Hatrack River Forum: D&D Probability Question

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Posted by Blayne Bradley (Member # 8565) on :

if I roll a 20 sided die 1 to 20 but if it lands on a one I have to reroll again whats the probability of the next roll landing on a 1?

I think theres a feat that lets me reroll a failed save on a 1 which is my only weakness for my epic level wizard in neverwinter nights.

Posted by Hobbes (Member # 433) on :

I'm not sure I can fully parse that question but in probability you've got to be careful about exactly how you ask the question. "If I roll a '1' the first time what's the chance I'll role a '1' the second time?" is very straight forward: 1/20. Dice (assuming proper conditions) don't have memory, it doesn't matter if you rolled '1' one hundred times before, there's a 1/20 chance the next roll will be a '1'. Though if that happened I'd check your dice.

Hobbes [Smile]

Posted by Blayne Bradley (Member # 8565) on :

my statistics class a few yonks back I think would disagree but i dont recall the details.

Posted by Puppy (Member # 6721) on :

Each time you roll a die, the chance of a 1 is 1/20. However, the event of rolling two successive 1's, with no other result in between, is more rare — I'm just not sure how that is calculated exactly. 1/20 times 1/20, I'd imagine? 1/400?

Would that mean the time I rolled three 1's on three d20's cast from the same hand at the same time was a once-in-8000-times occurrence?

Posted by Hobbes (Member # 433) on :

Puppy, yes but we must be careful with that. If we find that 20% of all mutual funds beat some particular index, and that one mutual fund in particular beat it 10 years in a row we might say: there's a (1/5)^10, or about 10^-6% chance that it would do that. However, there's more than one mutual fund, so the chance that a mutual fund does so is actually much greater. So it's accurate that the chance of rolling three ones in a row in three rolls is one in eight thousand, that doesn't give an accurate view. First, the chance of rolling any particular number three times in a row in three tolls is 1/400, second you'll roll a lot more than three times in period in which you would remember; over the course of your life, if you play a lot of games with dice, this should happen to you ... well a lot. Third, getting three of one number in a row is noticeable, but is that the only pattern you'd notice? What about '1, 2, 3', or maybe you're OCD enough to notice the prime sequence '7, 11, 13'. So then the question becomes how many noticeable patterns are there? And how do they interact? It actually increases the probability as one sequence can fill in another. Point being most things aren't as unlikely as we think they are, but we're built to find patterns so it's easy and common to create meaning in randomness.

(FYI Geoff, that wasn't directed at you, your comment was just the springboard for something I've been thinking about [Cool]

)

Hobbes

Posted by BlackBlade (Member # 8376) on :

As others have noted there's a difference between asking these two questions.

1: What are the odds of rolling 1 twice in a row?

2: If I just rolled a one, what are the odds of rolling one again?

The answer to the second question is 1/20 which seems to be the question you are asking.

Also, I'd wager AC, fortitude, reflexes, and athletics are all weaknesses your wizard has. [Razz]

Posted by Xavier (Member # 405) on :

quote:
my statistics class a few yonks back I think would disagree but i dont recall the details.

I had an argument with my older brother years and years ago about this once, and when we appealed to my father, he sided with my brother [Grumble]

.

They were both wrong, and your "statistics class" are also wrong, or more likely your memory is faulty. The universe doesn't say "you know, he rolled a 1 last time, lets choose a different result this time!"

Posted by Hobbes (Member # 433) on :

Fascinating probability questions (from a great book I got for Christmas):

Q: If a couple has two children, what is the probability they're both girls?

A: 1/4

Q: If a couple has two children and one is a girl, what is the probability they're both girls?

A: 1/3

Q: If a couple has two children and one is a girl named 'Mellisa', what is the probability they're both girls?

A: 1/2 (or slightly better depending on how you want to look at it)

I told this to a few people at a party once (I don't want to hear about it, I know I'm a nerd!) and got in a huge debate with like 4 guys at once over my answers. I won of course. [Wink]

Hobbes

Posted by BlackBlade (Member # 8376) on :

Hobbes:

quote:
Q: If a couple has two children and one is a girl, what is the probability they're both girls?

A: 1/3

Q: If a couple has two children and one is a girl named 'Mellisa', what is the probability they're both girls?

A: 1/2 (or slightly better depending on how you want to look at it)

OK I'll ask it, and I'll just tell myself I'm asking for all the hatrackers with more pride than me, how did 1/3rd become 1/2? What does having a name have anything to do with the probability increasing?

