classical probability-die rolling

[dboybbfs]

Hi,

My teacher said this would be a hard problem and boy was he true!

Find the classical probability of getting a total of 24 when 5 balanced dice are rolled. I also have to show all work but I don't even know where to start.

any help is greatly appreciated.

 

[galactus]

There are \(\displaystyle 6^{5}\) possible rolls with 5 die.

Now, count up how many different 5 number combinations of 1,2,3,4,5,6 you can use to sum to 24.

I can think of 5 real quick.

66633
64455
66444
66624
66453

 

[pka]

Find the classical probability of getting a total of 24 when 5 balanced dice are rolled. I also have to show all work but I don't even know where to start.

I normally do not just work a problem. But the sheer size of this one makes working out on your own quite impossible in my view.
Rolling five dice we can get a sum of 5 in one way: (1,1,1,1,1).
Rolling five dice we can get a sum of 6 in five ways: (2,1,1,1,1), (1,2,1,1,1), (1,1,2,1,1), (1,1,1,2,1), (1,1,1,1,2).
Rolling five dice we can get a sum of 7 in fifteen ways: example; (1,2,2,1,1) or (1,3,1,1,1) etc.

One can see how very difficult a counting job we have in this particular problem.
Without your knowing so advanced counting techniques, I see no hope of actually finding an answer.

Now it is easy to see that the smallest sum is 5 & the largest sum is 30. (WHY?)
It is also easy to see that the total number of possibilities is 6<SUP>5</SUP>=7776. (WHY?)
Thus, P(sum=5)= P(sum=30)=1/7776.

To find the number of ways we can get a sum=24 would almost impossible if you don’t know the following.
Look at the polynomial \(\displaystyle \L
\left( {\sum\limits_{k = 1}^6 {x^k } } \right)^5\).
When we expand that expression, it will contain each of these terms: \(\displaystyle \L
x^5 ,\;5x^6 ,\;15x^7 \;,\& \;205x^{24}\).
In other words, that expansion tells us that there is one way to get sum=5, five ways to get sum=6, 15 ways to get sum=7, and 205 ways to get sum=24.

You can finish this off.

 

[galactus]

Wow, generating functions. Quite clever, pka.

 

[JakeD]

After seeing pka's reply, I had to check out what my probability texts said on this. I found that Epstein in The Theory of Gambling and Statistical Logic uses the same polynomial expansion as pka shows to derive the formula

\(\displaystyle \L x = \sum_{j=0}^{M} (-1)^j {n \choose j}{s-6j-1 \choose n-1}\)

where \(\displaystyle \L s\) is the sum of the dice, \(\displaystyle \L n\) is the number of rolls and \(\displaystyle \L M = \mbox{floor}((s-n)/6)\).

For \(\displaystyle \L s = 24\) and \(\displaystyle \L n = 5\), this yields

\(\displaystyle \L x = {5 \choose 0}{23 \choose 4} - {5 \choose 1}{17 \choose 4} + {5 \choose 2}{11 \choose 4} - {5 \choose 3}{5 \choose 4} = 205,\)

the same result.

LATER: Oh, I see a computer algebra system such as QuickMath can be used to actually expand the polynomial and get all the coefficients very quickly. Nice!

 

[JakeD]

Listing out all the combinations that sum to 24 is not so bad.

55554    5
65544   30
65553   20
66444   10
66543   60
66552   30
66633   10
66642   20
66651   20
Total  205

The second column is the number of distinguishable ways to rearrange the 5 digits. It is calculated using the multinomial coefficient

\(\displaystyle \L \frac{n!}{n_1 ! n_2 ! \ldots}\)

where here \(\displaystyle \L n = 5\) and the \(\displaystyle \L n_i\) are the number of digits of each type. So for 55554 the number of rearrangements is

\(\displaystyle \L \frac{5!}{4! 1!} = 5.\)

To try the solve the problem this way you really have to know the total 205 to make sure you have all the possibilities. Too many chances for error otherwise. So while this approach to the problem is not impossible, it is difficult to do with any confidence that you have it right unless you already know the answer.