# Probability and Factorials Question

1. Aug 31, 2010

### Nick.S

Hi all. Does anybody know their stuff when it comes to math? More specifically calculating probability... i have a question about factorials and when they should be used.

If you have let's say 2 events,
the chance of event A is 1/2
the chance of event B is 1/3
you're just supposed to multiply to give you 1/6 for the chance that A and B will both occur

BUT if let's say the order in which those events occur doesn't matter, then from what i understand you have to divide by the factorial of 2 in this case

so 6 divided by (2!) = 3.

to me that doesn't make sense, how can the odds of two events both happening where the order isn't important be less than the product of both of them if the order was important?

or let's say you had 3 events

event A is 1/3 odds
event B is 1/2 odds
event C is 1/3 odds

that's 1/18 divided by 3! which is 3x2x1= 6

so 18 divided by 6 = 3

so now my odds are back to 1/3

where am i going wrong here?

2. Sep 1, 2010

### HallsofIvy

Because, writing A for "A happens" and A' for "A doesn't happen", B for "B happens" and B' for "B doesn't happen, then for the two events together, there are 4 possiblities if we ignore order:
AB, A'B, AB', A'B'

but if order is important we have 8 possiblities
AB, BA, A'B, BA', AB', B'A, A'B', B'A'

The probablity of an event happening is "number of ways it can happen divided by the total number of ways anything can happen". Since your "number of ways an event can happen" stays the same, but the "total number of ways anything can happen" is doubled, the probability is divided by 2.

Further, you say "that's 1/18 divided by 3!" and then say "so 18 divided by 6= 3 so now my odds are back to 1/3".

Be careful to distinguish between "(1/18) divided by 3!" and "1/(18 divided by 3!)".

When you did "18 divided by 6= 3" you were doing "(18 divided by 3!)" and got "1/(18 divided by 3!)" but what you really mean is "(1/18) divided by 3!" which is the same as (1/18)(1/3)= 1/54.

3. Sep 1, 2010

### Nick.S

Yikes I seem to be even more confused now!

Sorry if I sound like a simpleton here but this is all new to me and I'm trying to wrap my head around it as best i can.

What I can't seem to understand, is that for the 6/49 lottery, it works out like
1/49 x 1/48 x 1/47 x 1/46 x 1/45 x 1/44

and that leaves you with 10,068,347,520
at which point you apply the factorial for 6! because order doesn't matter and you arrive at roughly 14 million. which is where the odds 1 in 14million come from
so let's leave that for a second

this is the part where I'm getting confused.

now i have another set of events to figure out, all with different odds.
they are
1/2, 1/2, 1/3, 1/3, 1/4, 1/2

now let's say the first 3 events don't depend on order, and the final 3 need to happen sequentially.
as in event ABC could happen the other way around, CBA, BCA, CAB, etc. but events D E F must happen in that specific order
D needs to happen then E would need to happen then F would need to happen

Would i simply multiply all the events together to get the final odds of all of them happening? as in my odds are 1/288 ?
Or would i need to break them up and do 2 separate equations?
(1/2, 1/2, 1/3) order doesn't matter
(1/3, 1/4, 1/2) must happen in this order

OR do i need to apply a factorial to the first 3 since those events don't require a specific order - just like the lottery example?

I guess what I'm asking is how do i calculate the TOTAL probability of multiple events
(1/2, 1/2, 1/3, 1/3, 1/4, 1/2) where i know the odds for each individual event - given that the first 3 of them don't require a specific order and the last 3 do.

I hope my wording makes sense here, thanks for your help.
And again I'm really new at this and don't understand all the formulas so bear with me.

Last edited: Sep 1, 2010
4. Sep 2, 2010

anybody?

5. Sep 8, 2010

### Nick.S

bump for help

6. Dec 1, 2010

### Nick.S

Still stuck on this.

It's not a homework question it's just me trying to understand.
Please help it would be much appreciated. thanks.

7. Dec 1, 2010

### dav2008

Formulas aside, do you understand the difference between order mattering and not?

