posted
I have a probability conundrum that I (a mediocre math mind at most) cannot deduce. It's a simple question really. When rolling two die, what is the probability of rolling any particular roll. I ask, because in the last year, my family has got really big into backgammon.

Jatraqueros, I need some ayuda por favor.
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posted
There are 6 ways the first die can fall, and 6 ways that the second can fall. If you put it all together, there are 36 ways that one die can fall, and then a second fall.

To calculate the odds, you just need to figure out how many ways you can get the result you're talking about.

For example, there's only one way to get snake eyes: the first one has to be a 1, and the second has to be a one. So the odds of getting snake eyes is 1/36.

But to roll a 3, there are two ways to do it: Roll a 1 then a 2, or roll a 2 then a 1. The odds of rolling a 3 are 2/36 or 1/18.

To get 4, there are 3 options: 1 then 3, 3, then 1, or 2 then 2. Odds are 3/36 or 1/12.

posted
I don't know how backgammon is played. So it depends on what you are asking.

Does it matter if a 7 is rolled or a 4 and a 3? I'm pretty sure the same odds are there if it is a 1 1, as it is for a 4 3. But I am pretty sure a 7 has better odds than a 2 being rolled.
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posted
Yes. 7 is the most likely, followed by 6 and 8, followed by 5 and 9, followed by 4 and 10, followed by 3 and 11, followed by 2 and 12.
Posts: 16551 | Registered: Feb 2003
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Is calculated by determining how many possible combinations yield that result. There is only one possible way to get a 2 (1,1) or a 12 (6,6) while there are 6 different ways to roll a 7 [(1,6)(2,5)(3,4)(4,3,(5,2),(6,1)]. There are 36 possible rolls, so the odds of rolling a 2 are 1/36.

Edit: I knew full well I'd be ninja'd when I hit "Reply". Oh well.
Posts: 5656 | Registered: Oct 1999
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Is calculated by determining how many possible combinations yield that result. There is only one possible way to get a 2 (1,1) or a 12 (6,6) while there are 6 different ways to roll a 7 [(1,6)(2,5)(3,4)(4,3,(5,2),(6,1)]. There are 36 possible rolls, so the odds of rolling a 2 are 1/36.

Edit: I knew full well I'd be ninja'd when I hit "Reply". Oh well.

Tried talking my family and friends into letting me make a deck of cards for Catan based on that, instead of the dice. They wouldn't allow it.
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posted
In backgammon the particular roll of the dice is important, not just the total. The probability of each pair is 1/36. The probability of every other combination is 1/18.
Posts: 1794 | Registered: Jul 2002
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posted
Here's a page with a nice bunch of probabilities to consider when playing backgammon. Might have to bone up on some of these before I play with my dad again.
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