For a math class I was given the assignment to make a game of chance, for my game the person must roll 4 dice and get a 6, a 5, and a 4 in a row in 3 rolls or less to qualify. the remaining dice must be over 3 for you to win. my question though is how can I find out the probability of rolling the 6,5, and 4 in a single roll?
My thought was $\frac{4}{24} + \frac{3}{15} + \frac{2}{8} = 0.61$
Please tell me if this is correct or if I need to do it in another method.
Thank you!
On
Let's say the dice are blue, green, red, and yellow. Then an outcome may be specified by $(b, g, r, y)$. There are six possible outcomes for each die, so our sample space has $6^4$ possible outcomes.
For the favorable cases, there are two possibilities:
Case 1: Four distinct numbers are obtained, including 4, 5, and 6.
There are four dice on which a number other than 4, 5, or 6 could appear and three possible outcomes that die could show. There are $3!$ ways for the remaining three dice to display a 4, 5, or 6. Hence, there are $$\binom{4}{1}\binom{3}{1}3!$$ outcomes in this case.
Case 2: Only 4, 5, and 6 are obtained, with one of those numbers appearing twice.
There are three numbers that could be the one to appear twice and $\binom{4}{2}$ ways for two of the four dice to display that number. There are $2!$ ways for the remaining two dice to display the two numbers which each appear once. Hence, there are $$\binom{3}{1}\binom{4}{2}2!$$ outcomes in this case.
Total: Since the two cases above are exhaustive and mutually exclusive, the number of favorable outcomes is $$\binom{4}{1}\binom{3}{1}3! + \binom{3}{1}\binom{4}{2}2!$$
Probability: The probability that the numbers 4, 5, and 6 will appear if four fair dice are rolled is $$\frac{\dbinom{4}{1}\dbinom{3}{1}3! + \dbinom{3}{1}\dbinom{4}{2}2!}{6^4} = \frac{1}{12}$$ as drhab found using the Inclusion-Exclusion Principle.
On
You want the probability for obtaining 4,5,6, and one other number, or of obtaining 4,5, and 6, where one from these is rolled twice, from a roll of four independent dice.
Count the ways, add and divide.
$$\begin{align}&\dfrac{\binom 31 4!+\binom 31\binom 422!}{6^4}\\[3ex]=~&\dfrac 1{12}\end{align}$$
If I understand well then the question is: "if $4$ dice are thrown then what is the probability that $4$, $5$ and $6$ are among the results?" (please correct me if I am wrong)
Number the dice and let $E_i$ be the event that the numbers $6,5,4$ are rolled with the $3$ dice that have numbers in $\{1,2,3,4\}-\{i\}$.
If $E$ is the event described above then $E=\bigcup_{i=1}^4E_i$ and using inclusion/exclusion and symmetry we find:
$$P(E)=P\left(\bigcup_{i=1}^4E_i\right)=4P(E_1)-6P(E_1\cap E_2)=4\cdot3!\cdot6^{-3}-6\cdot3!\cdot6^{-4}=\boxed{\frac1{12}}$$
Note that $E_1$ occurs if the outcome is e.g. $(x,4,6,5)$ where $x$ can take any of the $6$ values.
Note that $E_1\cap E_2$ occurs if the outcome is e.g. $(5,5,4,6)$.
Note the $E_1\cap E_2\cap E_3$ and $E_1\cap E_2\cap E_3\cap E_4$ are both empty (i.e. cannot occur).