Math Is Fun Forum

How many different rolls of 6 dice are there?   6 dice with 6 different outcomes (1 through 6) each.  That would be 6^6 = 46656.

How many different ways to roll zero 6's?   6 dice with 5 different outcomes (1 through 5) or 5 ^ 6 = 15625.

How many different ways to roll one 6?   5 dice with 5 different outcomes (5^5 = 3125) plus one dice with one outcome (a six).  The six could be on any one of the six dice (6 choose 1).  So there are 3125 * 6 = 18750 ways to roll one dice.

Thats all you really need to know to compute the chances of rolling 1 six (18750 / 46656 =~ 40.2%) and the chance of rolling one or more sixes ((46656-15625) / 46656 =~ 66.5%) but let's continue anyway.

How many different ways to roll two 6's?   4 dice with 5 different outcomes (5^4 = 625).  And there are 15 different pairs of dice that the would roll the two 6's (6 choose 2).   625 * 15 = 9375. 

How many different ways to roll three 6's?   5^3 (3 dice with 5 outcomes each) * 20 (6 choose 3) = 2500

How many different ways to roll four 6's?  5^2 * 15 (6 choose 4)= 375

How many different ways to roll five 6's?  5^1 * 6 (6 choose 5) = 30

How many different ways to roll six 6's?  5^0 * 1 (6 choose 6) = 1

Looks good.   Very easy read.  Young students should be able to handle it as well as their parents who may need a refresher when it comes time to help out with the school work. 

In your example of the square root of -9, shouldn't you include -3i as a solution?

The one labeled Apples must be oranges because if it was Apples and Oranges as you suggest , then the crate labeled Oranges would contain oranges and that can't be since we were told that ALL of the labels were wrong.

In the case where 2 balls (A and B) weigh the same as 1 ball (C), the remaining, un-weighed ball (D) has to be either 2 or 4 kg.  Weighing D against C will then give one of the following results:       
1)  D is heavier so it must be 4 kg and C is 3 kg.  You can then weigh A against B to figure out which is 1 kg and which is 2 kg. 

2)  D is lighter so it must be 2 kg and C is 4.  You can then weith A against B to figure out which is 1 kg and which is 3 kg. 

I haven't even looked at the inequalities yet.   

I personally think you should be able to figure out the wieghts of each ball with ZERO weighings.   We're talking KILOGRAMS here!

I don't know what your country is, but I can get to it from the US.     Try again later.   Maybe the site was down temporarily.

I believe there's a mistake in the 3rd line:   Shouldn't it be a 2 rather than a 2p in the numerator?  So from there:

The last step was to get it the form necessary for the quadratic function.  I'm not sure if that's the best way to go from here or not.

An easy way to see 2 unsolvable ones:   Use the "Select Game" option and when it asks you to enter a number between 1 and 1000000, enter -1 or -2 instead.

Okay, here's a reasonable "proof" that the probabilty of the  sum of n dice is even is 50%.   

Roll n dice.  We're going to compute a "rolling sum", i.e. adding the dice together one at a time.   Look at the first die.   It has a 50% chance of being even (2, 4, or 6).   Add the second dice to it.   In the 50% of cases where the first die was even, there is a 50% chance that the sum will remain even (the second die is even) and 50% chance the sum will become odd (the second die is odd).  In the 50% of cases where the first die was odd, there is a 50% chance the the sum will remain odd (the second die is even) and 50% chance the sum will become even (the second die is odd).  So there is a 50% chance the sum of the first 2 dice will be even. 

Add the 3rd die to the sum of the first 2.   Follow the same logic.   There is a 25% of changing the running sum from even to odd, 25% chance of the changing the running sum from odd to even, 25% chance of the running sum remaining even and 25% chance of the running sum remaining odd.   So there is a 50% chance the sum of the first 3 dice is even.

You can keep repeating this process of adding another die to the running sum until you reach n dice.   The probabilty of the sum being even will always be 50%.   You could expand this to say that the probability of the sum of any n random numbers being even is 50%.

EDIT:   You can just skip to post #4 if you want.   Posts #2 and #3 are my ramblings while trying to figure it out. 

Rather than trying to compute the actual odds, let's think it through first.  It helps to know that the sum of opposites sides of a die always total 7.   Thus, 1 and 6 are on opposite sides of a die.  So are 5 and 2.  And 3 and 4. 

Let's say we have n dice.  Imagine you've listed all of the different ways of rolling a sum of T using those n dice.   Now think about the sums on the bottom.   The sum of the bottoms would be 7n - T.   So the odds of rolling a sum of T would be the same as rolling a sum of 7n - T.   Let's take a simple example of 3 dice for a total of 18.   There's only 1 way to roll an 18 and that's 6-6-6.   What's the sum on the bottom?   We know that on the opposite side of a 6 is a 1 so the sum would be 3 which is also 7n - T (21-18).   The odds of rolling a 3 is the same as rolling an 18.   

Likewise, and still using 3 dice, the probability of rolling a 4 is the same as rolling a 17.  Same goes for 5 and 16, 6 and 15, 7 and 14, 8 and 13, 9 and 12 and 10 and 11.   So for each odd sum, there is an even sum (7n-t) with the same probabilty.   Notice that for each set of numbers, one of the sums is even and one of them is odd.   So the chances of rolling an even sum is equal to rolling an odd sum?      Yes, but only if your using an odd number of dice. 

If you're using an even number of dice, this won't work.   For 2 dice, the odds of rolling a 2 is the same as rolling a 12.   But both of those sums are even so they don't "cancel" each other out nicely like they do for an odd number of dice.   

Hopefully my explanation for the odd number of dice makes sense.   I'm too tired to think about rolling an even number of dice right now.

Edit: Stanley beat me to it.    I need to get my LATEX syntax down!