If I flip a coin 2 times, you could ask two questions:

What is the probability that out of those 2 flips, 1 will be heads and 1 will be tails?

What is the probability that I will get a heads on the first flip and a tails on the second?

Let's look at the possibilities of the flips:

Flip 1: Heads // Flip 2: Heads
Flip 1: Heads // Flip 2: Tails * *
Flip 1: Tails // Flip 2: Heads *
Flip 1: Tails // Flip 2: Tails

I have color coded the two questions to show which scenarios match up. 2 out of 4 scenarios have 1 heads and 1 tails, but only 1 out of 4 scenarios have heads first and tails second, giving probabilities of 2/4 and 1/4.

8. Dec 1, 2010

### Nick.S

Yeah the 1 heads 1 tails example has a higher probability because the order doesn't matter. I get that.

But I'm still lost on when to apply factorials.

So back to my example if events 1/2, 1/2, 1/3 must all happen but order doesn't matter, would it be 1/12 and then you divide that by a factorial of 3?

Last edited: Dec 1, 2010
9. Dec 1, 2010

### dav2008

I think what's a bit confusing (and maybe I added to the confusion) is that there are basically four cases: permutations with repetitions, permutations without repetitions, combinations with repetitions, and combinations without repetitions.

Each case has a formula which you can apply.

My coin flip experiment had repetitions, while your lottery experiment didn't. Sorry about that. The coil flip was an example of combinations with repetition and permutations with repetition. The lottery is typically an example of permutations without repetition (although pick-3 games are combinations with repetition)

Here's a site I found that gives an overview of each case and gives you how to apply the formulas: http://www.mathsisfun.com/combinatorics/combinations-permutations.html

10. Dec 1, 2010

### Nick.S

Thanks for the site.

I'm still having trouble trying to understand how this would work in my scenario however.

Let's say you have event A, B, C. 3 events,

probability of A is 1/2.
probability of B is 1/2.
probability of C is 1/3.

what is the probability of all 3 occurring? Order doesn't matter, they are not dependant on one and other.

How would i calculate this and would i apply a factorial of 3! or not? This is where I'm stuck.

11. Dec 1, 2010

### dav2008

What do you mean by "Probability of A is 1/2"?

I will interpret it as the following: In a given time period (a day, let's say) there is a 1/2 chance of A happening, 1/2 chance of B happening, and 1/3 chance of C happening.

In this case if you look at a given day, there will be a 1/2*1/2*1/3 chance that all 3 will happen during that day. This has nothing to do with permutations or combinations.

Combinations/permutations are involved when you have some set and you're randomly picking members of that set and want to find out how many different ways you can pick things from the set.

12. Dec 1, 2010

### Nick.S

Yes! your interpretation is correct. Heh.

So yeah like let's say I'm asking "what's the chance of A and B and C all happening in no specific order"

event A has a 1/2 chance of happening
event B has a 1/2 chance of happening
event C has a 1/3 chance of happening

so you're saying i don't have to worry about permutation or factorials with this example?

i thought i did because the way they write the lotto formula it lists a bunch of fractions, multiplies them all and then divides by a factorial.

13. Dec 1, 2010

### dav2008

In the case of the lotto formula, you are considering a permutation without repetition. You are dividing by the factorial because you want to eliminate all the duplicates (i.e. if you draw a 24 and then a 43 its the same as 43 and then 24) In that case you are basically counting the number of different ways you can draw 6 lotto balls. Then to determine the odds you are counting possibilities and then saying probability = [number of states that are desired] / [number of total states]. If you bought one lottery ticket then there is one set of numbers that will win. If there are 10,000,000 different permutations of numbers, then your odds are 1/10,000,000

Your example has nothing to do with combinations or permutations. You are saying "Today, there is a 50% chance of a fire, 50% chance of a drive-by shooting, and a 33% chance of a dog giving birth. What are the odds that there will be a fire, a drive-by shooting, and a puppy giving birth today?" The answer would be 1/2*1/2*1/3. Factorials come into play when you are counting things.

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