Probability of 2 dice, sum < 2? please help! T_T?

First we find the number of outcomes that give you a sum of exactly 6. To get a sum of 6, the number on the first die must be strictly less than 6, thus there are five possible outcomes for the first die. For each of these, exactly one number on the second die will give you a sum of 6, so there are 5 ways to roll a 6. By the same logic, there are 4 ways to roll a 5, 3 ways to roll a 4, 2 ways to roll a 3, and 1 way to roll a 2. Therefore, the total number of ways to roll a die that is strictly less than 6 is 1+2+3+4 (the 5 is not included, since you said less than and not less than or equal to). This is ten possibilities, out of 36 total, for a probability of 5/18.

In general: Where n ≤ 7, the probability of rolling a number less than or equal to n is [k=1, n-1]∑(k)/36 (that is, the sum of the first n-1 numbers, divided by 36). If you remember gauss's formula, you may write this even more explicitly as (n-1)n/72. For n > 7, this formula does not hold, since there are only 5 ways to roll an 8, not 7 (since there are only six possibilities for the first die, and a roll of 1 no longer works). Thus, for n>7, the number of possiblities is 36 minus the sum of the first (12-n) numbers (since these would be the number of ways to roll numbers that are too high). Therefore, we have:

Probability that a number less than or equal to n is rolled = {(n-1)n/72 if n ≤ 7; 1 - (12-n)(13-n)/72 if n > 7}

There are 36 possible ways two dice can end up. I would suggest that you make a grid with the numbers 1 - 6 on top and 1 - 6 on the side. There will be 36 squares. Add the numbers in

the squares that meet. For example, put a 2 in the square that goes with 1 and 1. That way you'll have a way of counting what you need. For your problem, you have to look at sums of 2,3,4 and 5. Count the number of squares that contain those numbers. Then divide that number by 36 for your probability.

Look at www.geocities.com/mr_kaaaaa/dice.html . There's a decent chart there.

Less than 6

Die 1 Die 2

3 2

2 3

4 1

1 4

2 2

1 1

1 2

1 3

2 1

3 1

10 possible outcomes

out of 6 * 6 = 36 combinations

10/36 = 5/18

The answer is 4/11 or .3636(infinity)

There are 11 total outcomes: 2,3,4,5,6,7,8,9,10,11,and 12. Four of them are less than 6: 2,3,4,and 5

Number of favorable outcomes 4

over

total numbers of outcomes 11

More: Sorry, I probably didn't word this right. The formula for problem such as these are the (number of favorable outcomes over the total number of outcomes). There are six possible outcomes that the sum will be under six: 1+1, 2+2, 1+2, 2+3, 1+3, and 1+6). There are elevin outcomes in all. (I hope I got it right this time, it's been a while since I completed a probability problem.)

With two dice you have 12 sides because each die has 6 sides.

1+1, 1+2, 1+3, 1+4, 2+2, 2+3 are the only things less than six, so that is 6 combinations. I don't know from there because I learned it right before I retook my GRE, but does that help?

start with 6x6= 36--this is the total numbe rof possible outcomes.

One of those outcomes is the number 2 (1+1)

Two of those outcomes are the number 3 ( 1+2 and 2+1)

Three of those outcomes are the number 4 ( 1+3 and 3+1 and 2+2)

Four of those outcomes are the number 5 ( 1+4,4+1,2+3,3+2)

So, ten of the 36 outcomes total less than 6 so its 10/36..i think this is right

It does not take for ever.

Add the probabilties of each sum.

2=1/36

3=2/36

4= 3/36

5= 4/36

Total

10/36 = 5/18

First of all, I assume you mean 2 6-sided dice.

so you need to get 5 or less, there are 6 ways to obtain this result out of a possible (6x6) 36 roll combinations, so 6/36 or 1/6 or 16% chance.

school takes time forever and ever!!!