Given two ordinary six-sided dice, what is the probability of rolling
a 3 and a 4 in that order?

I know this formula is needed:
P(E)=n(E)/n(S)

I'm not sure how to work the problem out.  Any help is greatly appreciated.

---

Request for Question Clarification by mvguy-ga on 19 Nov 2006 17:20 PST

The question seems ambiguous.

One possibility is that you want to know the chance of rolling one die
and getting a 3, then rolling the other and getting a 4.

Another is that you want to roll both dice and get a total of 3, then
roll the two again and get a total of 4.

Another is that you want to roll both dice and get at least one 3,
then roll the two dice again and get at least one 4.

I'd be happy to answer the question, but I wouldn't want to spend my
time giving you the answer to the wrong question (nor would you want
to pay me to give you the answer to the wrong question).

Also, do you want that formula to be used to arrive at the answer? I
was planning on using a different method. If you want that formula
used, please indicate what P, E, n and S stand for.

Thanks!

---

Clarification of Question by unknownentity-ga on 19 Nov 2006 17:24 PST

Any formula that seems logical to arrive to the answer is welcome.  I
assume, that the question is asking what is the chance for the total
of both dice to equal 3, and then a four, in that order.

OK, I'll use the "brute force" method.

Basically, then, to restate the question, what you'll want is the
chance that the two dice will add to three on the first throw, then
add to four on the second throw. They are independent events, so to
find the probability both will happen in that order, you multiply the
two probabilities.

First of all, when you throw two dice, there are 36 possible results
(6 times 6), like this:

1 1
1 2 = 3
1 3 = 4
1 4
1 5
1 6
2 1 = 3
2 2 = 4
2 3
2 4
2 5
2 6
3 1 = 4
3 2
3 3
3 4
3 5
3 6
4 1
4 2
4 3
4 4 
4 5
4 6
5 1
5 2
5 3
5 4
5 5
5 6
6 1
6 2
6 3
6 4
6 5
6 6

You can see the above graphically on the following page:

Dice and the laws of probability
http://www.edcollins.com/backgammon/diceprob.htm

Of those, there are only two combinations that will give you a total
of three. They are that the first die will be 1 and the second one 2,
and that the first die will be 2 and the second one 1.

So the chance of getting a total of 3 by throwing two dice is 2/36 or 1/18.

Next, when you throw two dice, there are three ways that you can get a
total of 4. They are first die 1, second die 3; first die 2, second
die 2; and first die 3, second die 1.  So the chance that throwing two
dice will give you a total of 4 is 3/36 or 1/12.

To determine the chances of getting a total of 3 and then a total of
4, you multiply the probabilities. So you'd multiply 1/18 times 1/12,
to get a probability of 1/216 or 0.004629629629629629629629629...
(that's a repeating number there).

I think I have made it as clear as possible how I arrived at this
answer. But if I didn't, please let me know.

Sincerely,

Mvguy-ga

You're brilliant :-).

Are you talking about rolling a three with the first die and then a
four with the second die?  Or are you talking about rolling a total of
three with the two dice and then a four with the two dice?

In either case, the two events are independent.  So what you rolled
the first time has no bearing on what you will roll next.

So the probability is:

P(3 then 4) = P(3) x P(4)

It's a total of three, and a total of four I would assume.

There are six possible numbers on each of two dice, so the number of
possible rolls is 6^2 or 36.

You can roll a three with 1 2 or 2 1.

I trust you can figure out the rest.

Thanks for the kind comment and positive rating!