Posted by Hobbes (Member # 433) on :

When I read it, I honestly didn't believe it. Even after reading the explanation I still didn't, I had to figure it out myself (which is to say if this is a test of intelligence to understand it, I failed the test myself [Cool]

), but I'll give it a shot.

The first time you know nothing about the children, so let's create a list of possibilities and organize them by order of birth (B = Boy, G = Girl):

BB
BG
GB
GG

Very easy, and there's nothing that makes one more likely than the other; so there's four equally likely possibilities and only one that we want. 1/4

The second time we'll take that same list:

BB
BG
GB
GG

But now the first one can't happen given our extra information (there's one girl) so we take it out. Now we have:

BG
GB
GG

Again no one option is more likely than the next and only one is the one we're after: 1/3.

Finally the confusing one. The trick to this is how we create the list. My hint to getting it is to recognize that if we were specific about which child was a girl, the first or the second, the odds would've been 1/2 instead of 1/3 in the last one. If we said "the first child is a girl", or "what are the chances the other child is a girl" (which sounds so similar but is different!) then we'd be at 1/2. Naming the child is just a clever way of being specific about the child.

So back to our list, we'll add a third abbreviation: M - girl named Melissa.

BM
MB
GM
MG
MM

The last one is not nearly as likely as the other four (both girls named Melissa!) but with the information given in the question, the other four are just as likely. So if we eliminate the fifth option we have four options and two are what we're looking for: 1/2. If we choose to include the fifth at a reduced probability than it becomes > 1/2.

Hobbes [Smile]

Posted by scholarette (Member # 11540) on :

I thought that with the name, you need to know the probability of the name to calculate that exactly.

For the 1/3, if you know that one is a girl, you have the following options:
a) girl then boy
b) girl then girl
c) boy then girl

So, only 1 out of three possible outcomes give 2 girls.

ETA- Hobbes beat me to it.

Posted by scifibum (Member # 7625) on :

Hobbes, I haven't found a problem with the .5 answer to #3, but it's definitely a puzzler. Perhaps even more counter intuitive than the Monty Hall problem (But that problem is the one that taught me my intuition with probability is untrustworthy).

However, I don't think I can come around to the >.5 answer. I don't think MM is a distinct option. Known M + another M is a subset of MG, and another M + known M is a subset of GM. It doesn't represent an additional option.

Posted by Hobbes (Member # 433) on :

Interesting point SF, I'll have to think about it. In what I was reading they definitely didn't address that, though to be honest the whole explanation was lacking in my mind which is why I had to come up with this one. You very well may be right; for sure a good puzzler.

I remember when I heard the Monty Hall problem, that took some time! Now it seems so obvious but before you're given the key to it, or fully comprehend the key anyways, it's obviously wrong!

We, as humans, just really don't "get" probability. We're not good at it, and if it's takes minutes or hours and scribbling things down on paper to try and come up with the probability of having a couple of daughters what makes us think we have a clue about if it's safe to, say, let our kids walk to school instead of drive them or any of the other mundane, everyday choices were constantly making with little time to think about and yet could change our lives? I guess the only hope there is that it's worked out pretty well so far. [Big Grin]

Hobbes

Posted by Elmer's Glue (Member # 9313) on :

And just like that I've lost all interest in taking a probability/statistics class.

Posted by MightyCow (Member # 9253) on :

Sometimes, when I read stuff like this, it reminds me how stupid math can be [Razz]

Posted by T:man (Member # 11614) on :

Hee hee hee.

Posted by just_me (Member # 3302) on :

I'm not a probability expert, but I'm decent at it and just brushed up on it to take the FE test.

So, I don't buy it. Knowing the girl's name does absolutely nothing to change the probability as far as I can see. Naming her doesn't establish order, so that doesn't change things for example.

Of course, I reserve the right to be wrong, but someone's gonna have to be awfully convincing...

Posted by lem (Member # 6914) on :

This thread reminds me of the beginning of Rosencrantz and Guildenstern are Dead, the movie. I wish I could find the whole clip for their coin tossing.

quote:
I feel the spell about to be broken......well, an even chance.

lol.

Posted by Kwea (Member # 2199) on :

5%, regardless. It is an infinite set, with no total amounts of tries, and each time you pick up the die and roll it you have a 5% chance of rolling a 1.

