Is 99% enough?

My math is a tad simpler, but being 99% accurate, if we're going by moves, should mean that you'd be expected to blunder once over the course of two 50 move games. In a tournament, that certainly leaves an opening, especially if it's a big blunder you're making. As we can see, even if a big blunder is as rare as 1%, you still may have some problems.

Obviously this is just to get the point across -- not every move is equally easy to play well of course, depending on the position in front of you.

By the way, I wouldn't try to calculate your percentage of winning with a term  as ambiguous as mistake : they come in all shapes and sizes.

It depends on what do we assume to be 99%.

If that's the chance you'll mistake a move when it's your turn (or the value of Y), then your chance not to make a mistake in a 50 move game is 0,99^50=60,5%, so for two 50 move games you are likely 0,605^2=0,366=36,6% not to make a mistake in both games; 0,395^2=15,6% to make mistakes in both games; 100-36,6-15,6=47,8% to make a mistake in exactly one game. Your chance to make at least one mistake in both games is 1-0,99^100=15,6%+47,8%=63,4%.

But if 1% is not the mistake per move chance, but the average percentage of moves in your play that are not correct, it's another matter. It shows the value of X that gives one mistake on average. You make a mistake once in X=100 moves. From the formula Y^X=(chance for no mistake in X moves) => Y=(chance for no mistake in X moves)^(1/X).This would mean your chance not to mistake a move when it's your turn is Y=0,99^0,01=0,9998995.

So, if on average one of your 100 moves is wrong, then your chance to make a mistake per move is about 1:10000.

P.S. I have also made 2 other assumptions about game outcome:

- mistakeless play from A vs play with a mistake from B-> A wins;

- mistakeless play from A vs mistakeless play from B -> A and B draw.

If they are not correct, it should mean mistakes can't be connected with game outcome, and therefore avoiding them shouldn't be important.

My mistake. It's actually 1:10000 I'll correct it in the upper post. I have missed a digit when substracting 0,9998995 from 1, but all the rest is the same.

How did I calculate it:

- Your chance not to mistake per move = Y;

- Number of moves = X;

- I'll try to prove that your chance not to mistake within X moves = Z(Y;X)=Y^X, it's a function of Y and X.

Obviously when X=1 then Z(Y;1)=Y, or for one move sequence, you have Y chance not to make a mistake. That's by definition.

Let's assume that for X=N Z(Y;N)=Y^N. N exists, because N=1 is a solution.

Then for X=N+1:

Your mistakeless N+1 move sequences occur from your mistakeless N move sequences if you play a mistakeless N+1st move. Your chance that this N+1st move will be correct by itself is Y. So, from all your mistakeless N move sequences, Y part will continue with a correct move. The count of your mistakeless N+1 move sequences will be equal to the count of your mistakeless N move sequences, multiplied by Y, and so will be their chance. In other words:

Z(Y;N+1)=Z(Y;N)*Y=Y^N*Y=Y^(N+1)

So, Z(Y;N+1)=Y^(N+1)

and so on. For all integer values of X, it is true that Z(Y;X)=Y^X.

Now about the 1:100 and 1:10000 chances. We have X=100, Z(Y;X)=0,99, and we are searching for Y.

0,99=Z(Y;X)=Z(Y;100)=Y^100

Y=0,99^(1/100)=0,9998995

Your mistake per move chance = 1-Y = 1:10000.

I'm not going to dig through your math to find a mistake, but this just seems uttterly illogical.

Sorry if I have confused you, but maths is the only useful logic to apply here that comes to my mind. If you think it's wrong, I'll be happy if you say where the mistake is, so that we can come up with the truth. It shouldn't be difficult for you if you really think this is "utterly illogical".

Those are different events, so their probability shouldn't be equal unless there is some objective reason. The first probability is about a sequence of moves, and the second one - about a single move.

Let's say you have to play 20 moves, you have 10 possibilities on every turn, and every time one of the possibilities is wrong. What do you think your chance not to make a mistake for the 20 moves is, if your choice is random every time?

No, it's not 1-20*1/10=-1. You are not guaranteed to make 2 mistakes.

I would say it is 0,9^20=0,1216=12,16%.

To make things simplier: 5 moves, 10 possibilities, 1 wrong option every time.

Hint: if you don't trust calculating, you can use a random number generator (http://www.random.org/) for a 6 digit number and check how often the last 5 digits don't contain, say, 0. Or you can try some other way.

