I know that the probability of getting the 3 dice to be of the specific same number in a 3 dice roll is (1/6)*(1/6)*(1/6) = 1/216. What is the probability of getting the same specific number for 3 dice in a 5 dice roll?
Do we calculate that with adding the probabilities for 3 of the same dice with 4 and 5 of the same dice together?
Something like this:
P(three or more dice show the same number) = P(3) + P(4) + P(5)Which then would be:
P = (1/6)*(1/6)*(1/6) + (1/6)*(1/6)*(1/6)*(1/6) + (1/6)*(1/6)*(1/6)*(1/6)*(1/6)
P = 0.00552983539Is that the correct way of calculating the probability for such case?
$\endgroup$ 22 Answers
$\begingroup$Your question falls under the category of binomial probability. Since you're asking for 'at least 3...' you should look at the pdf of the binomial distribution.
Step 1: Why is it binomial? reference: Yale
1: The number of observations n is fixed.
2: Each observation is independent.
3: Each observation represents one of two outcomes ("success" or "failure").
4: The probability of "success" $p$ is the same for each outcome.Step 2: the equation:
$${n\choose k}p^k(1-p)^{n-k}$$
Step 3: setup the data:
\begin{align}n&=5\\ k&=3\\ p&=\frac16\\ 1-p&=\frac56\end{align}
Step 3: plug in
$${5 \choose 3}* \left(\frac{1}{6}\right)^3*\left(\frac{5}{6}\right)^2=0.032$$
Step 4: Cumulative probability:$P(X \geq x)$
You should plug (again) for $k=4$ and $k=5$ and sum them up. You'll get: $0.035$
Step 5: conclusion:
For any number:$0.035*6=0.21$.
$\endgroup$ 2 $\begingroup$No... it is not.
You must consider the number of ways you can get $3$ specific numbers (e.g., $1$) out of five rolls: $(111xx),(x1x11),\ldots$, and likewise for four appearances, or five appearances. This is governed by a binomial distribution.
For a given chosen number (e.g., a $1$), the number of ways you can get this is the probability of getting exactly three $1$s in known positions: $\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$. But how many sets of three positions exist? That is ${5 \choose 3} = 10$. But for this to give exactly three $1$s, the other two positions must be non $1$s. The probability of that is $\frac{5}{6} \cdot \frac{5}{6}$.
I'll let you multiply these out for yourself.
Next, if you allow getting exactly four $1$s, the above changes to $\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}$ and ${5 \choose 4}$. See why?
It is simple calculating the change of getting exactly five $1$s: $\left( \frac{1}{6} \right)^5$
Then add them up.
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