# Backgammon with negative value dice [closed]

What if we had a pair of dice for each player. 1 dice with positive values(1 to 6) and the other with negative values(-1 to -6) instead of both dice with positive values?

So the player who rolls the dice will have to move forward for the positive dice and move backward for the negative dice.

This can change the entire strategy of the game.

• Sorry. What's the question? Feb 6, 2014 at 16:47
• "This can change the entire strategy of the game" - surely this makes the game last much much longer? Try playing it on your own without an opponent and count the number of rolls needed to get your pieces home. Feb 6, 2014 at 17:57
– KSK
Feb 7, 2014 at 9:02
• @KSK, first off, welcome to the site! The problem is that you didn't pose a question in your post. This site isn't a regular forum where people come to discuss things. It's specifically a Q&A site, where people come to get answers to questions, and to help people by answering specific questions. Your post simply mentions an idea that you have, but it doesn't ask any specific question that people can answer. Feb 7, 2014 at 14:40
• @KSK, I don't think it's so much that people here are taking the games really seriously; just that the format of this particular system (Stack Exchange) is pretty particular in terms of its purpose. These series of sites are not meant for general discussion about things; but rather for people be able to get answers to questions they have. Feb 9, 2014 at 20:46

Do some statistical analysis on the possible rolls:

• The average value of all rolls will be zero.
• 41.667% (5/12ths) of all rolls will be negative.
• 41.667% (5/12ths) of all rolls will be positive.
• 16.667% (1/6th) of all rolls will be zero.

It seems like it would have a random walk type of effect, with no one being able to move very far (as @tttppp pointed out in his comment to the OP). Any positive moves will eventually be offset by negative ones, and vice versa.

• I ran a quick simulation of getting one piece 24 spaces and in 10 runs I saw it usually take between about 100-200 rolls (the lowest was 26). Feb 6, 2014 at 19:06
• I got MUCH larger numbers from my simulation. Over 100 tries, the mean number of rolls needed to get a total sum of 24 was 1,544,901; the max was 7,411,661; the min was 10,340. Feb 7, 2014 at 14:59
• If you assume 1 minute per turn, then for one player to get one of his pieces to the end, you will have to play for about 3 years non-stop. Feb 7, 2014 at 15:00
• Which translates to 90 years non-stop for a 2-person game, with average luck. Feb 7, 2014 at 15:01
• @GendoIkari I've just re-entered the code I used and added a loop: ideone.com/LMy49o There are quite a few rules that weren't specified in the question (e.g. what happens when going backwards past the start, what happens if you throw doubles), so I'm not too surprised that we have different results. Sample output from my code: "The average number of turns in 1000 games was 87 (min = 8, max = 631)" Feb 7, 2014 at 18:30