Tug-o-war Claims

[BMAD]

At a tug-o-war contest two teams are pitted against each other.  Team 1 has five players: Alice, Bob, Carol, Douglas, and Eric.  Team 2 also has five players: Frank, George, Harry, Ian, Jessica.  When both teams used all of their members, team 1 won.  When Team 1 and Team 2 each lost two players, it was a draw.  When two new mini-teams were formed with Alice and Frank competing against Bob and George, Bob and George won the contest.

 

Besides what was already given, what other claims could be made about individual or team strengths and of outcomes of other tug-o-war contests?

[dgreening]

Additional observations

 

We have established that B>F is required 

 

G can be larger smaller or equal to A

 

B and F can be smaller or larger than G and A

 

For any set of values  that fulfill all the  problem requirements [say A=2, B=5, F=3, G=1: from k-man], you can add a positive number to B and F and/ or you can add a positive number to G and A and the new numbers will still satisfy all the conditions. So any of the following would still work:

• A=2, B=5, F=3, G=1
• A=12, B=5, F=3, G=11
• A=2, B=95, F=93, G=1
• A=12, B=95, F=93, G=11

 

The key to this is that:

• B and F must be relatively close in value
• Similarly G and A must be relatively close in value

 

Taking it further, the key to combinations that will work is the difference between B and F compared to the difference between G and A.

 

We have already established that B>F and that G can be larger, smaller or equal to A.

 

So to satisfy the problem 

 

Abs [G-A] < [b-F].

 

If B-F = 1, then G  = A must be true

if B-F = 2, then G +1 = A or  G=A or G-1 = A must be true

...

 

[DejMar]

An assumption is made that the strength of a player's position in the tug-o-war chain has no bearing other than which side of the rope the player is on, and the strength of a team is the sum of the individual players' tug-o-war power.

Another assumption is that the two new mini-teams is composed of the two members lost to the original teams, and Carol+Douglas+Eric = Harry+Ian+Jessica.

Let the initial letter of the player's name represent the player's tug-o-war power. Then the following should be true:
A+B = F+G
A+C+D+E+F < B+G+H+I+J

Other than when forming teams of unbalanced numbers, I can not ascertain any other valid claims.

Edited December 11, 2014 by DejMar

[k-man]

Making the same assumptions that DejMar made...

 

Bob is stronger than Frank (B>F). It comes from the following two inequalities

A+B > F+G (this comes from the A+B+C+D+E > F+G+H+I+J and C+D+E = H+I+J)

A+F < B+G (mini-team result)

[DejMar]

Oops. I did make an error. I incorrectly read "It was a draw" for when all players used the teams, and for the subset of three members. K-man got it right.

[dgreening]

 

 +g+c+d+e>+g+h+i+j,>

Given

A+B+C+D+E > F+G++H+I+J               [1]

and

C+D+E = H+I+J                                    [2]

H+I+J   = C+D+E                                  [2a]

 

and

A+F < B+G                                           [3]

 

Therefore

[1] – [2] yields                                  

A+B> F+G                                            [4]

 

[4] + [2a] yields

A+B+H+I+J > F+G+C+D+E                   [5]

 

[3] +[2] and [3]+[2a] yield

A+F+C+D+E < B+G+H+I+J,   and        [6]

A+F+H+I+J   < B+G+C+D+E                 [7]

 

Comparing [3] and [4] tells us that

 

B>F

and/ or

G>A

 

Since either condition could cause this, we cannot reduce this further.Since either condition could cause this, we cannot reduce this further.

+g+c+d+e>+g+h+i+j,>

[k-man]

 

 

 +g+c+d+e>+g+h+i+j,>

 

Given

A+B+C+D+E > F+G++H+I+J               [1]

and

C+D+E = H+I+J                                    [2]

H+I+J   = C+D+E                                  [2a]

 

and

A+F < B+G                                           [3]

 

Therefore

[1] – [2] yields                                  

A+B> F+G                                            [4]

 

[4] + [2a] yields

A+B+H+I+J > F+G+C+D+E                   [5]

 

[3] +[2] and [3]+[2a] yield

A+F+C+D+E < B+G+H+I+J,   and        [6]

A+F+H+I+J   < B+G+C+D+E                 [7]

 

Comparing [3] and [4] tells us that

 

B>F

and/ or

G>A

 

Since either condition could cause this, we cannot reduce this further.Since either condition could cause this, we cannot reduce this further.

+g+c+d+e>+g+h+i+j,>

 

 

How can G>A cause [4] A+B>F+G?

Edited December 19, 2014 by k-man

[dgreening]

Sorry for the delay. I was out of touch over the holidays.

 

A+B> F+ G??]
 

style="color: rgb(40, 40, 40);">How can G>A cause [4]
A+B>F+G?

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">The easiest way to demonstrate this is by
applying number values for each person.

 

style="color: rgb(40, 40, 40);">Let’s use some undefined, normalized units [my
Engineering professors would be aghast].

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">Let’s say that C, D, E and H, I, J each have a
value of 10

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">For the four people in question (A, B, C, D)
let’s assign values (11, 10, 8, 12)

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">Rechecking the earlier
equations

 

style="color: rgb(40, 40, 40);">

 

style="color: rgb(40, 40, 40);">[3]    
style="font-family: calibri;">A(11) +F(8) = 19 < B(10) +G (12) =
22

 

style="font-family: calibri;">

 

style="color: rgb(40, 40, 40);">[4]    
style="font-family: calibri;">A(11)  +B(10) =21
     > F(8)      +G(12) =
20 

 

 

 

style="color: rgb(40, 40, 40);">And G > A

[k-man]

I wasn't questioning that it's possible for G to be greater than A. It's certainly possible, but you can't conclude that from A+B>F+G. A can be greater than or equal to G.

[dgreening]

You are right, I thought about it some more [even before I saw your post]

 

B must be > F

 

G can be less than, greater than or equal to A

[DejMar]

 

An assumption is made that strength and endurance is not altered between contests, no forces

other than a team's combined strength favors either team, and each player puts forward a best
effort to win for their assigned team. An additional assumption is made that the players of
the two mini-teams are the same lost players in which their contest results in a draw.

A+B+C+D+E > F+G+H+I+J (3)
    C+D+E = H+I+J     (4)
      A+B > F+G       (4)
      A+F < B+G       (5)
-------------------------
        F < G

[k-man]

An assumption is made that strength and endurance is not altered between contests, no forces

other than a team's combined strength favors either team, and each player puts forward a best

effort to win for their assigned team. An additional assumption is made that the players of

the two mini-teams are the same lost players in which their contest results in a draw.

A+B+C+D+E > F+G+H+I+J (3)

    C+D+E = H+I+J     (4)

      A+B > F+G       (4)

      A+F < B+G       (5)

-------------------------

        F < G

 

A=2, B=5, F=3, G=1: 

A+B > F+G: 7>4

A+F < B+G: 5<6