Binomial or Joint Probability. The probability of getting a 6 when rolling a dice is 1/6.

beobachtereb

beobachtereb

Answered question

2022-09-23

Binomial or Joint Probability
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 ] P [ D 2 = 6 ] P [ D 3 = 6 ] P [ D 4 = 6 ] P [ D 5 = 6 ] = ( 1 / 6 ) 5 = 0.000128601
Where P [ D i = 6 ] is the probability of the i t h 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 ! 5 ! ) ( 1 / 6 ) 5 ( 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?

Answer & Explanation

Cassie Moody

Cassie Moody

Beginner2022-09-24Added 10 answers

Step 1
The answer is that the probability of rolling {6, 6, 6, 6, not-6} on your dice is equal to 1 6 4 × 5 6 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 1 6 4 × 5 6 0.00064.
Step 2
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 k ) , the total probability of having k successes is ( n 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 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 .

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