Probability question.

[AstralFire]

From a pool of 12, 5 will be selected. Each choice is mutually exclusive. You cannot pick choice "A" (or any other choice) more than once.

What are the odds that any one given choice will be among that group of five?

This is a gaming related question, not homework, if anyone's curious.

[Eloel]

12 choose 5 = 12!/(7!*5!)

You can also think as you choose 5 objects
12*11*10*9*8, but that would make an ordered set. Divide that by 5!, since you don't care about the order.

Edit: Of course, that's the number of different permutations. I misread.
Probability is alot easier.
There are 12 spots to be in, 5 of which are chosen. 5/12.

[AstralFire]

Wait, did I really over think it that much?

[Brother Oni]

Wait, did I really over think it that much?

For the question as given, then probably.

You grab 5 items in one go from a bag, chance of a particular item being in those grabbed = 5/12.

Depending on how you make the picks or the uniqueness of the items, the probability can change.

[Douglas]

Wait, did I really over think it that much?
Yes. Yes, you did.

Allow repeat picks, or care about the order, and it gets a little more complicated. As is, it's just a straight 5/12.

[Wardog]

Wait, did I really over think it that much?

Yes - although if you still want to over-think it, you can do it this way:

On the 1st pick, the chance of not getting it is 11/12.
On the 2nd pick, the chance of not getting it is 10/11.
On the 3rd pick, the chance of not getting it is 9/10.
On the 4th pick, the chance of not getting it is 8/9.
On the 5th pick, the chance of not getting it is 7/8.

The chance of not getting it after 5 picks is 11/12 * 10/11 * 9/10 * 8/9 * 7/8
=55440/95040
=7/12

So the chance of not (not getting it within 12 5 picks) is 1 - 7/12
=5/12

[Brother Oni]

So the chance of not (not getting it within 12 picks) is 1 - 7/12
=5/12

I think you got a typo here as the chance of (not getting it in 12 picks) should be 0 since there's only 12 choices in total.

I blame it on the double negative. :smalltongue:

[thubby]

if the objects are actually being removed and the selection pool is decreasing then ya.
binomial distributions are a lot more algebra @,@

[Wardog]

I think you got a typo here as the chance of (not getting it in 12 picks) should be 0 since there's only 12 choices in total.

I blame it on the double negative. :smalltongue:

Oops! (And now corrected).

[AstralFire]

Can I just say I feel like a moron right now? So much for being proud of my math skills...

[warty goblin]

if the objects are actually being removed and the selection pool is decreasing then ya.
binomial distributions are a lot more algebra @,@

If you're drawing without replacement, that's a hypergeometric, not binomial. Binomial assumes independent draws with equal probability of success, both of which are violated by drawing without replacement.

[Astral Avenger]

I'm trying to work out the probabilities and average result for roll NdX best/worst Y, anyone know the equations for that offhand?

[Eloel]

I'm trying to work out the probabilities and average result for roll NdX best/worst Y, anyone know the equations for that offhand?

There's this (http://rumkin.com/reference/dnd/diestats.php), but no equations on my part.

[shadow_archmagi]

I'm trying to work out the probabilities and average result for roll NdX best/worst Y, anyone know the equations for that offhand?

http://anydice.com/ should do what you need.

[Traab]

I have a different probability question, its one that comes up a lot of the time on mmorpg message boards. When a mob has a specific chance to drop an item, for example, .1%, does it really matter how many times you try to get it to drop for your odds of getting it next time?

I say no because while statistically you SHOULD have gotten it in the last 999 attempts, that doesnt mean you are any more likely to get it on your 1000th kill than you were on your first try. Some try to argue that because it has a .1% drop rate, it almost HAS to drop at least once every 1000 kills, so the more times you try and fail to get it, the more likely you will get it next time. I think thats confusing statistics with probability, especially when you consider in the case of games like world of warcraft where there are millions of players killing that single spawn potentially a billion times or more overall in order to reach that average.

[warty goblin]

I have a different probability question, its one that comes up a lot of the time on mmorpg message boards. When a mob has a specific chance to drop an item, for example, .1%, does it really matter how many times you try to get it to drop for your odds of getting it next time?

I say no because while statistically you SHOULD have gotten it in the last 999 attempts, that doesnt mean you are any more likely to get it on your 1000th kill than you were on your first try. Some try to argue that because it has a .1% drop rate, it almost HAS to drop at least once every 1000 kills, so the more times you try and fail to get it, the more likely you will get it next time. I think thats confusing statistics with probability, especially when you consider in the case of games like world of warcraft where there are millions of players killing that single spawn potentially a billion times or more overall in order to reach that average.

This is a problem that comes up because people don't understand when they are or are not making a conditional statement. You are correct that, given your item did not drop on the last 999 kills, the probability of it dropping on your next kill is 1/1000, just as always. The other people are correct that the probability of getting a 1/1000 drop over 1000 kills is very high (about 63% in fact)*.

