- Thread starter Domski74
- Start date

- Joined
- Dec 30, 2014

- Messages
- 3,698

I would start off thinking how one die could not be smaller than the other two.

- Joined
- Dec 30, 2014

- Messages
- 3,698

30/72 is the same as 15/36 !!!!

I still do not understand the question. Is one die special (maybe red and the other two are green) and you want to know if you roll all three dice the probability that the red die is smaller than each green die? Can we please see your work showing us how you got 15/36? Please!?

Last edited:

- Joined
- Jan 29, 2005

- Messages
- 8,442

Thanks for the clarification. Rolling three dice given \(\displaystyle 6^3=216\) outcomes in the form of triples.

Now must consider the numbers, look at this list:

\(\displaystyle \begin{array}{*{20}{c}}{\text{max value}}&{\text{#}}&{\text{total}}\\\hline{6|}&{25}&{75}\\{5|}&{16}&{48}\\{4|}&9&{27}\\{3|}&4&{12}\\{2|}&1&3\end{array}\)

The table tells us that there are seventy-five triples in which six it the greatest value.

That that is: the triple \(\displaystyle (6,1,5) \) has a die with a value greater than either of the other two.

BUT that happens in three ways. Look at the set of pairs \(\displaystyle \{1,2,3,4,5\}\times\{1,2,3,4,5\}\) there are twenty-five pairs.

Each of those pairs can go with the six in three ways. (Can you tell why we din't to use the one?)

Note that we counted the triple \(\displaystyle (4,5,4)\) but not \(\displaystyle (4,4,1)\) WHY?

Add up the third column. What do you get?

That is, we count the triples having a an unique maximum.

What is the answer to this problem?

Last edited:

- Joined
- Nov 12, 2017

- Messages
- 4,443

Roll 2 dice, then roll another, what is the probability that this third die is more than either of the first 2. Is that saying the same thing? Thanks for the help

As I understand it, you roll one pair of dice, then want to know theOk so I mean greater than either of the other 2 not the combination.

That is, you want

P(third die > max of first pair).

So the answer will be

P(max of pair = 1)*P(third > 1) + P(max of pair = 2)*P(third > 2) + ... + P(max of pair = 6)*P(third > 6).

You can find the probabilities for the pair by looking at your table of 36 outcomes, and the probabilities for the third are simple. What do you get?

- Joined
- Jan 29, 2005

- Messages
- 8,442

Roll 2 dice, then roll another, what is the probability that this third die is more than either of the first 2. Is that saying the same thing? Thanks for the help

My previous post was written before you completely clarified the setup.As I understand it, you roll one pair of dice, then want to know theprobability that a third die is greater than either of the first pair.

However, the model is the same. The outcomes are still triples. But now the question is to count the triples in which the third entry is the unique maximum. This table tells us all:

\(\displaystyle \begin{array}{*{20}{c}}{\text{max value}}&{\text{# of triples}}\\\hline{6|}&{25}\\{5|}&{16}\\{4|}&9\\{3|}&4\\{2|}&1\end{array}\)

Now the sum of the second column is \(\displaystyle 55\).

That means that in rolling a die three times there are \(\displaystyle 55\) times that the third value is the unique maximum of the three.

So if we roll two red dice and then roll one white die, what is the probability that the value on the white die is greater than the values on either of the red dice?