1/12

The answer is there are 6 sides on each die, therefore, 6 times 6 is 36, giving us all the possible outcomes. You can roll an 11 one of two ways a, 5 and a 6 or a 6 and a 5. You can roll a 12 only one way. This gives us 3 possible outcomes of 11 or over out of 36. Which reduces to 1/12.

Casey's right, 3/36. Peter, it isn't an ordered problem but there are two rolls out of the 36 possible that can be 11 (6 on die a, 5 on die b, and vice versa). Add that to the one way you can roll boxcars (6 on die a, 6 on b), totally of 3 winning instances out of 36 possible.

OK, you are right Casey...I made myself believe that what matters is the final sum, so unless they come up with dice that have 12 sides...I'm wrong :) Thanks Grant

A single die has six possible sides it can land on. That means two die have 36 possible outcomes, but only 18 unique. The probability that the two rolls die sum to 11 is 1/18 He answer is 1/18, not 1/12.

*the answer is 1/18 for the sum of an 11 and 1/9 for 11 and over. The correct answer is 1/9. (5,6)-(6,5)-(6,6)-(6,6) 4/36=1/9

The job is a numbers Game. The only for a WIN is what is in the bank and that will not be spent! (so you factor your accounts by % at the bank), to throw a die with a probability, not a indefinite, witch means there is a meaning and a definition, it means the there is a precedent and in a accounting career, president does not court so 11 is offered witch you all know inherently know because you all calme 11 is there on the twice, 6 is off the board being that is how many sides there is and any number OF, AGAIN OF of [əv] PREPOSITION expressing the relationship between a part and a whole: 6 is 7 so it is 1 in 7 to the number of your choice. YOU WILL ROLL THE NUMBER OF YOUR CHOICE WITHIN 7 ROLLS.

"Dices"? Did you just say "dices"? Excuse me for answering your question with a question, but what was the highest grade of grammar school education you recall passing? If I got this job, is there any possibility, that is, any chance higher than zero, that you would be my boss? How did you get to your current position in this organization? Can I go now?

I'm with Jeff and Carlos- "dice" is the plural of "die". Geez. However, if the question actually is "what is the probability of getting 11 or greater on a single throw with a pair of dice," then the answer is 50%. It either will or will not be 11 or greater.

1) Same as Jeff - "dices" makes no sense 2) How many sides are there on each die? Don't assume that all dice have six sides.

All possible results when throwing 2 dices ( 2-3 combination IS THE SAME as 3-2, you know it if you ever played dice, I don't know what are the probability purposes Casey is talking about ): (1,1) (1,2) (2,2) (1,3) (2,3) (3,3) (1,4) (2,4) (3,4) (4,4) (1,5) (2,5) (3,5) (4,5) (5,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) so there are 21 combinations, only (5,6) and (6,6) are >=11, so the answer is: 2/21=9.52%

I think the question is oriented to find out if you are actually listening. You can not throw 11 AND over. You can throw 11 OR over. So, the answer is 0. Impossible.

If you are rolling two dice the chances of getting an 11 and OVER is 2/12= 1/6. You can roll a 6 and a 5 to equal 11 and you can roll a 6 and a 6 and get 12 which is OVER 11, therefore the answer should be 1/6.