OK, guys. According to http://www.chessandpoker.com/dice_strategy_guide.html the probability of rolling the following results with two six sided dice are as follows:

Result Probability
2 2.78%
3 5.56%
4 8.33%
5 11.11%
6 13.89%
7 16.67%
8 13.89%
9 11.11%
10 8.33%
11 5.56%
12 2.78%


Using these probabilities, how would I calculate the probability of rolling a 7 or greater?

I know that the probability of rolling an 11 and then a 12 would be the product of .0556 multiplied by .0278, but I forgot how to calculate the probability of rolling a 7 OR a 8 OR a 9, etc.

these are just probabilities. with one d6 your probability of rolling an exact number (any you want) never increases or decreases never mind how many times you roll the dice and what result you had before.

same goes for two six-sided dice. if you roll two six sided dice six times your result will not show 2x1, 2x2, 2x3, 2x4,2x5, 2x6 and any other combination of. so rolling a 7 or better still has a probability of 1/6th, whereas the "xyz"-factor (don't know the english word for that matematical expression) will lead you to one dice always having a greater number than 3 or better. making the other dice easier to reach target number of (7 or higher).

probability is thus still 1/6th. all other dice-rolling-system stuff is for die-hard-vegas-craps-players who try to find a system beating the bank.

Well, he wants the probability of 7 or better on 2D6
there are 6 chances for a 7, that's 1/6
there are 5 chances for 8
4 for 9
3 for 10
2 for 11
1 for 12

that's 21 of 36 possible outcomes.
or, roughly 58.3%
The odds are slightly in your favor, but only slightly.

Thanks, guys! I appreciate.