Probability of rolling a "6" k times out of 20 rolls of a die

[Addez123]

TL;DR Summary

When rolling a dice 20 times, whats the probability of getting 6 k amount of times?

The probability should be
## (1/6)^k * (5/6)^{20-k} ##

But the book says the answer is :
##
\begin{pmatrix}
20 \\
k \\
\end{pmatrix} * (1/6)^k * (5/6)^{20-k} ##

Because there are 20 over k different sequences, but the order doesn't matter?
I just don't understand why the 20 over k is there at all.

 

[PeroK]

Summary:: When rolling a dice 20 times, what's the probability of getting 6 k amount of times?

The probability should be
## (1/6)^k * (5/6)^{20-k} ##

But the book says the answer is :
##
\begin{pmatrix}
20 \\
k \\
\end{pmatrix} * (1/6)^k * (5/6)^{20-k} ##

Because there are 20 over k different sequences, but the order doesn't matter?
I just don't understand why the 20 over k is there at all.

Why don't you try rolling the die just twice? Calculate the probability of getting ##0, 1## and ##2## sixes (your way) and add these up. Does it add up to ##1##? If not, why not?

 

[Addez123]

Rolling twice, 2 sixes:
1/6 * 1/6 = 1/36
Rolling twice, 1 six:
1/6 * 5/6 = 5/36
Rolling twice, 0 sixes:
5/6 * 5/6 = 25/36

Probablity of any of these happening should be 1:
##P(B) = P(A0) \cup P(A1) \cup P(A2) = 1 ##
Why it doesnt, idk :/

 

[PeroK]

Rolling twice, 2 sixes:
1/6 * 1/6 = 1/36
Rolling twice, 1 six:
1/6 * 5/6 = 5/36
Rolling twice, 0 sixes:
5/6 * 5/6 = 25/36

Probablity of any of these happening should be 1:
##P(B) = P(A0) \cup P(A1) \cup P(A2) = 1 ##
Why it doesnt, idk :/

You have a total probability there of ##31/36##. Where is the missing ##5/36##?

 

[Addez123]

Is it that rolling one six can be done by rolling: 6 then not 6 or not 6 then 6?
So the probaility of one six is:
1/6 * 5/6 + 5/6 * 1/6 = 10/36?

 

[PeroK]

Is it that rolling one six can be done by rolling: 6 then not 6 or not 6 then 6?
So the probaility of one six is:
1/6 * 5/6 + 5/6 * 1/6 = 10/36?

Yes, and in this case the binomial coefficient ##\binom n k = \binom 2 1 = 2##. That's why it's needed in the formula you questioned.

 

[FactChecker]

Your answer would be what you would get if they asked for the probability of getting '6' on the first k rolls and then not on the remaining n-k rolls. So you need to also account for all the possible rearrangements of those results. That is the reason for the combination multiplier, ##_{20}C_k##.

 

[zinq]

When you roll a die 20 times, there are 6 possibilities on the first roll, 6 on the second, etc. The number of distinct sequences of 20 rolls is therefore 6 x 6 x ... x 6 = 6²⁰. These 6²⁰ events are disjoint from each other: If one sequence of 20 numbers shows on the die, then a different sequence does not.

For that reason, when you consider all the sequences you are interested in, you add their probabilities to get the total probability you are interested in. Now suppose the number 5 shows on the die only on the 1st, 3rd, and 5th rolls of the die out of 12 rolls. What is the probability of that? It's (1/6) x (5/6) x (1/6) x (5/6) x (1/6) x (5/6)⁷. (See why?) In other words it is (1/6)³ x (5/6)⁹. The binomial coefficient denoted by ₁₂C₃ (and which is pronounced "twelve choose three") counts exactly how many ways you can pick three objects from a group of twelve of them. So the probability that in 12 rolls of the die, exactly three of them are fives (and the other nine are something else, no matter what) is just ₁₂C₃ x (1/6)³ x (5/6)⁹.

Since _nC_p is always equal to n! / (p! (n-p)!), this works out to (12! / (3! 9!)) x 5⁹ / 6¹². The original problem will of course be similar.

 

[MikeeMiracle]

If I may throw a spanner in the calculations :)

You are also assuming perfectly weighted die/dice. A lot of times the holes representing the numbers are drilled / puched out of the die and then filled in. The 1 and 6 are at opposite ends so then 1 side ends up ever so slightly heavier than the 6 side. Same with the 5 & 2 sides.

Folk lore from my grand father in Cyprus who use to play backgammon multiple times daily at the local cafe (it's where the old boys hang out in the villages.) If you want to roll higher numbers you throw the dice harder as there is more chance of the weight having an effect, if you want to roll a small number you roll gently or drop the dice on the board so it does not roll so much before settling.

I have not tried to do any statistical analysis, but all the old boys I know play this way.