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Super Bowl Square probabilities

NoahFence

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NoahFence
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Mar 17, 2011
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OK, so we've all seen the websites that say getting this number + that number = good or bad.

This question is a bit different, and I was wondering if it would be a statistical curiosity. It is at least for me.

Keep in mind this actually happened here at work. Funny stuff:


User A buys 2 squares. She winds up with "4" Patriots, "2" Giants on one square. Ewww.

Her other square? "2" Patriots, "4" Giants.

Yep, she got the same number combination for both teams.

Now, just when you thought it was an interesting outcome, the next day they draw numbers for a totally different set of squares.

The woman who sits next to her got the exact same numbers for her 2 squares. "4" Patriots, "2" Giants on one, "2" Patriots, "4" Giants on the other.

What are the odds?
 
100%. It happened. :p

GO PATS!
logoPatriotsSmall.jpg


As to predicting this before it happened, and specifically picking these two women out before hand, those are long odds. As to the first coincidence, that has (if I understand it correctly) about a 2% chance of happening if you only buy 2 squares, so not that astronomical. As for some one else getting the exact same distribution with 2 squares, does that become 0.04% (multplicative?)? Not sure how the problem's constraints are set up.

The fact that she sits next to the other woman has nothing to do with it, it's a red herring since both of them are involved in the betting pools.
 
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Here's my poke at it:

Since the numbers can't be identical (the reverse wouldn't exist in that case) there should be less than 1/100 chance of picking a pair of scores that are opposites.

Woman A Pick 1: Doesn't matter what numbers as long as they aren't identical. Gets numbers X and Y.
P1 = 90/100

Woman A Pick 2: Must pick Y and X. There are only 99 squares left to choose from.
P2 = 1/99

Probability of picking two scores that are opposites:
P1 x P2 = 90/9900 = 1/110

Probability of the woman 2 getting the same pairs as woman 1:
It doesn't matter what woman 1 picked, the odds of woman 2 picking the same two pairs out as her are going to be the same no matter the scores.
P1 = 1/100
P2 = 1/99
P = 1/9900

Note that in reality the probability of an event like this being noticed is much higher than 1 in a whole bunch. It's not unlikely that someone in the game would get pairs if many people bought two tickets. If someone the next time got the same set of pairs, they'd notice the set of pairs and not comment on all of the other people they hadn't matched with. This probability relies on us stating ahead of time that we are matching woman 2's score to woman 1's, not that at least 1 person in the 30 (just to make up a number) who participated got pairs on day 1 and that on day 2 at least 1 person matched those pairs.
 
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