This seems like a contradiction, but it isn't. You are making a claim about the probability of an event occurring on the next kill. If drops are independent, then what happened previous kills gives you no information about the next, so the probability of a drop is 1/1000 as always. They are making an argument about the probability of an event occurring over a sequence of 1000 kills. These are completely different statements, but you can relate the two by thinking about whether this particular sequence of 1000 kills is one that includes the item dropping, or not. Now the previous 999 kills carry no information about what happens when you gank the next orc, so the probability of the item dropping is still 1/1000, which is exactly the same as if you worked through the probability of (1 drop in 1000 given 999 no drops)



*The average number of kills until the item drops is still 1000 though. Averages can be funny things. Whenever somebody hands you an average without a measure of spread, and preferably skewness as well, you should tell them that they fail at statistics forever, and will be chased by giant man-eating sigmas in their miserable, uncertainty plagued afterlife. This is particularly true if they tell you half the people you know are below average. If you know enough of these people, the proportion can be well in excess of 1/2.

[sktarq]

It makes no difference to the chance that you next spawn will drop an item with a .1% chance depending how many of that spawn you have killed. . . unless blizzard starts coding it for that. Just because you have killed 1000 doesn't mean that the player "deserves" that item. Remember some lucky sod will get that item the first time they meet the spawn. . . 1 in 1K people even. To get back to your average then somebody would have to kill 1999 of the spawn without getting the item at all (on average-2 going to 1.5K kills would work etc).

[Brother Oni]

Some try to argue that because it has a .1% drop rate, it almost HAS to drop at least once every 1000 kills, so the more times you try and fail to get it, the more likely you will get it next time.

These are the sort of people who should stay away from Las Vegas for their own good.

[warty goblin]

These are the sort of people who should stay away from Las Vegas for their own good.

Everybody should stay away from Vegas for their own good. Remember, if the casino's still in business, they've making more money than they're losing, and the only sane expectation is that you, personally, will lose money.

[Brother Oni]

Everybody should stay away from Vegas for their own good. Remember, if the casino's still in business, they've making more money than they're losing, and the only sane expectation is that you, personally, will lose money.

It depends on why you're going to a casino - if you're just going there to spend a little money and be entertained for a while, then it's fine. If you're going there with the intention to make money, then you should really rethink your decision.

[warty goblin]

It depends on why you're going to a casino - if you're just going there to spend a little money and be entertained for a while, then it's fine. If you're going there with the intention to make money, then you should really rethink your decision.

Or you can stay home, simulate a random number between 0 and 1, and if it's less than .99, set fire to the contents of you wallet. It's probably cheaper, it's a lot more convenient, and I strongly suspect that the probability of blowing the kids' college fund and taking a high interest short term loan with your knees as collateral from a man named Dogmeat is substantially lower.

And you get to play with fire.

[AstralFire]

Okay, I have an actual problem here.

There are two sets of six user-input variables, ranging from 0 to 31. A1, B1, C1, D1, E1, F1 and A2, B2, C2, D2, E2, and F2. At random, one letter pair (called X) is struck entirely and replaced with X', which is randomly chosen from 0 to 31. The surviving pairs each have one number eliminated from them (at random), so that, including X', there are six numbers remaining.

What's the general formula for any specific combination of six or fewer numbers to result? (Or fewer meaning that the end user does not care about one or more values.)

What if only three of the six pairs were chosen, with the remainder replaced by randomly generated values?

[warty goblin]

The user input values can be taken as constants, right? That is, we're only interested in the probability of strings like (A1 B2 C2 X E1 F1) given the user inputs, and not the probability of getting a specific string of numbers across all possible user inputs?

Assuming that's the case, and using the notation (K, x) for 'k choose x:

There are 6 choices for the single pair to delete, that's (6, 1) = 6.

There are 32 choices for the single number that replaces that pair, or (32, 1) = 32

For each of the remaining five pairs, there are 2 numbers, one of which is chosen, so (2, 1)^5 = 32.

All these are determined independently, so the terms are joined by multiplication. That gives the denominator = 6144. The probability of any particular sequence of six numbers is thus 1/6144.

For the choose three to remove, it's pretty much exactly the same, except that you have (6, 3), (32, 1)^3 and (2, 1)^3. The denominator will now be exceedingly large.


The ignoring digits problem is a little bit thornier, but not much. Let's suppose we don't care about k digits, 0 <= k < 6. I'll assume the digits we don't care about aren't chosen at random - .e.g we always ignore the last k, or the first k, or whatever.

The way to do this is to look at the number of digits I do care about, which is 6 - k. The question then becomes how many choices are there for each digit?

Either the deleted pair is in the digits we care about, or it isn't. There are (6 - k, 1) ways for the deleted pair to be in the digits we care about, and (32, 1) choices for X. There are a remaining 6 - k- 1 digits to be filled, each of which has (2, 1) possible values, so (2, 1)^(6 - k - 1).

For the case where the deleted pair does not fall in the digits we care about, there are simply (2, 1)^(6 - k) possible values.

The probability of getting a particular sequence of 6 - k terms we do care about is then [(6 - k, 1)*(2, 1)^(6 - k - 1) + (2, 1)^(6 - k)]/6144. Just choose your k, and evaluate.

