Cmon... You have proved yourself, VERY bored.

User was banned for this post

-Chac

(It took me less than 30 minutes)

NO TYPO PRIZES FOR THIS POST



The first figure is correct, however the second is horribly wrong.

What you have done is doubled the number of possible openers for a single player, and then doubled it to count the number of possible openers for both players. You need to square the value.

With grip:
Number of possible openers for 1 player= 4^21
Number of possible openers for 2 players=4^42
Number of possible openers for 3 players=4^63
Number of possible openers for 4 players=4^84

Without grip:
Number of possible openers for 1 player= 4^20
Number of possible openers for 2 players=4^40
Number of possible openers for 3 players=4^60
Number of possible openers for 4 players=4^80


Some of you probably need this explaining, because form your posts, i gather that your method of thinking is slightly incorrect.

If you take the first joint as being the only joint, there are four possibilities of state. Meaning that you have four possible openers. Thus we get 4^1

If you take the first and second joints as being the only joints, you have four possibilities for the first, for each of these posibilities, the second provides another four possibilities. This we get 4^2

Add a third joint into the mix and we have another four possibilities ontop of what we already have, giving us 4^3 possible openers.

With uke having 20 joints, this logic gives us the figure of 4^20 joint arrangements for from the start of the game for a single player.

Taking grip into account, we can realise that there are two possibilities of grip state for each hand, meaning that there are now (2x2) extra player arangements to go with each exsistent arrangement, giving us the figure of 4^21 possible openers.

If you now include another player, for every possible opener the first player can have (which is 4^21), the second player can also have as many openers. This means that the number of openers for one player is squared when calculating the number of starters between two players:

4^21 squared is 4^42

If you want to get really picky, you could say that this value is wrong, because it counts uke doing move a, and tori doing move b as a different combination as uke doing move b and tori doing move a, when infact the two combinations owuld be exactly the same, apart form the payer doing the move is swapped round.

How is that? He was only calculating the how many openers 2 players could do, not how many outcomes there could be for 2 different players after each turn.

How about we quit flipping shit over someone using a move. Openers, of all things.

Let me try and explain it a different way, if you have two players, you now have 20 extra joints.

So you have 40 joints, each having four different possible states:

4^40

You also now have 4 hands, each with two types of grip states:

2^4=16
16=4^2

4^40x4^2=4^42






Sorry to rain on your parade, but this is wrong as a result of you previous error.

IN a 10/120 match with no dq there are always 12 turns.

On the first turn, there are 4^42 turns

Starting the second turn, there are 4^42 possible things that have already happened, and for each of these situations, there are another possible 4^42 that can happen. so after two turns there are (4^42)x(4^42) possible outcomes (This is of course all counting if there is no dq, fracture or dm, as already said by cmon).

At the start of the thrid turn (4^42)x(4^42) possible things have already happened. For each of these things, another 4^42 things can happen, so after the third turn, there are (4^42)x(4^42)x(4^42) possible outcomes.

(4^42)x(4^42)x(4^42) can be written as (4^42)^3, which can be written as 4^126

Continuing the pattern, after the fourth turn there are a possible (4^42)^4 things.

...
...
...

After twelve turns, a possible (4^42)^12 things might have happened.

(4^42)^12 can be rewritten as 4^504

which is 2.74x10^303

your number only had 105 characters, the correct value would be about 3 times longer when written out.

Crazy idea > It would be great If someone creates a bot to test all these openers and calculate which are the most effective .

You can't calculate effectiveness for multiplayer, because the opponent will move. You COULD however make a bot that would keep uke limp and run through god knows how many possibilities, and create the ULTIMATE MADMAN for a given number of maximum frames.

well the problem would be the human factor.

The bot can test what is most effective (e.g hits UKE with the most damage and in fewest frames possible) but it cannot factor all the different openers that UKE can use,

Use 2 bots.