Is there a stop to Internet Backgammon cheaters?

You should read Tripp523 on May 22,2011.  He actually used a cheat program.  They do exist and are being used.

I do find many players who play an honest game and enjoy both winning and losing with a good player.  However I find very few player who can play or know the odds.  No one plays the advised opening moves (except the obvious like 61,31 etc) but no one plays the best moves for 21,23,25,32,34,36,41,45,51,46 etc, as published in board games and books on backgammon.

I have played BG for 30 years on the 'board' and have read many books on strategy and tactics.  You might enjoy "Backgammon for Blood" by Bruce Becker.

Can't help you with the 5:15 problem but if you suspect a cheater, then don't play - quit right away.  If enough people boycott them, maybe they will go away and we can have enjoyable games again, winning some and losing some.

See my 'how to spot a cheater'  that I wrote on page 1.

Happy hunting for non-cheaters.

It's not cheaters...the board going crazy is some glitch in the program and usually will continue to run properly even if you look and see they have a rock already home until you move to a position that is in the glitch and then the program will terminate.  There are no cheaters only people that don't understand backgammom.  I can tell a newbie usually on their first move or when they double me right away.  Backgammom is a game of strategy...taking risks...setting traps...thinking ahead.  Sure I get beat by newbies that get lucky with there rolls and basic moves but, if you get beat by someone that thinks and takes risks...that's not cheating.

Well, it seems that God sent you some lightingbolt and now you accept that getting a double or any combination is an odd of 1/36, in backgammon it does not make any sense talking about one dice, because you are always going to use two and those two ALWAYS will make a combination, so no matter what your book says, in BG you are always get 1/36 on every combination.

And is not the sum of probabilities !!!: 1/6 + 1/6 = 1/3

I think you have to dive into factorial when talking at least two factors in n combinations

Sir, you obviously are weak in Math and English and I am not here to teach you.   You do not understand the problem posed in English nor the Math solution.

Just to be fair, I will pose the problem just once more (and for the benefit of others).

DEFINTION: come out roll -- the first roll of one die

Let's say, today I play a game and my opponent rolls a 6 on the come out roll.

It doesn't matter if I roll or not.

That probability is 1/6 that he/she rolled that 6

Tomorrow I play again and we both roll the same number (any number) on our come out rolls.

That is a probability of 6/36 or 1/6.  (this is the same as rolling any double)

There are two events here.

Lets call today's event, event A.  Lets call tomorrow's event, event B

Then the probability, P, of either event A happening OR event B happening is...

P(A or B) = P(A) + P(B) - P(A and B),   where P(A and B) = P(A) X P(B)

P(A or B) = 1/6   +  1/6  -  (1/6 X 1/6)

                =  1/6   +  1/6  - 1/36

                = 30.5 %

To put it another way by enumeration...

NOTATION: The first number in parenthesis is my opponent's roll; the second is my roll.

All possibilities...

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Event A or Event B

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) (1,1) (2,2) (3,3) (4,4) (5,5)

P(A or B) = 11/36 = 30.5%

What actually happened...

Using a program I developed, I counted event A and I counted event B.

I played a total of 344 games.  Event A or event B happened 242 times.

The expected probability is 30.5%

I got 242/344 or 70.3%

This says that the rolls are not random.  So if either event A happens or event B happens, then your opponent is likely to be cheating.