Is there a stop to Internet Backgammon cheaters?

Anonymous
2010-04-09T18:01:10+00:00

So I have enjoyed playing Backgammon in the past, but nowadays is just IMPOSSIBLE.  I am sure many others have noticed this.  There are players out there that can control the dice at least for them, and some can even move more chips than the allowed by standard rules.

For example, not 10 minutes ago, I rolled 6 and 5, then the guy doubled me on his first roll, and got double three, then I rolled and I got 4 and 1, then he rolled and got 3 and 1, then I rolled and got 3 and 2, then he rolled and got double 5 eating my lonely chip in position 1 and covering position 3 from position 8.  Do you call that a lucky roll?  I call that an engineered roll, if you know Backgammon.

Yesterday, a guy moved 6 chips using a double 5.  That same guy rolled 4 consecutive pairs.

Yesterday, I had 5 out of 6 positions occupied in my side and I ate one of the other player's chips.  I did this repeatedly.  The guy just lost 1 roll out of 4 chips I ate.  He also had 5 of 6 positions occupied.  He ate a chip of mine twice.  I had to roll some 10 times before I could get out each time.

If you know Backgammon, you know this is NOT right.  Losers out there have found a way to cheat, and that ruins the fun for us, the ones with a brain.

Microsoft, please put an end to this.  Your Windows license is expensive, and I paid for it, and I expect to be able to enjoy all the features in the OS, and that includes the included Internet gaming.  Roll a patch that can end this blatant cheating!

MCP

Windows for home | Other | Gaming

Locked Question. This question was migrated from the Microsoft Support Community. You can vote on whether it's helpful, but you can't add comments or replies or follow the question.

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Anonymous
2010-04-10T13:55:17+00:00

Hi,

If you have any feedback’s, then I would suggest you to post here:

https://connect.microsoft.com/?wa=wsignin1.0

 Please respond back with the status of the issue and let us know if you have any issues.

 Azam – Microsoft

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  1. Anonymous
    2011-11-06T22:39:32+00:00

    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.

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  2. Anonymous
    2011-11-03T23:15:29+00:00

    Can a cheater or scam-er block you from playing???

    BC will not down load anymore. Can't figure it out. Wish I could do this to some of these cheaters.

    I get a error code ox8004fffe. Does anyone know what that means or how to fix it?

    HELP???

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  3. Anonymous
    2011-11-03T21:42:14+00:00

    I could not agree with you more.

    The cheating is so common it is unusual to get a player that does not cheat.

    The other day I got into a game with some one who was not a cheater. I could not believe great game BG is when its played as you would play a regular board game. I was lucky enough to play the same person (I think) 2 times. Gee what a pleasure.

    I get so sick of some players that insult your intelligence by wanting a double or nothing game.

    I sure don't mind double or nothing, but they are so stupid they roll a 6/6 or a 5/5 on the first roll every time. All of the other rolls are either 4/4 2/2 1/1 6/3 1/3  gee where are all of the 5/1 4/3 2/5 and a entire gambit of other combinations. And yes if they ever fall behind the doublets come in truck loads.

    I have never left a game unless I knew for sure the game was cheated.

    NOW though I have a new problem. As of 5:15 I cant get in the game. No matter how many times I try

    the "Looking for a player" will run it's self out and their doesn't  seem anyway to reset it.I keep getting the error code ox8004fffe

    I am not one of these super smart computer brains that knows how to interpret the code and remedy the problem.

    Help anyone?

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  4. Anonymous
    2011-11-03T19:55:08+00:00

    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

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