~~~ABCDEs - Ranking puzzle...~~~

after reading the first line, i just got completely confused and it gave me a headache. I am sure the committie will make fine pound for pound rankings. As long as they get a even pool of different perspectives, it should be a good system. Woulndt be surprsied if the list looked like the ring rankings honestly.

 

C was just unmotivated for B and D. He raised his game as needed against A.

 

Lack of motivation, doesn't come into consideration, when giving rankings. This isn't at Jungs office...

 

:thumbsup

 

I failed algebra.........LOL!

 

Sorry Decebal, but you haven't given me enough information. I am a stickler for details and one small thing makes this impossible for me to rank and tosses a ton of other variables into the mix.

You said COULD BEAT not DID BEAT in all of those scenarios. While I am going to take skill level into account, I will also take resume into account. Usually 50/50 and I go by actual results unless there is a blatant blatant robbery.

I cannot place any value on what COULD happen in those scenarios because some of those fighters may be better than the ones they would lose to but are bad style matchups, etc and have better resumes, etc. Sure I can quantify skill within reason, but that won't tell me who will win.

As to your second question, current form should be taken into account, but you must have reliable information on their current skill level. Therefore recent wins must also be taken into account as looking good against second rate fighters is not an accurate representation. And it's not about names, but about how good that fighter actually is and the circumstances of the fight. As in, I don't have Hopkins on my P4P list because he dropped off after retiring, and beating Wright at 170 isn't going to get him back on. And he didn't look good doing it, neither did.

 

Proof of (i) . The equality ck = c1+(k-1)a just proven with k = m+1 implies cm+1 = c1+(m+1-1)a = c1+ma.

We will prove
k
å
j = 1 cj = kc1+ 1 2 k(k-1)a
by induction on k ³ 1.

If k = 1 then Sm = åj = 1m cj = c1 and
kc1+ 1 2 k(k-1)a = 1c1+ 1 2 1(0)a = c1
Therefore the assertion is true for k = 1.

Now we want to show if the assertion
k
å
j = 1 cj = kc1+ 1 2 k(k-1)a
is true for k = m ³ 1 then it must be true when k = m+1. To this end, suppose
m
å
j = 1 cj = mc1+ 1 2 m(m-1)a
This and the equality
m+1
å
j = 1 cj = cm+1+ m
å
j = 1 cj
imply
m+1
å
j = 1 cj

=

cm+1+ é
ê
ë mc1+ 1 2 m(m-1)a ù
ú
û

=

[c1+ma]+ é
ê
ë mc1+ 1 2
m(m-1)a ù
ú
û

=

c1+mc1+ 1 2 m(m-1)a+ma

=

(1+m)c1+[ 1 2 m(m-1)+m]a

=

(1+m)c1+ 1 2 [m(m-1)+2m]a

=

(m+1)c1+ 1 2 [m(m-1+2)]a

=

(m+1)c1+ 1 2 [m(m+1)]a

=

(m+1)c1+ 1 2 [m(m+1)]a
This says that
k
å
j = 1 cj = kc1+ 1 2 k(k-1)a
when k = m+1. The principle of mathematical induction now applies. It gives the desired conclusion.

Proof of (ii) . We wish to show
k
å
j = 1 cj = 1 2 k(c1+ck)
Observe
1 2 k(c1+ck) = 1 2 k(c1+[c1+(k-1)a]) = kc1+ 1 2 k(k-1)a
Therefore the assertion
k
å
j = 1 cj = 1 2 k(c1+ck)
follows from
k
å
j = 1 cj = kc1+ 1 2 k(k-1)a

really its as simple as that.