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Rose: you are exactly right. You did get bonus point (2 for finding all your frogs and 1 for the remaining frog of your opponent didn't find). But you could only win if your opponent only finds two frogs. That's just plain unfair and the rule needs to be changed. It's a very bad rule. The idea of the game is to have fun and a challenge, but in the situation like in your game, you lose on your first move.
Something can be done to change this imbalance and should.
Zmenené užívateľom Papa Zoom (17. augusta 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.
Subjekt: Re: I have a question. Maybe it's been discussed
Zmenené užívateľom Papa Zoom (16. augusta 2008, 18:51:27)
coan.net: but I can't ignore the fact of how even the win/loss is for each player (50.2% vs. 49.6%)
the majority of games are played where the first player doesn't land on his own frog. This would account for the even stats above. When both players are able to get a "0" square, the game is relatively even. Even when I've guessed wrong later in the game and lost 3 points, I can still win. The game is still "on."
But, if you were to consider the stats of ONLY those games where player 1 is -5 points with 4 frogs to find, and player 2 is +5 points with 5 frogs to find, I'll guess that the stats are 20% or less for player 1 to win. Maybe even 10%.
Player 1 MUST find all remaining frogs (that's all 4 must be found) and will only achieve 15 points. Player 2 only needs to find 2 frogs to achieve a tie.
IF you had designed the game where player one started the game with a -5 but only had 4 frogs to find while player 2 had a +5 with 5 frogs to find, which side would you choose?
Maybe someone with a math background could provide the winning stats in a game like that. I'm not a math wiz. But once player 1 has shot his/her own frog, the game is no longer 50 50. It's probably more like 10-90. (I'm guessing here)
The game would be much more even if the very middle square was already a zero. Or even a random square at zero. Or even under the current rules, change the -10 points to just a -5. At least look into the statistics of that type of position. Even with a -5, the game still favors player 2. (because with 4 frogs to find player 1 still can only get 15 points while player 2 only needs to find 3 to get that 15).
Subjekt: Re: I have a question. Maybe it's been discussed
coan.net: But wouldn't it be better overall just to lose 5 points instead of 10? It's still gives the other an advantage (in points). Then, with a -5, you need to find the other 4 and you'd still only get 20 points. The other person has to find the same number, just 4, and will tie at worst. Seems more fair. Otherwise, when I lose 10 points to start the game is all but over. Sure, you could win, but it's so unlikely statistically. I think we ought to consider just making it a loss of 5 points. Keeps ya in the game and evens the odds a bit. Keep in mind that the opponent still has an advantage, but it's not so huge as it is now.
Subjekt: Re: I have a question. Maybe it's been discussed
joshi tm: It's a bad rule of the game. You basically lose on the first move. Game over. Not only are you ten points behind, but you can only get 15 points total. Statistically, you can't win this game. When a rule such as this one creates such a huge inbalance, something is wrong with the rule.
Subjekt: I have a question. Maybe it's been discussed
If I or my opponent lands on our own frog, we lose 5 points and the opponent gets 5. That's a 10 point advantage to the opponent. If that happens on the first move, I do a shoot move, hit my own frog, and am in the minus the equivalent of 10 points. Then if my opponent only finds three of his/her frogs, they'll have 20 points. I now need to find 5 frogs before my opponent finds just 4. It's a huge disadvantage. Wouldn't it be better to keep the odds more even by just giving a -5 and nothing to the opponent?
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