Die don't have a memory chip.

BEFORE you roll any of them it would be a 1/400 chance. However, don't fall for the gamblers fallacy, which would be to bet a huge bet after the second time saying it wouldn't happen again. EACH time you roll it is a 5% chance it will happen.

The preceding times have NO impact on the probability of any other roll.

(What BlackBlade said)

Posted by King of Men (Member # 6684) on :

Naming the girl changes the sample size drastically; you go from "All two-child parents with at least one girl" to "All two-child parents with at least one Melissa". That must be a change by a factor of several hundreds. Why would you expect probability to remain constant over such a dramatic drop?

Posted by King of Men (Member # 6684) on :

Touching the original question, what people have been saying is true for ideal dice (not for real ones, but it's a really excellent approximation) but may not be true for the Neverwinter Nights random number generator. It's likely not so random as all that.

Posted by Blayne Bradley (Member # 8565) on :

I mean in D&D if I roll a 1 on my fortitude/will saves against say a Beholder I will be petrified, I think there's a feat that allows a reroll to see if I still roll a 1, if I don't roll a 1, then the next time I roll a 1 it gets to be rerolled again.

I'm trying to figure out if this probability is worth it.

Posted by Kwea (Member # 2199) on :

http://en.wikipedia.org/wiki/Gambler's_fallacy

[ May 12, 2009, 11:52 AM: Message edited by: Kwea ]

Posted by fugu13 (Member # 2859) on :

KoM: their RNG is most likely pretty good (and for all I know they use a good source of entropy to re-seed periodically), so if there's any variation in the (perceived) probability, it should be infinitesimal.

Posted by fugu13 (Member # 2859) on :

BB: if you could state the feat more clearly (what you wrote could be interpreted at least two or three different ways), it would be possible to determine what the change in probability of success is due to the feat.

Posted by King of Men (Member # 6684) on :

quote:
Originally posted by fugu13:
KoM: their RNG is most likely pretty good (and for all I know they use a good source of entropy to re-seed periodically), so if there's any variation in the (perceived) probability, it should be infinitesimal.

You would think so, but game RNGs are occasionally really crappy, for some reason. (Not all of them, obviously; NN may not be one of them.) Some companies insist on writing their own, and aren't smart enough to do a good job. There was even one company that deliberately wrote one with the gamber's fallacy in it, so that a roll of 1 genuinely did make you less likely to roll low next time.

Blayne, you are going to have to ignore your ADD and use more than thirty seconds to post your question. Take a deep breath and use some dang punctuation; you are nowhere near as cool as Yahtzee and anyway, even he wouldn't put questions about probability in his reviews and expect people to answer.

Posted by swbarnes2 (Member # 10225) on :

I still don't buy the 1/3. I don't think that all GG, BG and GB are equally likely.

I think naming the kid Melissa makes it clear...If Melissa has one and only one sibling, what are the odds that it's a girl? 50%. The name doesn't make a difference.

Posted by King of Men (Member # 6684) on :

quote:
I don't think that all GG, BG and GB are equally likely.

Yeah, you're wrong. I revoke your posting-in-math-related-threads privileges. Go find some fluffy subject to post about.

Posted by King of Men (Member # 6684) on :

Here, I wrote a Monte Carlo simulator. Run it and see for yourself:

code:

#include <iostream>
#include <sys/time.h>
timeval one = {0, 0};

int main (int argc, char** argv) {
  gettimeofday(&one, 0);
  srand(one.tv_usec);

  int twoBoys  = 0;
  int twoGirls = 0;
  int boyGirl  = 0;
  int girlBoy  = 0;

  for (int i = 0; i < 1000000; ++i) {
    bool firstIsBoy  = (0 == rand() % 2);
    bool secondIsBoy = (0 == rand() % 2);

    if (firstIsBoy) {
      if (secondIsBoy) twoBoys++;
      else boyGirl++;
    }
    else {
      if (secondIsBoy) girlBoy++;
      else twoGirls++;
    }
  }

  std::cout << "Two girls : " << twoGirls << std::endl
            << "Girl, boy : " << girlBoy  << std::endl
            << "Boy, girl : " << boyGirl  << std::endl;

  return 0;
}

My results:

Two girls : 249741
Girl, boy : 249958
Boy, girl : 250606

Equally likely, dangit!

