Coaching records

[RebelBruiser]

Kiffin - 0.364
Croom - 0.356

titus should probably sign up for a new screen name. His point does still stand though that the winning percentages are pretty close, even though he did a McClurian job of calculating them.

 

[Maroon Eagle]

[No message]

 

[DerHntr]

round down one and up the other and you get .36 for both. jeez.

 

[LTblows]

The poster moved the decimal two places too.

 

[Johnson85]

then you get what HD6 stated.

/hope you were being sarcastic and i missed it

I hope it's not me, and I'm not even looking at Stans' record, but I'm pretty sure dividing his wins by his losses will be better than 100%.

 

[DerHntr]

[No message]

 

[pDigital32Dawg]

But I will credit you for providing me with great laughter over your error.

 

[BR549.sixpack]

asdf

 

[Johnson85]

Again, not looking at the record, maybe being dense, but I can't see how dividing Stansbury's win-loss record the same way titus did to Crooms's works out to less than 100%

 

[DerHntr]

you were right. the sarcasm passed you. my bad. veiled to heavily.

 

[titus.sixpack]

my math sucks, but the point remains valid.

I'm glad that I brought some enjoyment into your afternoon and I will say that my math makes as much sense as many of the posts that I see on this board. Mine is obviously wrong, but some of you are subjectively wrong.

 

[pDigital32Dawg]

So many of us have tried so few have accomplished.

 

[titus.sixpack]

I will try my best not to top it. I should have said "Hey! It looks like Kiffin & Croom's loss - win record are similar. Maybe Croom should get a chance to coach at UT".

I'm leaving for the basketball game now....will try not to get lost.

 

[UpTheMiddlex3Punt]

Your comparison of the two coaches at least uses two numbers calculated the same faulty way. Also, one can prove that:

if (a/b)=(c/d), then a/(a+b) = c/(c+d)
(I will not bore you with the details, but cross-multiplication and eliminating like terms are your friends)

Also, if you have (a/b) and (c/d) terms that are close to each other, then your a/(a+b) will be close to c/(c+d), depending on how you define close.

Pythagoras just 17ed up entirely on comparing percentages, one calculated the traditional way and one calculated his own way.

 

[o_1984Dawg]

nm