# Rolling 5 dice probability

• May 26th 2009, 08:43 PM
krzyrice
Rolling 5 dice probability
In the game Yahtzee, five fair 6-sided dice are rolled.
a. What is the probability of rolling exactly three 1s?
b. What is the probability of rolling exactly three of any number?
c. What is the probability of rolling three, four, or five of a kind?

This question has taken me completely by surprise. The book never talked about any of this question and it has taken me hours of thinking to find nothing.
• May 26th 2009, 09:30 PM
noob mathematician
Quote:

Originally Posted by krzyrice
In the game Yahtzee, five fair 6-sided dice are rolled.
a. What is the probability of rolling exactly three 1s?
b. What is the probability of rolling exactly three of any number?
c. What is the probability of rolling three, four, or five of a kind?

This question has taken me completely by surprise. The book never talked about any of this question and it has taken me hours of thinking to find nothing.

Hi krzyrice,
For part i), we know that the probability for 1 is 1/6 and (non-1) is 5/6, hence we can try to use binomial equation:
$P(\text {exactly three 1s})={5 \choose 3}(\frac{1}{6})^3(\frac{5}{6})^2$
• May 27th 2009, 06:54 AM
Soroban
Hello, krzyrice!

These are from the topic "Binomial Probability."
You should be familiar with the formula.

Quote:

In the game Yahtzee, five fair 6-sided dice are rolled.

a. What is the probability of rolling exactly three 1s?

noob explained this nicely.

$P(\text{exactly three 1's}) \:=\:{5\choose3}\left(\frac{1}{6}\right)^3\left(\f rac{5}{6}\right)^2 \:=\:\frac{250}{7776} \:=\:\frac{125}{3888}$

Quote:

b. What is the probability of rolling exactly three of any number?

Since there are six choices for the value of the 3-of-a-kind,
. . the probability is six times the answer in part (a).

$P(\text{3-of-a-kind}) \:=\:6\times \frac{250}{7776} \;=\;\frac{1500}{7776} \;=\;\frac{125}{648}$

Quote:

c. What is the probability of rolling three-, four-, or five-of-a-kind?
We already know that: . $P(\text{3-of-a-kind}) \:=\:\frac{1500}{7776}$

Four-of-a-kind: 4 of one number, 1 of another.
There are 6 choices for the quadruple.
. . $P(\text{4-of-a-kind}) \;=\;6\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\ right) \;=\;\frac{150}{7776}$

Five-of-a-kind: 5 of one number.
There are 6 choices for the quintuple.
. . $P(\text{5-of-a-kind}) \;=\;6\left(\frac{1}{6}\right)^5 \;=\;\frac{6}{7776}$

Therefore: . $P(\text{3-, 4- or 5-of-a-kind}) \;=\;\frac{1500}{7776} + \frac{150}{7776} + \frac{6}{7776} \;=\;\frac{1656}{7776} \;=\;\frac{23}{108}$

• May 27th 2009, 02:35 PM
krzyrice
Thank you so much to the both of you for this help! I had a feeling it would involved 1/6 and 5/6 but wasn't sure if I was going to do it correctly. Thanks again.