I can bet on: 59,05% of the time.

Original statement: "So, if on average one of your 100 moves is wrong, then your chance to make a mistake per move is about 1:10000."

This seems to go against basic probability. When "one in ten-thousand" is said, it literally means that you'd be expected for this thing to happen once every ten thousand times, not every one hundred times. Having one of your moves wrong out of 100 is consistent with a 1% probability.

I think Ozzie realized what he was trying to say:

"What Glex is showing is that a per-move mistake probability of 1:10000 gives a 1% chance of making a mistake over 100 moves."

Although this is correct, we're not looking for what makes it 1% likely to have a blunder; we're looking for how long it takes before you'd expect a blunder. So, if you have a 1% chance to make a blunder (instead of 1:10000), you would be expected to make one every 100 moves.

I think it's better to say "how permissive about mistakes can one afford to be". In other words, to what extent should one make sure he doesn't make a mistake when choosing a move between the legal possibilities.

The first question is what is your opponent's mistake chance. It has to be compared with your one in order to find out some game outcome probabilities, and in this way - mistake chance influence on the result.

The second question is what is the corelation betwen mistake chances (both players' ones) and game length.

Here is what I think:

If your move no-mistake chance is A, and your opponent's one is B, then for X moves you are likely A^X not to make a mistake, and your opponent is likely B^X not to make one.

The chance you both won't make a mistake within X moves is C=(A*B)^X.

The chance your opponent will make a mistake within X moves and you won't is D=A^X*(1-B^X).

The chance you both will make a mistake is E=(1-A^X)*(1-B^X).

After an opponent's mistake which is the first in the game, if you don't make one until the end, you win (that should be obvious). So, once the opponent makes the first mistake, if you can make sure not to make a mistake, you will win, but the matter is how many moves it will take you. If you need Y moves to win a position if your opponent has blundered within X moves, then you have A^X*(1-B^X)*A^Y=A(X+Y)*(1-B^X) overall chance to win due to exploiting opponent's mistake without making one. That's the case with "White to play and win" or "Black to play, checkmate in 5" situations.

Of course, you may not find the correct way, and make a mistake yourself. Then you go into the E=(1-A^X)*(1-B^X) case.

C=(A*B)^X can describe some endgame situations where a draw is possible in addition to whole games.

As a conclusion:

If you play against a more "careful" than you opponent (A<B) you can increase your result expectation by:

- reducing game length (in term of moves) - by early quick exchanges or by offensive play aiming for a checkmate;

- trying not to reach a complex board position;

- reducing candidate moves' number for both players (and in this way reducing mistake chance difference) - by exchanges; closing the position; etc;

- increasing your opponent's chance to blunder - playing openings he's not familiar with, using decoys, etc.

If you play against a less"careful" opponent:

- increasing game length - less exchanges, slow play going for an endgame;

- trying to achieve a complex board position;

- increasing candidate moves' number - less exchanges; open game;

- playing well-studied openings that you are comfortable with;

- being familiar with endgame solutions to finish off your opponent once he blunders.

You can't determine how many moves it will take you to make a mistake. You can say how many moves will make it possible to reach a given probability that a mistake has been played by you.

For example, you can search for how many moves you need to play to be 50% sure that at least one of them is wrong.

With 1% mistake per move chance, that would be 69 moves, since 0,99^69=50%. In other words, if you find a 69-move game played by you, there is 50:50 chance you will have blundered in it (of course that's provided mistake distribution is random).

Many master level games are finished within 40 moves. With 1% mistake chance per move, the chance to blunder in a 40 move game is 33,10%. So, if you play 40 move games only, you will win or draw at least 67% of the time. The win/draw proportion will depend on your opponents.

The matter is how can we approximate game length with players' mistake chances (if that's possible), so that we can have the connection between player mistake chances and game outcome without using game length as an independant variable.

"You can't determine how many moves it will take you to make a mistake."

That's true, but that's not precisely what probability is meant to do. If you flip a coin ten times, you just might not get 5 tails and 5 heads. But if you stretched that out to a million times the pattern will show that one possibility is not really more likely than the other.

With 99% accuracy, sometimes you will go 400 moves without making a mistake; others you make 5 mistakes in those 100 moves. But there is a logical pattern that will eventually show up.

99% ?