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).