How do you calculate the probability without using the complement?

[tjosan]

Let's say I'd like to calculate the probability of getting at least one 4 when rolling two dice.

That's 1 minus the probability of not getting any 4, i.e 1 minus the complement, 1-(5/6)^2.

But how would I calculate without using the complement?

 

[fresh_42]

36 possibilities, 6+5 positive constellation.

 

[tjosan]

Well yes. But say I have 100000 dice. Would take a while to sum up all the positive constellations then. Isn't there any... Well analytical ways of doing this?

Edit: basically I am trying to find a way to end up with equation like this (sqrt(11)/6)^2 somehow.

 

[fresh_42]

Sounds like $(1-p^n)=(1-p)\cdot (p^{n-1}+p^{n-2}+\ldots +p+1).$

 

[etotheipi]

If you roll $n$ dice, then each possible outcome in the sample space could be a tuple $(x_1,x_2,x_3,\dots, x_n)$ where $x_1, x_2, \dots, x_n \in [1,2,3,4,5,6]$. All outcomes are equally likely with $p = 6^{-n}$. You want to know many ways to pick the $x_i$ such that there is at least one $4$, without using the complement.

If you have $k$ 4's, then you have ${n\choose k}$ ways of inserting just these 4's in the tuple, and $5^{n-k}$ ways of inserting the remaining numbers, i.e. there are $5^{n-k} {n\choose k}$ possible outcomes for which there are $k$ 4's. Then you sum over all $k$ from $1$ to $n$, $$N = \sum_{k=1}^n 5^{n-k} {n\choose k}$$So the probability of getting at least one $4$ is$$P = 6^{-n}\sum_{k=1}^n 5^{n-k} {n\choose k}$$

 

[Dale]

Well analytical ways of doing this?

Yes, use the complement. Refusing to use the complement is strange

 

[tjosan]

Yes, use the complement. Refusing to use the complement is strange

"Anakin, if one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi." /Palpatine

@etotheipi & @fresh_42: thanks for the help

 

[PeroK]

Well yes. But say I have 100000 dice. Would take a while to sum up all the positive constellations then. Isn't there any... Well analytical ways of doing this?

There's no problem. You just calcuate the probability of getting $n$ 4's where $n$ goes from $1$ to $100000$. And add them all up.

It just quicker to do $1 - p( n=0)$.

 

[etotheipi]

"Anakin, if one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

Did you ever hear the tragedy of Darth Tjosan The Wise? I thought not. It's not a story the mathematicians would tell you. It's a Sith legend. Darth Tjosan was a Dark Lord of the Sith, so powerful and so wise he could use the summation notation to influence the midichlorians to create an expression without using the complement…

 

[FactChecker]

"Anakin, if one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi." /Palpatine

That is a good attitude as long as you don't expect all the alternatives to be simple. It is often the case that the compliment is much simpler than the original problem. You can see that this is one such case. You asked later if there was a simpler, direct, analytical solution and the answer was that it gets messy.

There often is a simpler approximation using Poisson, Gaussian, or other continuous distributions. You will see that very often. Suppose your original problem had asked for the probability that the number of 4's is greater than 1/10th of the total 100,000 dice. Then both the original and compliment might be very messy to calculate directly. Approximations using continuous PDFs would be very useful in that case.

 

[Dale]

That is a good attitude

Good attitude? He’s a Sith Lord!

 

[Vanadium 50]

"I don't want to do this the easy way."
"Fine, do it the hard way."
"But that way is hard!"

 

[fresh_42]

"I don't want to do this the easy way."
"Fine, do it the hard way."
"But that way is hard!"

It was more like:
> "I don't want to do this the easy way."
< "Then count."
> "I don't want to count."
< "Use this formula."
< "Or use that formula."
< "Or do it easy."
> "That's dogmatic."
< "Fine, do it the hard way."
> "But that way is hard!"

 

[gleem]

There are 36 possible events that can occur when you through two dice. If you are interested in at least one 4 appearing, then the events that you are interested in that satisfy this is a 4 on the first and non on the second, non on the first and one on the second and a 4 on both. So the probability of these occurring is

1/6 X 5/6 + 5/6 X 1/6 + 1/6 X 1/6 = 11/32

 

[tjosan]

Thank you for the replies.

I merely wanted to understand why the complement is used by seeing the actual implications of not using it.

I already knew you could count all the possibilities which is why I asked for an analytical approach. I did not have any expections of it being easy though.

I was just curious, because everywhere I looked it said that you should use the complement without providing any reason apart from it being easier. It's the same thing with complicated integrals:
"Use this and that substitution then solve it". Yes, but why? Show me what I'm up against if I refuse!

Thanks though, and may the force be with you.

 

[fresh_42]

I merely wanted to understand why the complement is used by seeing the actual implications of not using it.

This is mathematical standard. The question "what if not" is the most asked question of all: explicitly or silently.

In your example it is, because it is easier to phrase the statement $\text{ NOT IN }A_1\cup A_2\cup \ldots \cup A_n$ than to deal with the $\text{ OR IN }\text{ NOT }A_1\cap \text{ NOT } A_2\cap \ldots \cap \text{ NOT } A_n$ statement, which naturally requires a lot of cases. It is less a dogma than it is set theory and logic. If you draw a Venn diagram then you will see the fundamental difference between those sets.