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.

Tetrahedrons are 4-sided figures, aren''t they?

1. The chances of one number appearing on one die is 1/4. Since there are two dice, you square the answer and so there''s a 1/16 chance that they''re the same.

2. I''m not sure about this one, but I''ll say there''s 3/8 chance that two are the same.

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

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Well, the first die, we don't care what it rolls, as the second die can match it regardless. So P(Die1) = 1.

For the second die to match the first, it gets a 1 in 4 chance of matching it, so P(Die2) = 1/4.

So, P(Match) = P(Die1) * P(Die2) = 1/4.

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

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This one is tricky if you do it with an equation, but there is a shortcut...

Each die has 4 sides. Thus, there are 4⁴ = 216 possibilities of rolls. The only way of never getting a match is to have {1, 2, 3, 4} on the dice in any order. Thus, we'll use permutations to calculate that.

The possible non-matches is equal to ₄P₄ = 4! / 0! = 4! = 24. Thus there are 216 - 24 = 192 ways they can match. Thus your odds of having AT LEAST two dice of four match is 192/216 = 96/108 = 48/54 = 8/9.

~ Dragonus

Edited by - Dragonus on August 28, 2001 1:47:31 PM

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Original post by gph-gw
1. The chances of one number appearing on one die is 1/4. Since there are two dice, you square the answer and so there''s a 1/16 chance that they''re the same.

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That''s not right. If you look at all the possible combinations, you''ll realize {1,1}, {2,2}, {3,3}, and {4,4} are valid possibilities, and there are 4² = 16 permutations to it, so 4/16 = 1/4.

You were thinking about a specific case, as each of the 4 possibilities have a 1/16 chance. If you had multiplied by the number of possible answers, you''d have had 1/4.

~ Dragonus

I think the logic Dragonus used is correct but there is a mistake in the math. There are 256 total possibilities. The probability works out to 29/32 for the second question.

1. The answer is 1/4 like you thought ... the reason ... the first die is rolled ... and there is a 100% chance that it lands on SOME number ... then the second die ... has an equal chance to hit each number (1 in 4) ... so no matter what number it is ... there is a 1 in 4 chance it will match the second die.

2. For this kind of problem i often attack it like this: the first die roll is A ... there is a 1/4 chance that the second die roll matches it ... so with just 2 ... it''s one in 4 ... now ... IF that fails ... then there are 2 different numbers A and B ... and the third die has a 2 in 4 chance of matching one of those .... ... but if that fails ... there are 3 numbers .... so then the last die has a 3 in 4 chance of matching ond of them .... so expressed as math that is:

prob of 2 matches = chance of second die matching + chance of second die failing but third matching + chance of 2 and 3 failing but fourth mathcing ... so those expressions are ...

A. chance of second mathcing = 1/4
B. chance of second failing = 1 - 1/4 = 3/4
C. chance of second failing and third matching = 3/4 * 2/4 = 6/16 = 3/8
D. chance of second and third failing = 3/4 - 3/8 = 5/8
E. chance of second and third failing and fourth matching = 3/8 * 3/4 = 9/32
F. chance of ANY matching = A + C + E = 1/4 + 3/8 + 9/32 = 29/32

This whole thing could have been solved like this also ... you can see that the only way there can be NO matches is four each number to come up once ... so ....
the first die can be any number = 1/1
the second die must not match = 3/4
the third die must not match either = 2/4 = 1/2
the fourth die must not match all 3 = 1/4
multiply the chances = 3 / (4*2*4) = 3/32
and then invert this to get the chance the DO match = 1 - 3/32 - 29/32

there you go.

And I came up with 216 how again? It is 256. Slip of the finger I guess...

So (256 - 24)/256 = 232/256 = 29/32. My mistake.

~ Dragonus

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Original post by Dragonus

Original post by gph-gw
1. The chances of one number appearing on one die is 1/4. Since there are two dice, you square the answer and so there''s a 1/16 chance that they''re the same.

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That''s not right. If you look at all the possible combinations, you''ll realize {1,1}, {2,2}, {3,3}, and {4,4} are valid possibilities, and there are 4² = 16 permutations to it, so 4/16 = 1/4.

You were thinking about a specific case, as each of the 4 possibilities have a 1/16 chance. If you had multiplied by the number of possible answers, you''d have had 1/4.

~ Dragonus

Whoops: it''s been a while since I''ve done probablility and I''ve been known to make probability errors (lots more than other types of math).

If the question was the chances it''ll land on a particular pair of numbers, I''d be right.

Heh. Nope, you''d still be wrong. It''d have to be something like: what''s the chance of a 3 on my red die and a 2 on my blue one.

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Original post by Anonymous Poster
Heh. Nope, you''d still be wrong. It''d have to be something like: what''s the chance of a 3 on my red die and a 2 on my blue one.

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So that means there''s a 0% chance of me being 100% right?

Today certainly isn''t my lucky day. (Pun intended)