Playoff Spot magic number?

[YankeeStripes]

we have 10 games left and a 5.5 game lead.

we have 10 games left and they (detroit) have 9 games left....

what does that make our magic number?

[hardrain]

You just have to look at the WC standings. The Magic number is 5.

http://mlb.mlb.com/mlb/standings/wildcard.jsp

[yankeebot]

The formula is the number of games left to be played, add one, then subtract the number of games ahead in the loss column of the standings from the closest opponent

or

You can start with the number of games for a full season (162), add 1, subtract the number of wins your team has then subtract the number of losses the 2nd place team has.

[Kluivert4Ever]

The magic number (5) is now the same for both the Mariners and the Tigers.

[hardrain]

The formula is the number of games left to be played, add one, then subtract the number of games ahead in the loss column of the standings from the closest opponent

or

You can start with the number of games for a full season (162), add 1, subtract the number of wins your team has then subtract the number of losses the 2nd place team has.

Or you can look at the standings

[4bronxbombers]

Or you can look at the standings

Much easier for those of us that hate math.

[Murcer1]

Much easier for those of us that hate math.

[hardrain]

Much easier for those of us that hate math.

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

[Kluivert4Ever]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

And now in English?

[hardrain]

And now in English?

In English that means Yanks beat Red Sox

[Kluivert4Ever]

In English that means Yanks beat Red Sox

Ahhh, got ya

[4bronxbombers]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

Thank you. That was very helpful.

[hardrain]

Thank you. That was very helpful.

No problem! That's why I'm here.

Here is a behind-the-scenes look at how the Joba Rules were originally designed:

[Maynerd]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

I like pie.

[aeromac76]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

Hey I am an engineer, don't make me have to think..

[yankeebot]

I like pie.

Pie? Did someone mention pie?

[Janttje]

American Pie?

[BillBuckner]

Math sucks. This is why I majored in History.

[spyglass]

American Pie?

A long, long time ago I can still remember how that music used to make me smile...

[Rob Rocco]

"Pie"=3.14159265358979323846...

I think Joba prefers cake.

[Sam18]

Pie? Did someone mention pie?

I thought I heard pie!

[yankees135]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

Nice cryptographics system, but I'm not sure how RSA will help you to compute the magic number. BUT I can imagine the numbers P and Q being thought of as MAGIC numbers in that case ;-)

[NYKforever]

[4bronxbombers]

Math sucks. This is why I majored in History.

Math sucks. That's why I majored in English.

[4bronxbombers]

A long, long time ago I can still remember how that music used to make me smile...

And I knew if I had my chance.......

[ajra21]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

spot on.

[nh_snowman]

Nice cryptographics system, but I'm not sure how RSA will help you to compute the magic number. BUT I can imagine the numbers P and Q being thought of as MAGIC numbers in that case ;-)

Someone beat me to it, I knew it was RSA.

[yankees135]

Someone beat me to it, I knew it was RSA.

Glad to know I'm not the only math or computer person on the board ;-)

[Huktonfonix]

It's easy! Just Find P and Q, two large prime numbers.
2) Choose E such that E is less than PQ, and such that E and (P-1)(Q-1) are relatively prime, which means they have no prime factors in common. E does not have to be prime, but it must be odd. (P-1)(Q-1) can't be prime because it's an even number.
3) Compute D such that (DE - 1) is evenly divisible by (P-1)(Q-1). Mathematicians write this as DE = 1 (mod (P-1)(Q-1)), and they call D the multiplicative inverse of E.

Hughes2.50, is that you??