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Math riddle for you guys

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Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: Goosemaster
Originally posted by: chuckywang
Originally posted by: Goosemaster
531441

way less than that.

has anyone gotten it yet?

Someone posted the correct answer but hasn't explained.

So it's either 13, 5 or 96? Those are the only ones I see with no explanation that you haven't shot down yet.

Check out Random Variable's answer.
 
Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: Goosemaster
Originally posted by: chuckywang
Originally posted by: Goosemaster
531441

way less than that.

has anyone gotten it yet?

Someone posted the correct answer but hasn't explained.

So it's either 13, 5 or 96? Those are the only ones I see with no explanation that you haven't shot down yet.

How could it be 5? If you only made 5 submissions you could only be right on a maximum of 5 games...right?
 
Originally posted by: KingofCamelot

How could it be 5? If you only made 5 submissions you could only be right on a maximum of 5 games...right?

That's what I was wondering 😉

Edit: You're reading the question wrong
 
On your submission sheet, do you submit your guesses for all the games on one sheet, or is one game considered a submission?
 
it should be all games on one sheet, picking the winner for each game
edit: 13 choices per sheet (39 tick boxes if there is a box for a tie as well)
 
Originally posted by: cRazYdood
On your submission sheet, do you submit your guesses for all the games on one sheet, or is one game considered a submission?

All 13 games are considered ONE submission.
 
The probability of getting 13 games right is (1/3)^13, while the probability of getting 12 games right is 13*(1/3)^12*(2/3); therefore, the probability of getting at least 12 games right is (1/3)^13+13*(1/3)^12*(2/3) = 1/59049. So if you submit 59049 entries, there will theoretically be a 100% chance of getting first or second place.
 
Originally posted by: Random Variable
The probability of getting 12 games right is (1/3)^13, while the probability of getting 12 games right is 13*(1/3)^12*(2/3); therefore, the probability of getting at least 12 games right is (1/3)^13+13*(1/3)^12*(2/3) = 1/59049. So if you submit 59049 entries, there will theoretically be a 100% chance of getting first or second place.

Unless one of the quarterbacks is murdered or drives off a cliff while getting a blowjob from some cheerleeding title-costing slut...
 
Originally posted by: chuckywang
All 13 games are considered ONE submission.

well in that case, wouldn't you just need three submissions? one submission all wins, one all ties, one all losses? does that not cover everything?
 
Originally posted by: sash1
Originally posted by: chuckywang
All 13 games are considered ONE submission.

well in that case, wouldn't you just need three submissions? one submission all wins, one all ties, one all losses? does that not cover everything?

or do you need 12 games right on one submission?
 
Originally posted by: sash1
Originally posted by: chuckywang
All 13 games are considered ONE submission.

well in that case, wouldn't you just need three submissions? one submission all wins, one all ties, one all losses? does that not cover everything?

no no....read up on probability....

13 teams..3 posibilities each, but not all could win...some might tie...you are guessing the combination, not who wins....
 
Originally posted by: Random Variable
The probability of getting 13 games right is (1/3)^13, while the probability of getting 12 games right is 13*(1/3)^12*(2/3); therefore, the probability of getting at least 12 games right is (1/3)^13+13*(1/3)^12*(2/3) = 1/59049. So if you submit 59049 entries, there will theoretically be a 100% chance of getting first or second place.

Note that 59049 = 3^10. The fact that there are 13 games is immensely important. Consider one submission. How many of the 3^13 results are within one game of it? Well, you can be off in any one of the 13 different games in two different ways. Therefore, one submission can account for 1+2*13 = 27= 3^3 results. Therefore, you need at least 3^13/3^3 = 3^10 submissions. All that's left to show is that you can cover all the results in this way without overlap. (This is non-trivial)

This is immensely important in coding theory. There are codes called Hamming Codes, which can correct one error. There is a code called the (13,10) Hamming Code that takes in 10 ternary bits, and tags on 3 more ternary bits (the resulting string of length 13 is called a codeword) in such a way that if one of the bits is wrong, it can correct that bit to the right bit. You can show that every ternary bit string of length 13 is within one bit of those 3^10 codewords. Such codes that have this property are called Perfect Codes. There are two kinds of Perfect Codes: Hamming Codes and Golay Codes.

That's a little long winded, but I find this stuff fascinating. 🙂
 
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