[Creed]

Dang, had this clip ready to fire before opening the topic:

http://www.youtube.com/watch?v=Zr_xWfThjJ0

[AstralFire]

1) Thank you for helping.
2) Am I... multiplying the numbers inside of the parenthesis? I think so, just want to check.
3) The values not cared about can be anywhere in the set.


The user input values can be taken as constants, right? That is, we're only interested in the probability of strings like (A1 B2 C2 X E1 F1) given the user inputs, and not the probability of getting a specific string of numbers across all possible user inputs?

Assuming that's the case, and using the notation (K, x) for 'k choose x:

There are 6 choices for the single pair to delete, that's (6, 1) = 6.

There are 32 choices for the single number that replaces that pair, or (32, 1) = 32

For each of the remaining five pairs, there are 2 numbers, one of which is chosen, so (2, 1)^5 = 32.

All these are determined independently, so the terms are joined by multiplication. That gives the denominator = 6144. The probability of any particular sequence of six numbers is thus 1/6144.

For the choose three to remove, it's pretty much exactly the same, except that you have (6, 3), (32, 1)^3 and (2, 1)^3. The denominator will now be exceedingly large.


The ignoring digits problem is a little bit thornier, but not much. Let's suppose we don't care about k digits, 0 <= k < 6. I'll assume the digits we don't care about aren't chosen at random - .e.g we always ignore the last k, or the first k, or whatever.

The way to do this is to look at the number of digits I do care about, which is 6 - k. The question then becomes how many choices are there for each digit?

Either the deleted pair is in the digits we care about, or it isn't. There are (6 - k, 1) ways for the deleted pair to be in the digits we care about, and (32, 1) choices for X. There are a remaining 6 - k- 1 digits to be filled, each of which has (2, 1) possible values, so (2, 1)^(6 - k - 1).

For the case where the deleted pair does not fall in the digits we care about, there are simply (2, 1)^(6 - k) possible values.

The probability of getting a particular sequence of 6 - k terms we do care about is then [(6 - k, 1)*(2, 1)^(6 - k - 1) + (2, 1)^(6 - k)]/6144. Just choose your k, and evaluate.

[warty goblin]

1) Thank you for helping.
2) Am I... multiplying the numbers inside of the parenthesis? I think so, just want to check.

No, (6, 2) is shorthand for 'choose 2 things from six without replacement, order doesn't matter.' Mathematically (k, x) = k!/(x!*(k - x)!), which reduces to k when x = 1. Here the ! means factorial, so x! = x*(x - 1)*(x - 2)*...*(2)*(1). 0! is defined to be one.


3) The values not cared about can be anywhere in the set.

That's not the same thing as chosen randomly. My calculations are still valid a sequence (CDDCDD) as they are for (CCDDDD), writing C for Care, D for Don't Care. The question is whether, when you generate the sequence, if you randomly decide which digits to care about or not. If you don't, then my calculations are valid.

If you do choose which digits you care about at random, then you just multiply the numerator by the number of ways to choose the digits you care about, using the previous notation that's (6, 6 - k).

[AstralFire]

Ooh! Factorials! I KNEW that they'd be involved in this at some point. I was just waiting and waiting.

:D

Okay, I'm going to try to actually apply this in Excel. Thank you. I don't quite get your response to my last, but maybe it'll make more sense when I'm examining the notation with a proper understanding.

[warty goblin]

Ooh! Factorials! I KNEW that they'd be involved in this at some point. I was just waiting and waiting.

:D

Okay, I'm going to try to actually apply this in Excel. Thank you. I don't quite get your response to my last, but maybe it'll make more sense when I'm examining the notation with a proper understanding.

Basically the question is whether the digits you care about are always the same (e.g. the first, fourth and fifth) or chosen randomly each time you generate the sequence (so the first time you look at the first, second and sixth, the next and the second, fourth and fifth for example).

[AstralFire]

Not chosen randomly, but the user will choose differently on each usage.

[Traab]

Dang, had this clip ready to fire before opening the topic:

http://www.youtube.com/watch?v=Zr_xWfThjJ0

I loved the monty hall test. I had heard about it before until I watched mythbusters not only explain the solution but also actively test it out so we could see that it all played out exactly how the statistics said it would.

[huttj509]

I loved the monty hall test. I had heard about it before until I watched mythbusters not only explain the solution but also actively test it out so we could see that it all played out exactly how the statistics said it would.

Though there is one key facet that tends to be left out of setups of the question: Is the Host required to always open a door, and offer the chance to switch?

In general, this is assumed, but often not explicitly stated, which can change things entirely. If you only get a choice to switch if you initially chose the car, switching always loses.

It's also interesting how many arguments among experts a straightforward problem like that can cause, similar to the "can an airplane take off on a treadmill?"

[Brother Oni]

It's also interesting how many arguments among experts a straightforward problem like that can cause, similar to the "can an airplane take off on a treadmill?"

Interesting that you note some conditions left out of the Monty Hall question, since there's three interpretations to the setup for that question (http://blog.xkcd.com/2008/09/09/the-goddamn-airplane-on-the-goddamn-treadmill/).