User Name: Password:
New User Registration
Moderator: coan.net 
 Frog Finder

Frog Finder & variants (Frog Legs)


List of discussion boards
Mode: Everyone can post
Search in posts:  

17. August 2008, 22:53:41
Papa Zoom 
Subject: Case in point
Modified by Papa Zoom (17. August 2008, 23:01:58)
In a current game I am playing, my opponent has found 3 frogs.  One is upside down and as a result, she has -5 points.  However, I have found 3 frogs and currently have 20 points.  That's a 15 point spread.  This is a typical position.  I can't lose now.  and in nearly all of my games I always found at least 3 frogs.  It would be rare that someone would find all 5 and the opponent only find 2 frogs.  (yes, it happens but that's not a argument in favor of the current scoring situation - people win the lottery too - so?)

Lets say my opponent had found all 5 frogs.  That's 15 points.  Plus a bonus of 2.  That's 17.  Plus a bonus of 1 point for each of my remaining frogs.  Here's the breakdown.

Player 1   
5 frogs = 15 points plus 2 + 5 = 22.  That's the most you can get.  And only if I find no frogs.
If I find one frog it's player 1 = 21 and I'll have 10.
If I find two frogs player 1 = 20 and I'll have 15.

Keep in mind, player one must find all 5 frogs for the above scores.

If I find 3 frogs (the most likely thing that would happen statistically)
Player one gets 19 points and I'll have 20. 

Essentially, the game in this situation is designed for player one to find 4 frogs before I find three.  That's just not a good scoring rule.

game over.

With only a 10 point deficit it goes this way.  Consider player one lands on his/her frog.  Gets a -5 points.  But the opponent gets no points. then..

player 1     5 found frogs = 15 plus 2 bonus plus 5 = 22.  Same scores as above.  Player one's scores aren't affected. 

So, if player two finds

0   then it's 22 to 0
1  then it's 21 to 5
2  then it's 20 to 10
3  then it's 19 to 15
4 then it's  18 to 20.

This is more fair.  It basically eliminates player 1's first frog.  Now it's a 4 to 4 race.  I need to find just 4 frogs to win.

Only if Player 1 finds all 4 remaining frogs before I find 4, can he win. 
 

The best solution is to have a random zero square to avoid anyone landing on their frog on the first move.

As it is now, the current rules for points is flawed.  I don't mean to offend anyone but instead of shallow answers, how about some statistical facts?  Like I said, I'm not a math wiz so maybe I'm wrong.  But frankly I think I bring up a legitimate point about the game and it's just being dismissed with opinions, not statistical facts.  It's worth looking into.

  



Date and time
Friends online
Favourite boards
Fellowships
Tip of the day
Copyright © 2002 - 2024 Filip Rachunek, all rights reserved.
Back to the top