Random Statistics Question (sort of FFXI related)

[Byrd]

I always update the Dynamis drops for my linkshell, and a couple of weeks ago, I was trying to figure out the statistical probability of a certain scenario arising. I think I got it, but I know there are people here better versed in statistics than I. (The Monty Haul thread got me thinking about this). Perhaps this belongs in one of the FFXI threads, but it's not really an FFXI question so much as a statistical one.

The bottom line is, we've been doing Dynamis for almost two years, and have been to Sandy around 25 times. The weird part is we've gotten at least four pieces (most jobs more) of AF2 for every job in Sandy, except for RDM, which has seen zero. I was trying to calculate the odds of this.

First, I have to presume that the chances of each particular piece of AF2 are equal. That might be a large assumption, but I'm making it, and that's not really what my question is.

I was trying to calculate the odds that this would happen to any job. I first started off by trying to calculate the odds that every job would have two AF2 drop before any particular one had seen one. My thought is that some particular job has to be last (in this case, RDM), and the odds of each other job dropping a second one before RDM should be even to the odds of a particular job being the last to drop (1/11). This accounts for the fact that every other job has to drop a second AF2 (1/11 chance per AF2) before RDM gets a first drop (1/11 chance). In short, the odds of a particular job being dropped are equal to the odds of a particular job being the last to drop.

So, the odds of this scenario playing out seem to me to be (1/11) for the second set of AF2, (1/11) for the third and (1/11) for the fourth. (1/11)^3 or 1 chance in 1,331. This is only the chance for this scenario to happen, regardless of job. For it to be a particular job, the odds would increase to 1 in 14,641 as you would have to take the first set of AF2 into account as well.

Does this sound right, or am I crossing my hairs?

[Miji]

So, the odds of this scenario playing out seem to me to be (1/11) for the second set of AF2, (1/11) for the third and (1/11) for the fourth. (1/11)^3 or 1 chance in 1,331. This is only the chance for this scenario to happen, regardless of job. For it to be a particular job, the odds would increase to 1 in 14,641 as you would have to take the first set of AF2 into account as well.

Does this sound right, or am I crossing my hairs?

This seems to be the odds of the same AF piece dropping in the same run. It's been a few years since I took the class we handled this stuff in (my disclaimer), but here's how I think it goes:

Assumptions:
Equal chance of each AF piece dropping in any order.
4 drop every run (just based on some average to make the math easy, you could change the 4 to any number and figure it out just the same).
Infinite supply of each AF piece (or at least 4 of each).

The number of different ways you can fill up the 4 slots with an AF is a multiset combination "11 multichoose 4", where 11 are the number of different AF pieces and 4 is the number that drop on a run. "11 multichoose 4" breaks down to the binomial coefficient "11+4-1 choose 4" = "14 choose 4" = (14!)/(10! 4!) = 1001. So that's 1001 total different ways for the AF pieces to drop.

Now we have to separate the cases where a single particular piece appears at least once in the combination. After some manipulation, this ends up being "11 multichoose 3" (the 3 being our original 4 minus 1), since we can just pick any 3 combinations of gear and add the RDM piece for example and have a set that we don't want to count, noting that the "11 multichoose 3" also includes the option of having multiple RDM pieces drop. Either way, "11 multichoose 3" = "13 choose 3" = 286 possible combinations of 4 AF pieces that include at least one RDM.

That leaves us with 1001 total combinations, of which 286 include at least 1 RDM piece. So that's (1001-286)/1001 = ~71.43% chance that you'll get no RDM piece per run.

This seems kosher, but like I said, it's been a while.

[Keno]

So, the odds of this scenario playing out seem to me to be (1/11) for the second set of AF2, (1/11) for the third and (1/11) for the fourth. (1/11)^3 or 1 chance in 1,331. This is only the chance for this scenario to happen, regardless of job. For it to be a particular job, the odds would increase to 1 in 14,641 as you would have to take the first set of AF2 into account as well.

Does this sound right, or am I crossing my hairs?

