Tossing dice?

Question

Hi.
When you draw 4 cards from it can be computed by 52C4 but how do you use the same notation to compute the probability of getting some number on a dice? I know the chances are 1/6 but how do I use the C notation.
Thank you.

Answer 1

The 52C4 is a counting formula, not probabilistic as such. You may need the counting result to compute a probability, but not in itself.

With a die, you're not counting on a toss but direcly computing the chance of a given face's value (or perhaps more, like an even number).

Answer 2

Suppose you want to know the number of ways of drawing 4 red cards from a deck of 52 cards. There are 26 red cards so if you draw 4 cards without replacement you can compute the probability directly as

26/52 * 25/51 * 24/50 * 23/49 = 26!48!/(22!52!)

Using "C" notation you can also write it as

26C4/52C4 = 26!/(4!22!) / [52!/(48!4!)] = 26!48!/(22!52!)

Now with dice it becomes a little different since you are not choosing anything. However, the "C" notation represents binomial coefficients. So in the above problem the "C" notation represented the number of ways of selecting m items out of n. But for dice you would use it in a slightly different way.

The binomial theorem says that

(a+b)^n = nC0 a^n b^0 + nC1 a^(n-1) b^1 + ... + nCn a^0 b^n

Now suppose we let a=p and b=1-p.

Then we know that [p + (1-p)]^n = 1

Using this we can find the probability of getting m 3s when rolling n dice.

Pr[rolling m 3s] = nCm p^m (1-p)^m where p=1/6

the equation works like this

Pr[roll m 3s with n dice] =
ways n dice can be arranged into m 3s and n-m non-3s *
probability of rolling exactly m 3s with m dice *
probability of rolling exactly n-m non-3s with n-m dice.

This is just the m+1st term from the expansion of
(p + 1-p)^n where the 1st term is the probability of rolling exactly 0 3s, the 2nd term is the probability of rolling exactly 1 3, the 3rd term is the probability of rolling exactly 2 3s, etc.

And since [p + (1-p)]^n = 1 for any p then all these probabilities will add to 1 as we should expect.

Now if you roll just 1 die

[p+(1-p)]^1 = 1C0 p^0 (1-p)^1 + 1C1 p^1 (1-p)^0
[p+(1-p)]^1 = 5/6 + 1/6 = 1

[p+(1-p)]^1 = probability of rolling exactly 0 3s + probability of rolling exactly 1 3 with 1 die.

Not much of a problem. Now you want to know the probability of rolling 1 3 out of 3 dice

Pr[roll 1 3 with 3 dice] = 3C1 (1/6)^1 (5/6)^2 = 34.7%

or in general

Pr[roll m 3s with n dice] =
nCm *
probability of success^number of successes *
probability of failures^number of failures

Eseentially this is the same framework as the card problem but from a slightly different direction since with cards you are drawing without replacement but with dice you have replacement since you can roll the same number many times.

Hope this helps.

Answer 3

The C notation does not work with dice (at least not in the way you're thinking). The reason is that you're not "choosing" anything from anything.

Each toss of a die is independent of each toss before it. Not so with drawing cards. Whenever you draw one card, it changes the probabilities to follow. They are simply not the same kind of problem. Rolling dice is like drawing cards and replacing them.

Answer 4

The probability of rolling any particular number on a fair die with s sides is simply 1/s.

If you have a six-sided die, the probability of rolling a 5 is 1/6. The probability of rolling a five is also 1/6.

You do not use combinations for this type of calculation.

Answer 5

52 "choose" 4 is not the same as 6 "choose" x - that is probability, not combinatorics...