The probability of getting a 6 when rolling a dice is 1/6.
Say I have 5 dices in front of me, the probability of 5 of them being a 6 is equal to:
$P[D_1=6]\cdot P[D_2=6]\cdot P[D_3=6]\cdot P[D_4=6]\cdot P[D_5=6] = (1/6)^5 = 0.000128601$
Where $P[D_i=6]$ is the probability of the $i^{th}$ dice being a 6
Now, why is it that I get the same answer by seeing this problem as 5 Bernoulli trials each with a probability of success equal to 1/6 and by calculating the desired probability via the binomial distribution, namely the probability of 5 successes out of 5 trials:
$P[5|5] = 5!/(0!\cdot5!)\cdot(1/6)^5\cdot(5/6)^0 = 0.000128601$
I'm trying to understand two things:
why I don't get the same result for 4 successes out of 5 trials compared to just doing (1/6)^4 if what I wan to know is the probability of 4 out of my 5 dices being a 6.
Why are the results for 5 out of 5 the same? Is it a coincidence? Is one of them more or less valid than the other?
Thanks!
The answer is that the probability of rolling {6, 6, 6, 6, not-6} on your dice is equal to $\frac{1}{6^4} \times \frac{5}{6} \approx 0.00064$, but it's not the only way to get 4 6s. There's also {6, 6, 6, not-6, 6}, {6, 6, not-6, 6, 6}, {6, not-6, 6, 6, 6} and {not-6, 6, 6, 6, 6}, each of which has the same probability of occurring. So there are a grand total of 5 ways it can happen, resulting in a total probability of $\frac{1}{6^4} \times \frac{5}{6} \times 5 \approx 0.003215$.
In general, if the probability of success is $p$ and the probability of failure is $1-p$, then since the number of ways to arrange $k$ successes amongst $n$ events is $n \choose k$, the total probability of having $k$ successes is ${n \choose k}p^k(1-p)^{n-k}$, which is the Binomial probability distribution.
The reason your calculation works for 5 successes from 5 dice is because there is exactly 1 way to do so: {6, 6, 6, 6, 6} so ${n \choose k} = 1$ and $n-k = 0$ so $(1-p)^{n-k} = (1-p)^0 = 1$, so both those terms disappear from the calculation, leaving you with just $p^n$.