This seems to be the odds of the same AF piece dropping in the same run. It's been a few years since I took the class we handled this stuff in (my disclaimer), but here's how I think it goes:

Assumptions:
Equal chance of each AF piece dropping in any order.
4 drop every run (just based on some average to make the math easy, you could change the 4 to any number and figure it out just the same).
Infinite supply of each AF piece (or at least 4 of each).

The number of different ways you can fill up the 4 slots with an AF is a multiset combination "11 multichoose 4", where 11 are the number of different AF pieces and 4 is the number that drop on a run. "11 multichoose 4" breaks down to the binomial coefficient "11+4-1 choose 4" = "14 choose 4" = (14!)/(10! 4!) = 1001. So that's 1001 total different ways for the AF pieces to drop.

Now we have to separate the cases where a single particular piece appears at least once in the combination. After some manipulation, this ends up being "11 multichoose 3" (the 3 being our original 4 minus 1), since we can just pick any 3 combinations of gear and add the RDM piece for example and have a set that we don't want to count, noting that the "11 multichoose 3" also includes the option of having multiple RDM pieces drop. Either way, "11 multichoose 3" = "13 choose 3" = 286 possible combinations of 4 AF pieces that include at least one RDM.

That leaves us with 1001 total combinations, of which 286 include at least 1 RDM piece. So that's (1001-286)/1001 = ~71.43% chance that you'll get no RDM piece per run.

This seems kosher, but like I said, it's been a while.

Assuming you've done 25 runs, and taking what Miji said as fact (just for the sake of arguement, im too tired to do it properly), and a 71.43% chance of RDM drop per run would give these statistics for a chance of no RDM over 25 runs:

We can set this up as a binomial distribution, where pi (chance of success) = .7143, s (the number of successes), and n (the number of trials). The forumla is:

{(n!)/[s!(n - s)!]} * (pi^s) * [(1 - pi)^(n - s)]

Therefore when we enter in our data, s=0, n=25, pi=.7143.

The left side of the equation comes out to

25!/25! = 1

, the next part

(pi^s; .7143^0) = 1

,

and therefore the only reliant part of the equation is

[(1 - pi)^(n - s)] = [(1-.7143)^(25 - 0)] = .2857^25 = 2.499E^-14

To the layman, the odds of that happening is .000000000002499%.

The problem is Miji didn't calculate the odds correctly of RDM AF dropping, but yeah thats for another day...

[Miji]

I'm curious now, so what the odds are of a particular job's AF not dropping? Looking over it again, it appears to be correct based on the assumptions, but it's been a few years since I last touched the stuff.

[Devek]

It isn't random though..

There is some factor.. you really notice it when you do it for years with the exact same people taking the same path etc..

There is just that one AF you get a lot of.

[Keno]

I'm curious now, so what the odds are of a particular job's AF not dropping? Looking over it again, it appears to be correct based on the assumptions, but it's been a few years since I last touched the stuff.

See the problem is that you did the AF as "4 drops per session, equal chance of dropping, so calculate like that"...

Thats entirely wrong calculation of the odds.

In actuality there are a certain amount of mobs that can drop AF, and each one has an equal chance to drop an AF piece. It then goes into a randimization to choose which AF drops from the mob.

It's sort of like this:

Mob death
-Calculate randomization, if (random) > 920, drop AF
-Else; no drop
-Recalculate randomization, if (random) falls within 1-75, BST, 76-150, BRD, 151-225 NIN, etc. etc.

Yes I'm well aware those aren't actually the drop rates, and I'm well aware that that may not be exactly how AF is decided on what drops, but for the sake of this circumstance it will be easiest to use.

So in this case it's something like 80/1000 chance each mob has to drop AF (1000-920), then its 1/15 of that to drop a certain piece. Because these both need to happen in conjunction for RDM AF to drop, we multiply the numbers together.

So on any given mob, the chance RDM AF will drop is 80/15,000, or 2/375, which is approximately .5% chance on each mob. Assuming you clear it everytime for farming, there are approx. 100 mobs (could be way off, just a guestimate).

Each mob (except the NMs) has an equal chance to drop RDM AF at .5% chance. So we add .5% 100 times, which gives us about a 50% chance RDM AF will drop each run.

