# Thread:

1. ## Very Very Urgent!! Plz Help

a) how many times must a die be thrown to be sure that the same number occurs twice?

b) How many times must two dice be thrown to be sure that the same total occurs at least six times?

c) How many times must n dice be thrown to be sure that the same total score occurs at least p times.

2. Originally Posted by xXxSANJIxXx
a) how many times must a die be thrown to be sure that the same number occurs twice?
There are six different possible outcomes, so it's quite possible that none of the six first throws are the same, but the seventh throw has to be the same as one of the others.

b) How many times must two dice be thrown to be sure that the same total occurs at least six times?
Two die can give any combination between 2 and 12, so theres 11 combinations. To get the same combination six times, you have to realize that each of those eleven combinations can possibly happen five times before one happens six. Therefore you have to roll the dice 11x6+1 times to be sure of the answer.

c) How many times must n dice be thrown to be sure that the same total score occurs at least p times.
Try this with the information above and tell me if you can't get it.

3. Originally Posted by xXxSANJIxXx
a) how many times must a die be thrown to be sure that the same number occurs twice?
If you throw a die and it land on a 1 what is the probability it lands on the one the next throw?
Simple 1/6.

What about on the second time. Consider the probability of it not landing on 1. Then the probability is $(5/6)^2$. Thus, the probability that it would is,
$1-(5/6)^2$

For three it is $1-(5/6)^3$
This pattern continues.
You need an $n$ such that,
$1-(5/6)^2\geq 1/2$ for that is the meaning of "likely".
Basic computation shows that number is $n=4$

4. Originally Posted by ThePerfectHacker
$1-(5/6)^2\geq 1/2$ for that is the meaning of "likely".
Basic computation shows that number is $n=4$
"likely" is different than "sure"

5. Originally Posted by Quick
"likely" is different than "sure"
In that case it is not possible to be sure.
I assumed he meant likely, as with many probability problems.

6. Originally Posted by ThePerfectHacker
In that case it is not possible to be sure.
I assumed he meant likely, as with many probability problems.
It's possible to be sure, I showed it in my previous post.

7. I misunderstood the question.
I assumed the thing meant that I should find when of getting the same number again as on your first rule.
---
Quick, you might not be ware but you used the "Pigeonhole Principle".
Next time you want to sound smart and rely on it say "According to the pigeonhole principle...".
And if you really want to sound smart say "According to Dirichelet's Pigeonhole Principle...."

8. i cant get the last one

9. Originally Posted by ThePerfectHacker
If you throw a die and it land on a 1 what is the probability it lands on the one the next throw?
Simple 1/6.
Quicks solution is right.

In a simple system where it is only the probability of occurrence that
we need worry about, that the probability of a 1 occurring is 1/6 does
not guarantee that you will ever see a 1, or if you have seen one that
you will ever see one again.

(you will with probability 1, but that is not the same thing as
it will definitely happen - the details of why are too technical
for this thread).

RonL

10. Originally Posted by Quick
There are six different possible outcomes, so it's quite possible that none of the six first throws are the same, but the seventh throw has to be the same as one of the others.

(b)
Two die can give any combination between 2 and 12, so theres 11 combinations. To get the same combination six times, you have to realize that each of those eleven combinations can possibly happen five times before one happens six. Therefore you have to roll the dice 11x6+1 times to be sure of the answer.
Look at part (b) and (c) together. The only difference is that 6 is replaced with p. So look at Quick's answer to (b) and where he put 6 put p.
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