Who can answers this Probability challenge?

Question

About 1000 years ago, a bishop in what is now Belgium determined that there are 56 different ways in which three dice can fall provided one is interested only in the overall result and not in the outcomes of the individual dice.

a.) Find the number of ways in which three dice can all come up with the same number of points.

b.) Find the number of ways in which two of the three dice can come up with the same number of points while the third die comes up with a different number of points.

c.) Find the number of ways in which all three dice can come up with a different number of points.

d.) Use parts a), b), and c) to verify the bishop's calculation that there are altogether 56 possibilities.

I find this condition: "...there are 56 different ways in which three dice can fall provided one is interested only in the overall result and not in the outcomes of the individual dice." a bit confusing.

So, can anyone give light to this problem?

Answer 1

With two dice, there are 36 possible outcomes (6x6),

but if order is disregarded there are fewer (21), since

(4,3) is regarded the same as (3,4).

The number 21 can be derived from:

half the non-pairs (36 - 6)/2 = 15 + 6 pairs.

With three dice instead of 216 different outcomes there are fewer

if you disregard which die has which number.

Let's say the 3 dice are red, green, and blue.

Then if you take into account the colors,

there are 6 x 6 x 6 = 216 ways for them to come up.

But if you ignore the colors, which is what this saying,

then there are fewer results because R6 G4 B3 is now

regarded as the same as R4 G3 B6 etc.

That can be done several different ways,

one of which was given in the first answer.

Here is another one.

After you throw the dice, put them in ascending order

of number of spots.

So if you throw 3 5 2, change it to 2 3 5.

Now we can count as follows:

Second die equal or higher to first, third equal or higher to second.

First second third

1 1 1-6: 6

1 2 2-6: 5

1 3 3-6: 4

1 4 4-6: 3

1 5 5-6: 2

1 6 6-6: 1

Total 1 + 2 + 3 + 4 + 5 + 6

Following the same pattern,

if first is 2, we have 1 + 2 + 3 + 4 + 5

first is 3: 1 + 2 + 3 + 4

first is 4: 1 + 2 + 3

first is 5: 1 + 2

first is 6: 1

1 + 3 + 6 + 10 + 15 + 21 = 56.

Another way is to account for all 216 ways in groups:

All the same: 6 ways (1 1 1, 2 2 2, 3 3 3, etc)

Two + one: 6 x 5 choices of numbers, 3 positions for singleton: 90

All different: 6 x 5 x 4: 120

Total of 216.

But two+ones counted 3 times each, reduce to 30

All different counted 6 times each, reduce to 20

6 + 30 + 20 = 56.

The two dice case can also be counted the first way:

First is 1, second is 1-6 (6)

2 and 2-6 (5)

3 and 3-6 (4)

4 and 4-6 (3)

5 and 5-6 (2)

6 and 6 (1)

1+2+3+4+5+6 = 21.

=

Answer 2

The phrase means that you have 6 ways to roll each die, in which you get a combination of dots in the end.  Use the formula for combinations without replication with n = 6 and k = 3 (6 sides to a die, and 3 dice):

(n + k - 1)!/(k!*(n-1)!)

It will be 56.

In each question it asks you for the probability in which each scenario occurs.  What are the chances that if you roll all the dice at once, they will all come up with 1 dot?  I believe the phrase is also telling you the event of rolling all three dice and the outcome is not mutually exclusive.

If you label the dice as die A, die B, and die C, the first question is asking: "What is the probability of A = 1 and B = 1, and C = 1?" or in probability talk P(A and B and C).  The probability of rolling a 1 on each die is 1 in 6, or 1/6.

Remember for events that are not mutually exclusive:

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

and

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

Answer 3

Assuming we're talking about six-sided dice...

a)  There are only six ways for this... either they all come up with 1, or all with 2, or all with 3... etc.

b)  There are 6 x 5 = 30 ways for this.  Six possibilities for the "two with the same," and then for each of those, five possibilities for the other one (eg, if there are two 1's, then the singleton can be 2, 3, 4, 5, or 6).

c)  There are 6 choose 3 = 20 ways for this.  In a group of six possible choices, to choose three of them (with none repeated) is 6 choose 3, which is (6x5x4x3x2x1)/[(3x2x1)x(3x2x1)] = 20.

d)  Now just add up the different possibilities: 6 + 30 + 20 = 56