Posted by BlackBlade (Member # 8376) on :

quote:
Originally posted by fugu13:
BB: if you could state the feat more clearly (what you wrote could be interpreted at least two or three different ways), it would be possible to determine what the change in probability of success is due to the feat.

I just reread the entire thread trying to figure out what I'd said that would engender that statement. I will never get used to folks calling Blayne 'BB'.

Posted by Blayne Bradley (Member # 8565) on :

quote:
Originally posted by fugu13:
BB: if you could state the feat more clearly (what you wrote could be interpreted at least two or three different ways), it would be possible to determine what the change in probability of success is due to the feat.

I am having difficulty finding it as I'm not at my computer at home.

Posted by MrSquicky (Member # 1802) on :

Hobbes's two kids thing relies on the difference between talking about generalized sets and specific individual items.

When you're just talking about two kids, one of which you know is a girl, you're talking about a generalized set, so your sample space is

BG
GB
GG

but when you have information that makes one of them a specific item - or in this case person - then it is (S = specific person)

SB
SG

This information can be anything that distinguishes between the two children, names, which is older, hair color (if you know it is different), etc.

As swbarnes pointed out, if you knew that Melissa had 1 sibling, it's easy to see that Melissa having a sister is 50% likely.

It's because you have no way of distinguishing which child is referred to by saying "at least one is a girl" in the first case that the probability of two girls is 1/3.

Posted by Nighthawk (Member # 4176) on :

KoM: Your example has to make the assumption that RAND_MAX is an odd number.

It usually is, but still. [Smile]

Posted by swbarnes2 (Member # 10225) on :

quote:
Originally posted by King of Men:

quote:
I don't think that all GG, BG and GB are equally likely.
Yeah, you're wrong.

Of course that's the result of your script, but I don't think your script is calculating the right thing for this question.

I don't see why naming the kid changes the result. Any girl with exactly one sibling has a 50% chance of having a sister.

There aren't just three conditions, there are four:
Boy-the girl we are told is there
Girl we are told is there -Boy
Girl we are told is there -another girl
another girl-girl we are told is there.

If we assert that the elder is a girl, then it's obvious that the answer is 50%. If we assert that the younger is a girl, then its obvious that it's 50%.

So what is the third scenario that you think is happening to drag down the overall average of second girl to 33%?

Posted by MrSquicky (Member # 1802) on :

swbarnes2,
It is specifically that you don't know whether the girl you are told about is older or younger that brings the 33% in.

Posted by King of Men (Member # 6684) on :

Yes, but, hang on a moment. What is the difference between these two questions? (swbarnes, I misunderstood what you were arguing against, sorry.)

quote:
If a couple has two children and one is a girl, what is the probability they're both girls?

If a couple has two children and one is a girl named 'Melissa', what is the probability they're both girls?

Posted by MrSquicky (Member # 1802) on :

The chance that the girl we're told about is the youngest and has a sister is 1/4 (3/4 that she isn't). Likewise for if she is oldest.

So, the chance that picking one of them is going to be (1/4)/(3/4) or 1/3.

edit: Actually, wait...that's right, but I'm not explaining it correctly.

Posted by King of Men (Member # 6684) on :

Well. I admit to scratching my head, but the answer does appear to be different. I assign a 1% chance of a given girl being named Melissa, like so:

code:

#include <iostream>
#include <sys/time.h>
timeval one = {0, 0};

int main (int argc, char** argv) {
  gettimeofday(&one, 0);
  srand(one.tv_usec);

  int twoBoys  = 0;
  int twoGirls = 0;
  int boyGirl  = 0;
  int girlBoy  = 0;
  int twoMelissas = 0;
  int melissaGirl = 0;
  int girlMelissa = 0;
  int melissaBoy  = 0;
  int boyMelissa  = 0;

  for (int i = 0; i < 1000000; ++i) {
    bool firstIsBoy  = (0 == rand() % 2);
    bool secondIsBoy = (0 == rand() % 2);

