Thanks alot that was just what needed and well explained. Shit you replies were pretty quick, I should of looked back sooner.

quote:

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Original post by jeffakew
Hi I having difficulty with 2 probability questions it would be great if anyone could help me out a bit, ao anyway here they are:

1) Two tetrahedral dice are rolled together what is the probability that the scores on the dice are the same?

2) Four tetrahedral dice are rolled together what is the probalbilty that the scores on at least two of the dice are the same?

Well I got the first question its(1/4) but I'm not 100% sure, but the second question has really got , any help or ideas would be great.

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In elementary probability the likelyhood of any oneof two events occuring is:

P(A or B) = P(A) + P(B)

The likelyhood of both events occuring is...

P(A and B) = P(A)P(B)

The probability that one die is of the required score is 1/n, where n is the number of sides on your die.

So the answer to question 1 is (1/n)(1/n) = (1/n)/(n/n2), where n2 is n-squared (if my mental algebra serves me correctly!!).

The answer to question 2 is simply...

m*P(A)P(A), where m is the total number of dice.

I sincerely hope this is correct!! I gotta live up to my name!! :D

Regards,
Mathematix.

Edited by - Mathematix on September 1, 2001 9:58:41 PM

Edited by - Mathematix on September 1, 2001 10:23:09 PM

quote:

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Original post by Mathematix

Original post by jeffakew
Hi I having difficulty with 2 probability questions it would be great if anyone could help me out a bit, ao anyway here they are:

1) Two tetrahedral dice are rolled together what is the probability that the scores on the dice are the same?

2) Four tetrahedral dice are rolled together what is the probalbilty that the scores on at least two of the dice are the same?

Well I got the first question its(1/4) but I'm not 100% sure, but the second question has really got , any help or ideas would be great.

---

In elementary probability the likelyhood of any oneof two events occuring is:

P(A or B) = P(A) + P(B)

The likelyhood of both events occuring is…

P(A and B) = P(A)P(B)

The probability that one die is of the required score is 1/n, where n is the number of sides on your die.

So the answer to question 1 is (1/n)(1/n) = (1/n)/(n/n2), where n2 is n-squared (if my mental algebra serves me correctly!!).

The answer to question 2 is simply…

m*P(A)P(A), where m is the total number of dice.

I sincerely hope this is correct!! I gotta live up to my name!! :D

Regards,
Mathematix.

*************************************************************

Also note that the solution above to question 2 give the answer to only two having the required score! To say that ay least two have the required score would be…

m*P(A)P(A)+m*P(A)P(A)P(A)+m*P(A)P(A)P(A)P(A)…

which is equal to (if f(n) = m*P(A)^n, where P(A)^n is the probability of event A occuring to the power n.)

f(n) = f(n) + f(n+1), where n is the number of dice with the same score and n + 1 <= m.

Lets explain my findings.

Why m (the total number of dice)? In order to refine the probability calculation, you also need to include the number of dice that CAN actually have the score. For each dice that cannot have the score, or doesn't exist, you decrement m therefore reducing the probability that the said number of dice will come up with the score.

Why n (the number of dice with the same score)? We can have either two, three, four, etc. dice with the same score.

Why sum all these? Because we need the probability of either two OR three OR four, etc. dice of having the same score.

Hope this answers!

Regards,
Mathematix.



Edited by - Mathematix on September 3, 2001 4:58:09 AM

Mathematix - if you actually took more than one line to work out Q2, you''ve been wasting your time

There are 24/256 ways of NOT getting any duplicates. Therefore there are 232/256 ways of getting duplicates.

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Original post by Beer Hunter
Mathematix - if you actually took more than one line to work out Q2, you''ve been wasting your time

There are 24/256 ways of NOT getting any duplicates. Therefore there are 232/256 ways of getting duplicates.

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What do thse numbers mean?? How were they calculated??? Is there any reasonig to this???

Regards,
Mathematix.

I'm a bit rusty on my probability, but I've got to agree with Beer Hunter here. I'm gussing this is where he got his numbers:

There are 4^4 total possible rolls. This is easy.

There are 4! ways of getting no duplicates. You have 4 possibilities for the roll of the first die, 3 possibilities for the second that will give you no duplicates, 2 for the third dice to get no duplicates, and the last die must be the only unrolled number. 4*3*2*1 = 4!

Therefore the probability of getting no duplicates is (4!)/(4^4) or 24/256.

Therefore the probability of getting at least one duplicate is 1-(4!)/(4^4) or 232/256.

oops and I forgot to say 232/256 = 29/32 which is the answer Xai gave.

Edited by - Dobbs on September 3, 2001 3:55:39 PM

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Original post by Dobbs
I''m a bit rusty on my probability, but I''ve got to agree with Beer Hunter here. I''m gussing this is where he got his numbers:

There are 4^4 total possible rolls. This is easy.

There are 4! ways of getting no duplicates. You have 4 possibilities for the roll of the first die, 3 possibilities for the second that will give you no duplicates, 2 for the third dice to get no duplicates, and the last die must be the only unrolled number. 4*3*2*1 = 4!

Therefore the probability of getting no duplicates is (4!)/(4^4) or 24/256.

Therefore the probability of getting at least one duplicate is 1-(4!)/(4^4) or 232/256.

oops and I forgot to say 232/256 = 29/32 which is the answer Xai gave.

Edited by - Dobbs on September 3, 2001 3:55:39 PM

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you guys are perfectly right. There I go with my over-analysis again!


Regards,
Mathematix.

quote:

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Mathematix said:
In elementary probability the likelyhood of any oneof two events occuring is:

P(A or B) = P(A) + P(B)

The likelyhood of both events occuring is...

P(A and B) = P(A)P(B)

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Just a comment:
I think you should have mentioned that the events A and B must be mutually exclusive in order for the axioms to be true.
Although, of course, they are mutually exclusive in this particular case.

Anyway, I didn't mean to be arrogant!


/Mankind gave birth to God.

Edited by - silvren on September 20, 2001 3:23:23 PM

quote:

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Original post by silvren

Just a comment:
I think you should have mentioned that the events A and B must be mutually exclusive in order for the axioms to be true.
Although, of course, they are mutually exclusive in this particular case.

Anyway, I didn''t mean to be arrogant!


/Mankind gave birth to God.

Edited by - silvren on September 20, 2001 3:23:23 PM

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Thanks Silvren. I forgot to mention that very important bit!

Regards,
Mathematix.