Then again my estimates could be off, it could be less than a 8% drop rate or it could be more. There could be less or more than 100 mobs. Either of these will influence the rate.

But this is a more accurate way to calculate the odds of getting RDM AF.

If you wanted to calculate it the way you did then it would be best calculated as a binomial distribution (again), with odds of success (1/15) and n trials of 4 and s successes of 0. Leaving you with [1-(1/15)]^(4), which is approximately 76% of not getting any of a single (in this case RDM) AF on a single run.

Alternatively that means you have a 24% chance of obtaining a certain AF (RDM in this case) each run. Redoing the odds with that, and another binomial distribution with 0 successes and 25 trials, the odds of getting 0 RDM AF is .105% chance.

So you hit the unlucky jackpot. Only 1 out of every 1,000 linkshells who do Dynamis Sandy 25 times will not get RDM AF.

Keep in mind, these are statistics, which can hardly be used practically. Because theres a chance that you will not get RDM theres a chance no one will, and alternatively theres a chance that every piece will be RDM, etc. etc. I'm sure you understand the limitations of statistics, so I'll stop now.

[Miji]

See the problem is that you did the AF as "4 drops per session, equal chance of dropping, so calculate like that"...

Thats entirely wrong calculation of the odds.

In actuality there are a certain amount of mobs that can drop AF, and each one has an equal chance to drop an AF piece. It then goes into a randimization to choose which AF drops from the mob.

I see where we're differing now. He's not asking for an estimate before the run; he's asking after the fact. After the fact, we know how many dropped in a single run, and we don't have to use a random variable trying to predict how many dropped anymore. It's more accurate to find the odds of a particular run in which you know how many items dropped (which is what the OP is asking I believe), so you aren't stuck guessing drop rate and number of mobs. The 4 drops per session is beside the point with respect to the calculations, as he could just pick how many dropped and use the same method to discover what the odds were that a particular job didn't drop.

As for the drops not being random, I can't see them coding dynamis drop rates to be more complex than picking a random number after a mob that is capable of dropping AF dies. I never did much dynamis though, so I'm kind of curious what kinds of AF pop up more in each zone. Anybody else see similar results?

[GSDragoon]

http://img105.imageshack.us/img105/8...roprateif0.jpg

[Byrd]

To clarify:

I'm not really interested in the odds/run, that depends on too many variables including: number of mobs killed, TH available, moon phase (?), etc.

What I'm basically looking to calculate is:

Given a certain number of equally probable events, what are the odds that all events except one occur four (or more) times before the one particular event occurs once?

Here's what I'm thinking.

n=number of possible, equally likely events (in this case, number of different AF2 that drop in the zone, in Dynamis San D'oria, 11)
d=desired events (the last AF2 to drop, in this case RDM, but could have happened with any other job)
x=x number of times that the repetition occurs (in this case, complete sets of the other 10 AF2)

My thought for the equation: (n/d)^x or in this case (1/11)^3

[Miji]

To clarify:

I'm not really interested in the odds/run, that depends on too many variables including: number of mobs killed, TH available, moon phase (?), etc.

What I'm basically looking to calculate is:

Given a certain number of equally probable events, what are the odds that all events except one occur four (or more) times before the one particular event occurs once?

Here's what I'm thinking.

n=number of possible, equally likely events (in this case, number of different AF2 that drop in the zone, in Dynamis San D'oria, 11)
d=desired events (the last AF2 to drop, in this case RDM, but could have happened with any other job)
x=x number of times that the repetition occurs (in this case, complete sets of the other 10 AF2)

My thought for the equation: (n/d)^x or in this case (1/11)^3

What I put down should get you there. The (1/11)^3 gives you the odds of the same piece dropping 3 times in a row. What I posted is basically the odds of not getting a particular piece on a past run from which you know 4 pieces dropped.

[Maguspk]

/whispers:

Discrete Mathhhhhh!

it's your friend.

[Byrd]

The (1/11)^3 gives you the odds of the same piece dropping 3 times in a row.

Yes. But doesn't that also give you the same odds of the same piece being the last one to drop three times in a row? Four times in a row if you're going for a particular job.