    if (firstIsBoy) {
      if (secondIsBoy) twoBoys++;
      else {
        if (0 == rand() % 100) boyMelissa++;
        else boyGirl++;
      }
    }
    else {
      if (secondIsBoy) {
        if (0 == rand() % 100) melissaBoy++;
        else girlBoy++;
      }
      else {
        if (0 == rand() % 100) {
          if (0 == rand() % 100) twoMelissas++; // Assumes independence in names, probably wrong
          else melissaGirl++;
        }
        else {
          if (0 == rand() % 100) girlMelissa++;
          else twoGirls++;
        }
      }
    }
  }
  std::cout << "Two boys  : " << twoBoys  << std::endl
            << "Two girls : " << twoGirls << std::endl
            << "Girl, boy : " << girlBoy  << std::endl
            << "Boy, girl : " << boyGirl  << std::endl
            << "Melissa and girl : " << (melissaGirl + girlMelissa + twoMelissas) << std::endl
            << "Melissa and boy  : " << (melissaBoy + boyMelissa) << std::endl;

  return 0;
}

And behold, there are as many Melissas with boys as Melissas with girls:

Two boys : 250036
Two girls : 245053
Girl, boy : 247831
Boy, girl : 247138
Melissa and girl : 4939
Melissa and boy : 5003

Posted by fugu13 (Member # 2859) on :

swbarnes2:

Your choices aren't disjoint, while they are in the 'Melissa' case (well, the last MM choice actually isn't unless you make the earlier choices G other than Melissa. If you assume uniform distributions, that's unnecessary, and you just leave the MM term out as it is already accounted for).

Starting from the beginning:

We have a large number of families, all with two children. Each sex is equally likely and independent, so there are approximately equal numbers of boy-boy, boy-girl, girl-boy, and girl-girl families. If we assert one member of the family is a girl, that gets rid of one equally sized group (say there are around 1283 of each). Out of the 3849 families left, in 2566 the other child is a boy, and in 1283 the other child is a girl. One in three.

Posted by King of Men (Member # 6684) on :

Ah. I think I get it. Given that one girl exists, the chance that the other is a girl is one-third; I think we all agree on this. But if there are two girls, then you have two chances for one of them to be named Melissa! (Ignore the tiny chance that they share a name - this is even more unlikely on cultural grounds than on probabilistic ones.) So the probability of the given facts is (chance of two girls (1/3) times chance at least one of them is named Melissa (2*1%)), versus (chance of one-each (2/3) times chance the single girl is Melissa (1%)), and we can see that these are equal. Hence 50-50.

Posted by Sterling (Member # 8096) on :

I'm reminded of the last game of Settlers of Catan I played. "8" came up once, late in the game, after two "12"s, three "2"s, and a seemingly interminable string of "4"s.

I sometimes think SoC was designed largely to burn down probability's house while doing unspeakable things to its pets.

As to the original question- I agree, 1 in 400 before the first roll, 1 in 20 after the first 20 is rolled, but noting that the physical nature of the die and the roller probably drop some (difficult to accurately establish) bias into the equation.

Posted by MEC (Member # 2968) on :

I did some math and the Melissa thing makes sense to me now.

Posted by swbarnes2 (Member # 10225) on :

Grinstead and Snell's Introduction to Probability

pp175-177

“Much of this section is based on an article by Snell and Vanderbei”

“This problem and others like it are discussed in Bar-Hillel and Falk. These
authors stress that the answer to conditional probabilities of this kind can change
depending upon how the information given was actually obtained. For example,
they show that 1/2 is the correct answer for the following scenario.

Mr. Smith is the father of two. We meet him walking along the
street with a young boy whom he proudly introduces as his son. What is the
probability that Mr. Smith's other child is also a boy?
As usual we have to make some additional assumptions. For example, we will
assume that if Mr. Smith has a boy and a girl, he is equally likely to choose either
one to accompany him on his walk. In Figure 4.13 we show the tree analysis of this
problem and we see that 1/2 is, indeed, the correct answer.

It is not so easy to think of reasonable scenarios that would lead to
the classical 1/3 answer. An attempt was made by Stephen Geller in proposing this
problem to Marilyn vos Savant. Geller's problem is as follows: A shopkeeper says
she has two new baby beagles to show you, but she doesn't know whether they're
both male, both female, or one of each sex. You tell her that you want only a male,
and she telephones the fellow who's giving them a bath. \Is at least one a male?" she asks. “Yes," she informs you with a smile. What is the probability that the
other one is male?
The reader is asked to decide whether the model which gives an answer of 1/3
is a reasonable one to use in this case.”

Posted by MightyCow (Member # 9253) on :

With the feat, you have to take into consideration more than just probability.

If you fail the save, you're dead. Game over. If you get another save, that's another chance at life.