Optimal Strategy for this schoolyard game - (Charge, block, shoot) I encountered this game when I was a kid (we called it Street Fighter back when it was all the rage) and recently saw it again with my nephews playing the same game with a different name and slightly different rules.
The basic game is an RPS-style game where each participant selects one of the following actions per round.


*

*Charge

*Block

*Fireball (uses up 1 charge)

*Super Fireball (Uses up 5 charges) 


Anyone who gets hit by a fireball while charging is dead.  Blocking cancels fireballs thrown at you and two fireballs fired at each other also cancel each other out.  Super fireball goes through blocks and overpowers regular fireballs to automatically kill the opponent unless he super fireballs as well.
I was wondering what the optimal strategy was, if any. During which rounds is it best to fire/block? Is it better to go for the super blast, or to catch your opponent unawares with a well-timed regular fireball? 
What would be the numbers for the 2-player case? How will this increase in complexity as the number of players increase as well?
Edit:  What if the number of required charges for the super fireball is increased/decreased?
 A: Here is a very incomplete answer which might help use make some progress for two player version. 


*

*I'm going to ignore super-fire-ball. 

*I'm going to make an assumption which simplifies things a lot. If the charges are ever imbalanced, the player with more charges can force a win. (I have some serious doubts about this.)  
In this case, equilibrium play involves charging whenever neither have a charge and equally randomizing otherwise. 
In the first round, it is weakly dominant to charge (this doesn't rely on my assumption. Having fewer charges cannot possibly increase your probability of losing against any strategy). 
In the second round, 
If 1 blocks and 2 uses fireball, then 1 eventually wins (by assumption). 
If 1 blocks and 2 blocks, then the game simply starts over from the same state. 
If 1 blocks and 2 charges, then 2 eventually wins (by assumption) 
If 1 charges and 2 uses fireball, then 2 wins
If 1 charges and 2 blocks, then 1 eventually wins
If 1 charges and 2 charges, the game continues with both at 2 charges. 
If 1 uses fireball and 2 blocks, then 2 eventually wins
If 1 uses fireball and 2 charges, then 1 wins
If 1 uses fireball and 2 uses fireball, then the game returns to the original sate. 
Notice that no matter what 1 does, the game either ends in a win for 1, a win for 2, or continues (with each having equal charges). These are the exact same outcomes possible in every round of a best-of-one rock-paper-scissors game. The only equilibrium is to randomize equally over all three options (ignoring super-fire-ball) in every round unless neither player has a charge in which case, both charge. 
A: I'm new here, I hope it's not bad form to post a partial answer to your own question!
Thank you CommonerG for that analysis.  Your post got me to think about this even more.  
It seems that there are the following game states, and changes to another state based on the results of the previous round.


*

*Both players have no charges.  (Initial State)

*Both players have at least one charge. (Charge parity)

*Player 1 has at least one charge and Player 2 has no charges. (Charge advantage)

*Inverse of State 3 (Charge disadvantage)


The reason I characterized the game states is that I kept thinking about Assumption# 2 and kept thinking that while Player 2 is disadvantaged, he can still force the game back to parity or the initial state and still have a chance.  So I thought, what is that chance?  How much of an advantage is it?
I guess this kind of makes it more interesting.  Given that the game is symmetrical from the initial state, it can still reach a point of asymmetry (unlike RPS which is symmetrical at all times).  Maybe we could develop some sort of optimal mixed strategy for an advantaged player and a disadvantaged player.  
So here’s my thoughts so far.


*

*Initial state - In the initial state, both players will always charge. 

*Charge parity
Assuming each player has an equal chance to pick each action, each of these has a 1/9 chance of happening.

Now it’s my turn to assume something that I’m not sure about.  I think the number of charges over another doesn’t matter (i.e. 1-charge, 2-charge, 3-charge advantage is all the same and won’t affect the player’s strategy, except in the miniscule case that he builds up 5.)  The only difference is that the state will transition to a still-disadvantaged state if the player digs himself out of it.


*Charge advantage / disadvantage
In this state, the advantaged player will never block because his opponent cannot shoot.  The disadvantaged player will never shoot.

*Parity with some changes in the 3rd (Fire, block) and 9th (fire, fire) squares.  I think? 
And then the disadvantage chart would be the opposite.

Here’s what I will do.  I will try to run scenarios where P1 and P2 pick either action 50% of the time in advantage/disadvantage scenarios.  Then compute win % assuming that, and revise the probabilities accordingly in order to eventually get an optimal course of action.
