Ted "Rocky Bowl Father" Skaarup --
The most bizarre thing has happened in the Rocky Bowl and I must reach out to the smartest person in the group for an answer. It is a task I can only ask someone who will truly get to the mystifying probability of it ever happening once or again.
If you go to the Master score sheet you will find two entries EXACTLY the same: Payton "Pay Way" Redelsheimer and Jimmy "the Greek" Malevitsis. We have confirmed by reviewing each entry form (Both sent in several days before the deadline) and have confirmed with each bowler that their entries were correct.
What I would like to know is the odds of this happening, it is not a mistake. Not only were the games selected the same but the EXACT values placed on each game. "Pay Way" is a approximately 14 yrs old living in Kansas and the "The Greek" is 60 living in Seattle. If you have time and the interest ,I would love to hear your perception of this bizarre event to publish on the blog.
Have at it my old bowler friend.
Other Rocky Bowlers are also free to chime in ...
-- Rocky
"Underworked" wrote ...
ReplyDeleteIt has been way too long since I took any type of probability class in school. Although the odds seem pretty low. However, they are better than if each game were actually a coin flip with a 50-50 outcome in every game. In other words, if Vegas had every game as a "pick-em game with no spread. In addition, I was wondering how the points were distributed among the games. If each bowler had forward weighted their picks by applying 35 points to the first game and 1 point to the final game, the matching result would not be as extremely rare as it seems. For instance, it would not surprise me to see two identical entries from people with little time to spend on their picks and choosing just to go with all the favorites and assign points in order, either 1-35 or 35-1.
However, if the games were assumed to be a coin flip and the points were figured randomly, I am guessing some math genius in the Rock Bowl could calculate the incredible odds of having two identical entries. Anyone up to the challenge?
Skaarup wrote --
ReplyDeleteThe problem is easy to figure out if you just work it one step at a time:
PART I:
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Mugway and The Geek decide to pick 35 games in order, each with a 50%-50% chance of winning. There are only 4 possible outcomes:
Mugway The Geek
======= =======
PICK A A AGREE
PICK A B DISAGREE
PICK B A DISAGREE
PICK B B AGREE
The odds of them picking the same team is (2/4) or 0.5 (50%).
Picking ONE game in agreement is therefore 50%.
Picking TWO games in agreement means they have to 1) pick the first game the same, then 2) a second game the same, or 50% x 50% = ¼ = 25%
This goes on with 1/(2^n) as the probability where (n) is the number of games they pick.
In this case, 1/(2^35) = 1/34,359,738,368
If they picked one game a second until they matched it would take (on average) 65,372 years to pick the same 35 games.
Now, it gets tougher…
PART II:
===========================
Assigning a POINT RANK to each team (1-35) means that for any given game they have to agree on a Point Value. Assuming this is completely random, each team would have a (1 of N) selection of numbers [where N is the number of still available to be assigned].
So, for a given game, say the FIRST game, Mugway could pick one of any 35 numbers… then The Geek could pick, again, one of 35 numbers… so the probability is
(1/1)*(1/35) = 1/35th.
Now this is a stumbling block for some as they would think that it should be (1/35) * (1/35) but that’s not correct. If Mugway goes first and picks a number, any number, say 35… then the only “unknown” is the second player, The Geek, picking exactly that same number… or in this case, 1/35. [Note: the (1/1) is Mugway’s pick… that is, he will CERTAINLY pick a number, no matter what numbers are left. That assigns a probability of 1.0 or 100%]
The SECOND game would have only 34 remaining numbers… so the probability of picking the same number is (1/34)… and the next (1/33)…
So the probability of picking MATCHING NUMBERS would be:
1/35 * 1/34 * 1/33 *….. 1/4 * 1/3 * 1/2 * 1/1 = 1/35! = 1/10,333,147,966,386,144,929,666,651,337,523,200,000,000
This is a larger number than the number of atoms in the universe (including black holes, dark matter, and unseen gravitational wells).
Now, the probability of BOTH events occurring is a achieved by multiply each INDEPENDENT EVENT together, or:
1/34,359,738,368 * 1/10,333,147,966,386,144,929,666,651,337,523,200,000,000 = impossible *TILT*
Steve Long would have a better chance scoring with Morgan Fairchild than this happening by chance. Or Tony Romo making it to
the Superbowl this year.
=======================
In biblical studies, if you’re into ancient texts, there is the now discredited “Q-Theory” (from the German Quelle: “Source”) in
which the Synoptic Gospels are all supposed to come from the same “base” text.
This would account for much of the agreement across the Gospels... along with the disagreement across Gospels.
So my best guess would be the following:
1. Mugway and The Geek used a common source – the “Q” theory
2. Steve Long typed them in wrong <<< – best possibility
It's a Rocky Bowl Miracle, after perusing the Web, it has been discovered that there is a 3rd exactly matching selection sheet in the wild. It can be found here ... http://collegefootballpoll.com/games_preview_121313_confidence_picks.html
ReplyDeleteNow the $1,000,000 question, who copied who? My guess is that Dog Payton was the original author.
What you say Jimmy "The Greek" Malevitsis?
the fact that im a redelsheimer and dont know who Pay Way is, is very weird.lol. still great catch. May the best Bowler win
ReplyDelete