# Let's Play Dice!

1. Oct 19, 2003

Ok, a friend at another forum posted this challenging probability problem.

Mike throws 20 6-sided die on the table. He then adds up the results of all the dice, resulting in a number between 20 and 120. What are the chances the total number is equal to or greater than 100?

How, in detailed steps, would one go about solving this???[?]

2. Oct 19, 2003

### quantum

I'll be honest, I have no clue in hell... However by the sounds of the question I think this is better suited for the homework section. You sure it was your friend, and not a mathematics textbook that ponders the answer?

3. Oct 19, 2003

4. Oct 21, 2003

### HallsofIvy

Staff Emeritus
The "detailed steps" are long and tedious. I don't think you are paying us enough!

One way to get "above 100" is to get 120. To get that, all dice must land on 6. The probability of that is (1/6)20.
Another way is to get 119. To get that all dice except one must come up 6 and the one die must be 5. The probability of that is
20C1(1/6)20= 20(1/6)20.
Another way is to get 118. One way to get that is for all dice except one to come up 6 and that one die to come up 4. The probability of that is 20C1(1/6)20=
20(1/6)20. Another way to get 18 is for 18 of the dice to come up 6 and the other two to come up 5. The probability of that is
20C[/sub]2[/sub](1/6)2= 190(1/6)20.
Adding those gives a probability of 210(1/6)20 of getting 118.

Now continue like that down to 101. Believe me, they get really messy really fast!

5. Oct 26, 2003

I found another way. If one takes the polynomial

(x+x2+x3+x4+x5+x6) and raises it to the power of however many dice there are, the coefficient of the sum you wish to get will be the numerator in the probability.

So if we want to know the probability of getting the sum of 120, we look for the coefficient of x120 when that is all factored out. It turns out to be 1. So then we divide by 620 and we have the probability. Do find the probability of a sum 100 or higher, simply sum up the coefficients and then divide it by 620. It seems to work out pretty good!

6. Jan 7, 2004

### sam2

What about using the Central Limit theorem?

X(1) = The outcome from dice number 1, it can be 1,2...6

We want Y = X(1)+X(2)+... X(20). This should be approximately normal in distributoin. Mean = 20 *3.5, variance = 20*2.92.

We want P(Y>100)

Sam

7. Jan 7, 2004

### NateTG

The odds are:

$$\frac{137 846 528 820}{3 656 158 440 062 976}$$

8. Jan 7, 2004

So, the method I posted works.

9. Jan 7, 2004

### NateTG

Yes.

In this case it simplifies out to
2d choose d

It's also easy to spot a pattern if you work your way up throug varying numbers of dice.

Last edited: Jan 7, 2004