have to,... NO LESS SHOULDN'T. HOWEVER, this can potentially get REALLY BAD. Let's say, with the VERY next pick out of the hat, one of the 3-ticket guy's numbers gets selected. So, on the next pick, the 5-ticket guy has a 38% chance (5 out of 13), and the 2-ticket guy has a 15% chance (2 out of 13), a 23% difference. ACTUALLY, THOUGH, the 5-ticket guy is supposed to have a WHOPPING 63% chance (5 out of 8), and the 2-ticket guy is supposed to have a 25% chance, i.e., the 5-ticket guy is SUPPOSED to have a 38% advantage over the 2-ticket guy, not just a 23% advantage. FUCKIN' YIKES!!!
So, because Steve, the 1-ticket guy, was picked out first last year, there was no damage to the percentages of the second pick,... but ultimately this method makes the percentages better the fewer the numbers/tickets in the hat you have, i.e., this method favors, even if only by a handful of percentage points,... drum roll, please,...  the guy with 1 ticket in the hat. ANYFUCKINWAY,... the 3-ticket guy, Pak, was picked next, which did mitigate the damage. However, the percentages, though ever so slightly, do start to bend toward the lower guy. To wit (though, of course, not necessarily as a direct cause), DB, the 4-ticket guy, was selected next.
And, for those of you who think, "Aw, no big deal, what's really the effect?" Look, first of all, Markie, who ended up vying for a championship last year, had, by way of a trade with me, positioned himself to have the BEST chance at the #1 pick (with all the according percentages on his side, which he didn't totally get,... but, we got "lucky",... if you want to call it that,... it wasn't that bad). He ended up with the fourth pick. That almost certainly affected the outcome of our championship 23 weeks later. BUT, that isn't the most curious effect. I, who was the 5-ticket guy, picking now fourth in the second round took MUHSIN MUHAMMAD!!! WE ALL KNOW that our championship ultimately, in some sense, came down to THAT pick!!!  AND, thank God it wasn't a lot, but that pick was awarded to DB over me with a miniscule, but not less than an integer, 2 percentage-point advantage he wasn't supposed to have.
Again, the numbers were, thankfully, negligible last year (though your Championship is now tainted on two fronts Doug,... eh-hem,... T-Rave).  BUT, let's not make the same mistake again this year. WE MUST HAVE THE "GET-TOGETHER GUYS" PULL ALL THE TICKETS/NUMBERS OF THE GUY WHO THEY JUST PICKED.
And, again, in closing, as per my message immediately above, I will take care of the spreadsheet this year,... for no reason other than to make sure there's no jumping the gun this year. Thank you for your understanding and cooperation.
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Philosopher King
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Re: Draft Lottery suggestions...
« Reply #10 on: Aug 18th, 2007, 9:39pm » |
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on Aug 18th, 2007, 4:25pm, DirkDiggler wrote:Steg-
I would disagree with your logic above in the fact that once a team is drawn, it does not matter if they are selected again.
So for example, if the 5 team is drawn first..... there would be 14 tickets still in the bag. BUT, only 10 of those tickets are competing. So every ticket has a 1 in 10 chance of getting drawn. The guy with 1 ticket has a 1 in 10 chance and the guy with 4 has a 40% chance of being in a draw that MATTERS. The other 4 tickets become irrelevant. If a 5 guy ticket is drawn AGAIN, the odds are STILL 1 in 10 for the guy with one ticket and 4 in 10 for the guy with the 4 tickets.
It is kind of in comparison to a penny being flipped 99 times and coming up heads. On the 100th flip the odds are still 50% chance of coming up heads. |
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Touche... You've caught me in a gambler's fallacy...  I just worked out a few "conditional probabilities", and you're right. The calculations get more complicated, for sure, but, at the end of the day, the percentages are the same. Even though it seems that the percentages shift favorably/unfavorably as I've described above, they, in the end, do NOT. Nevertheless, for some psychological reason, if, in a contest, I had 2 tickets in a hat and my opponent 1, I wouldn't want any extra, "meaningless" tickets being added, just because IT REALLY DOES FEEL LIKE they would dilute/hurt my chance of victory vis-a-vis the first pull (percentage-wise). But, bottom line, the conditional probability remains the same as the simple probability, namely 66% and 33%, respectively. It just gets more "chopped up", so to speak. But,... hey,... that's why it's called a gambler's fallacy...  OUCH! 
So, the pull method from last year is good to go. There is no reason to have to weed through and pick out the leftover tickets.
And, I, furthermore, apologize to you, Steve, for the insinuation that you were not playing it totally aboveboard here.  I'm obviously still sore at you on "other" fronts. That's not to say that that effected my thinking about the math here. That came into play only after the (erroneous) "fact". I was just wrong in my calculations. But, while this wasn't a witch hunt, I did go out of my way to take a swipe at you, which was wrong of me. I'm sorry.
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Philosopher King
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Re: Draft Lottery suggestions...
« Reply #11 on: Aug 19th, 2007, 4:55am » |
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By the way, for those of you who might be a little thick about this like me (probably nobody),... here's a further, rather simple clarification... Even though, when additional meaningless ticks are "in the hat", one's chances of winning on whatever given pull (from the hat) decrease, SO IS one's chances of losing. Take the simple example of three ticks in the hat, two for X and one for Y, so X has a 66% chance of winning and Y a 33% chance. Now, just using X here,... if one extra meaningless tick is added to the hat, yes, X's chances of winning have been reduced from 66% to 50%, but X's chances of losing have also been proportionally reduced from 33% to 25%. By "proportionally", I am just pointing to the fact that a 16-percentage-point drop from 66% is, percentage-wise, equivalent to an 8-percentage-point drop from 33%.
Okay,... enough... 
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