MOM 2025

[RogerT]

Posting this to help Redditchbulldog.

Please vote for your MOM in the usual way 1. 2. 3. from any source.

[distantdog]

1. Butterworth - great handling in awful conditions and 2 tries.

2. Dean - tough to stop close to the line. Hopefully he gets plenty of tries this season.

3. Parsons - his extra pace really brightened us up when he came on.

In truth, the whole team deserve credit for playing in truly miserable conditions. 

[Keep The Faith]

1. dean /Parsons 

2. Greensmith 

3. Walshaw/butterworth 

[phildog]

1 Butterworth, has looked at Hooley and is in that mould, always in great positions on attack m, throws a great pass, finishes well when it's his turn, now a great catcher of the ball in all conditions.

2 Woods, he has improved already this season, finding himself more room to run with the ball, like my #1 has a great cut out passing game.

3 Moore constantly breaking down the defence yesterday, moved through the pack as usual with no effort and is constantly talking and directing.

Edited January 27 by phildog
Spelling

[seen it before]

woods

Butterworth

Parsons 

[RogerT]

This topic now closed.

[phildog]

Totals v Newcastlee;

Butterworth 9 vote

Dean, Parsons, Woods 5 votes 

Greensmith 2 votes

Walsham, Moore 1 vote

So Butterworth 3 points, Dean, Parsons, Woods 2 points, Greensmith 1 point

[bromleybulldog]

3 hours ago, phildog said:

Totals v Newcastlee;

Butterworth 9 vote

Dean, Parsons, Woods 5 votes 

Greensmith 2 votes

Walsham, Moore 1 vote

So Butterworth 3 points, Dean, Parsons, Woods 2 points, Greensmith 1 point

Thought we gave points to the top three not top five !

[phildog]

If players are equal on votes then they have to get points!

[bromleybulldog]

48 minutes ago, phildog said:

If players are equal on votes then they have to get points!

Of course, but we're awarding points for the top 3, so points for the Newcastle game should be 3 Butterworth  and 2 each for second equals. So 2 each for Dean, Parsons and Woods.  Greensmith came 5th in the voting, so sorry, Olly but zero points

[phildog]

I will try and check that with Redditch but he's rather difficult to get hold of.

[phildog]

We've taken onboard Bromley's point. Going forward, any time there is a tie on votes points will be awarded at the level they occur ie 2 players tieing in 1st place will get 3 points, then if 1 player wins the 2nd place vote he will get 2 and no points will be awarded for 3rd. Should there be one winner in 1st, 1 in 2nd and 2 (or more) in 3rd EACH will receive 1 point. Hope that satisfies. Therefore the 1 point awarded v Newcastle has been removed.

[bromleybulldog]

1 hour ago, phildog said:

We've taken onboard Bromley's point. Going forward, any time there is a tie on votes points will be awarded at the level they occur ie 2 players tieing in 1st place will get 3 points, then if 1 player wins the 2nd place vote he will get 2 and no points will be awarded for 3rd. Should there be one winner in 1st, 1 in 2nd and 2 (or more) in 3rd EACH will receive 1 point. Hope that satisfies. Therefore the 1 point awarded v Newcastle has been removed.

Phil, your first example above is still wrong. Think about the points as if they were doughnuts you were handing out to the winners. You have six doughnuts (3+2+1)  Take the following as an example, it's an extreme one, I know, but shows you what happens when you give out the doughnuts in your method

Two players gain 6 votes

Three players gain 4 votes

Five players gain 2 votes

If you give 3 doughnuts to each of the first two and then two to each of the second group, you've already given out 12 doughnuts, so you're out of pocket and already thinking b*gger this, someone else can buy the doughnuts next week, In your first example, the player who comes 'second' in your words should actually get only one doughnut as he gained less votes than the two players in first. He's actually in third place not second.The five guys in 'third' obviously go hungry. In my example above you end up giving out 9 doughnuts  (2x 3 )+ (3x 1), so you're still out of pocket doughnut -wise but not as badly!

 

 

 

 

Edited February 12 by bromleybulldog

[georgeb1]

The players will get fat

[Sir Jekyll Stocking]

10 hours ago, bromleybulldog said:

Phil, your first example above is still wrong. Think about the points as if they were doughnuts you were handing out to the winners. You have six doughnuts (3+2+1)  Take the following as an example, it's an extreme one, I know, but shows you what happens when you give out the doughnuts in your method

Two players gain 6 votes

Three players gain 4 votes

Five players gain 2 votes

If you give 3 doughnuts to each of the first two and then two to each of the second group, you've already given out 12 doughnuts, so you're out of pocket and already thinking b*gger this, someone else can buy the doughnuts next week, In your first example, the player who comes 'second' in your words should actually get only one doughnut as he gained less votes than the two players in first. He's actually in third place not second.The five guys in 'third' obviously go hungry. In my example above you end up giving out 9 doughnuts  (2x 3 )+ (3x 1), so you're still out of pocket doughnut -wise but not as badly!

 

 

 

 

If we keep with Bromley's example, and want to keep within our doughnut budget,

The first to players on 6 votes each share the (3+2=) 5 doughnuts for first and second places - so each would get 2.5 doughnuts.

The three players sharing the third position on 4 votes would get a third of a doughnut each.

Hope this helps. 

[bromleybulldog]

11 hours ago, Sir Jekyll Stocking said:

If we keep with Bromley's example, and want to keep within our doughnut budget,

The first to players on 6 votes each share the (3+2=) 5 doughnuts for first and second places - so each would get 2.5 doughnuts.

The three players sharing the third position on 4 votes would get a third of a doughnut each.

Hope this helps. 

Absolutely correct.I just thought that starting talking about fractions of doughnuts would confuse people even further.

[Sir Jekyll Stocking]

Deleted

Edited February 13 by Sir Jekyll Stocking

[phildog]

So basically what you're saying is that if 2 players tie in 1st place they get all the donuts? Anyways, I think there's a bit over-complicating the dough for the ....

[bromleybulldog]

50 minutes ago, phildog said:

So basically what you're saying is that if 2 players tie in 1st place they get all the donuts? Anyways, I think there's a bit over-complicating the dough for the ....

No, two players tieing for first get 5 doughnuts 3+2.

[phildog]

That means 2.5 it's each? No....no ha'pennies!

[bromleybulldog]

34 minutes ago, phildog said:

That means 2.5 it's each? No....no ha'pennies!

Nah! I'd give them both 3. The main thing I think is to avoid giving points to a player who might have the third highest number of votes but has three or more players with more votes- if that's clear.   So two players on 6 votes, two on 5 and 2 on 4 votes would result in 2x3, 2x1 and 2x0. No one said it was going to be easy!! At least this method allows us to avoid half votes, one third votes or 16% of a vote! It's just that some weeks you might have to buy more doughnuts!

Edited February 16 by bromleybulldog

[RogerT]

My head is aching fathoming this out, that's why I started the post off but not the mechanics of ot.

[fredm]

The easiest way is to declare any joint declarations as non admissable.  It should be three names listed, first second and third only.

[RogerT]

44 minutes ago, fredm said:

The easiest way is to declare any joint declarations as non admissable.  It should be three names listed, first second and third only.

Agreed.

[phildog]

So 2 blokes play out of their skin come joint 1st by a street and get nothing? I'm losing the entire thread of this. It'll be done to best of our ability.