Poll: A little math problem

[werepossum]

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Just back in this quagmire for a second this was explained wrong and I think it can help with the main problem.

you cannot look at the first nine coins in a set and still treat them as a set. In this example you have reduced yourself to only two options coin 10 is heads or tails leaving a 50-50 chance.

For this to remain a theoretical set no knowledge must be known about individual coins. Say if someone told you at least nine of those coins were heads. Then you would have 11 options. First all coins are heads OR coin 1 is tails OR coin is tails 2 OR coin 3 etc. Then the chance of all the coins being heads would be 1 in 11.

This is equivalent to both puppies male OR first puppy female OR second puppy female. Then the chance of both male is 1 in 3

Patently untrue. Mathematicians, scientists, and engineers constantly work with sets whose values are completely or partially known. Even if you have identified one or all puppies, you can still count them as a set - it's just that set probabilities no longer apply once you've identified a particular set item's value. While there is nothing wrong with your math, it represents a third set formation, not either of the two I discussed. My sets were a random set of ten coin tosses, of which the first nine (not any nine as in your example) came up heads; and a set of ten formed with the knowledge that the first nine (not any nine as in your example) are heads. In the latter case, the unlikely event of tossing nine heads in a row happens before you form your set. In the former case, each heads becomes more and more unlikely. I probably phrased it badly, but the first set was meant to represent the possibility that you have randomly thrown ten heads in a row before you know you have thrown nine heads in a row. The second set shows the change in the relative odds as you gain more information either by forming the set with more known information OR by discovering information about the set and/or its items.

My point with the two coin sets was that to show that when you form a set affects what possible probability distributions that set can take as much so as does what you subsequently learn about the set or its individual items. Each possible probability distribution has an unchanging absolute probability determined solely by the set construction - by the number of items and the possible values each can take. The relative value of each possible probability distribution is its absolute probability divided by the sum total of all possible probability distributions. When nothing is known about a particular set, the relative probability of each possible probability distribution is its absolute probability because the sum total of all possible probability distributions is one. As we learn information about the set OR about its individual items, we can eliminate some probability distributions that are possible for the set construction BUT which have been proved false for our particular set. The absolute probability of each possible probability distribution for the set construction remains unchanged because it is solely a function of the set construction, but the relative probability of each possible probability distribution for our particular set changes because by eliminating some possible probability distributions we reduce our divisor. This is true with both combinations and permutations.

To go back to our pup example, by learning there is at least one male we now have a 25% chance of a male/male combination and a 50% chance of a mixed pair combination. However our divisor has changed to 75% because we have eliminated the possibility (25% probability distribution) of a female/female combination. Thus the relative probability distributions remaining for our particular set are 25%/75% (or 1 in 3) for an all-male combination and 50%/75% (or 2 in 3) for a mixed pair combination. You cannot change the dividend without reforming the set; you can only change the divisor as you eliminate possible probability distributions. The problem is much the same for permutations, except that the 50% chance of a mixed pair combination becomes two separate 25% chances for male/female and female/male.

To see again how odds shift, let's look at the permutations. When I have two random pups, I know I have a 25% chance of having two male pups. When I learn that I have at least one male pup, my odds of having two increase to 33% because the 25% of having two females drops out of my divisor. It's still applicable to the set as a function and if I repeat this test a large number of times I'll get female/female pairs 25% of the time, but it's been proved false for my particular set. Now my chance of having a pair of males is 25%/(25%+25%+25%) = 33%. When I learn that one identifiable member of the set is male, my chance of having two males increases to 50% because one 25% permutation (that the identifiable member of the set is female whilst the other is male) drops out. Now my chance of having a pair of males is 25%/(25%+25%) = 50%. When I learn the other pup is male, my chance of having two males increases to 100% because I have all the knowledge I need to eliminate every other possible set probability distribution from this particular set.

Anyone who hasn't should read the Marilyn vos Savant link. http://en.wikipedia.org/wiki/Marilyn_vos_Savant
Besides shedding light on this problem, she's a fascinating person in her own right.

Those who remain absolutely convinced that 50% is the correct answer should definitely go into mathematics as a career, as you are smarter than every actual mathematician. Think of the opportunities!

 

[werepossum]

Okay, here's the problem. There's a difference between the sexes of a pair of puppies being unknown because we know nothing about the puppies, and the sexes of a pair of puppies being unknown because we know nothing about which pair was pulled from a pool of four known pairs, composed of one all male pair, one all female pair, and two mixed pairs.

What people are doing is they are confusing the permutation of possible results for two unknown puppies with a listing of the pool of puppy pairs from which a pair will be drawn because they both look the same.

In order for set probability to play a role, we need to know the composition of the pool from which the pair were drawn. WE DO NOT KNOW THAT! In order to know that the question would have to look like this:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair they are one of four pairs that the breeder had, and the breeder picked the pair from a pool that was made up of one all male pair, two mixed pairs, and one all female pair. 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 him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

See the kind of additional information you need to start using set probability? You can't just pretend because of a table of the possible permutations that you're picking from a pool of 4 pairs or 100 pairs like werepossum was talking about or 574543985770 pairs--until you get information that the pair of puppies was picked from a certain pool of pairs and you know something about the distribution of pairs in that pool, you can't just start using set probability.

Like I've been saying for a while now, you can't just pretend every symbol in a math problem refers to something in the real world. People have been reifying the possible permutations of two unknown events into actual pairs of puppies, and that is why they think they can just eliminate the F/F and rebalance the probability.

In short, the people who think it is 33% are mistaken because they think the information that at least one puppy is male means that the unknown pair was drawn from a pool of known pairs consisting of two mixed pairs and one all male pair, when what it really means is that a pair with at least one male was drawn from a pool of unknown pairs.

But you don't even need set mathematics to solve this. By your own example there are three out of four pup pairs that fit the description of "at least one male", but only one of those three sets will give you two males. What is the chance you'll draw the ONE of the THREE pairs of pups that gives you two males? The chance is ONE in THREE or 33%. You can't get any simpler than that.

 

[Aurora219]

The correct solution to this kind of question is quite simple.

If one is male and you wish to know if the other is male as well, you grab the damn phone and ASK.

 

[FrcknFrckn]

I'll explain as best as I can, using coin flips as an example.

The problem breaks down to this:
Your friend flips 2 coins.
You ask if at least one coin came up heads.
Your friend says "Yes".
What are the chances that both coins came up heads?

Ok, so I think we can all agree that the odds of any one coin flip being heads is 0.5. So:

- Probability of a single flip:
- P(H) = 0.5
- P(T) = 0.5

Thus the probability of the friend flipping heads THEN heads again is 50%x50%, or 25%. Same logic for heads then tails, tails then heads, or tails then tails.

- Probability of each 2-flip combination
- P(H->H) = 0.5x0.5 = 0.25
- P(H->T) = 0.5x0.5 = 0.25
- P(T->H) = 0.5x0.5 = 0.25
- P(T->T) = 0.5x0.5 = 0.25

So, the total probability of 2 heads is 0.25, same as for 2 tails. But the probability of flipping one of each is the sum of the probability of heads THEN tails and the probability of tails THEN heads.

- H T --- Prob.
- 2 0 --- 0.25
- 1 1 --- 0.25+0.25 = 0.5
- 0 2 --- 0.25

Now, since we know at least one coin was heads, we can ignore the third row, leaving us with:

- H T --- Prob.
- 2 0 --- 0.25
- 1 1 --- 0.5

Which, when normalized, becomes:

- H T --- Prob.
- 2 0 --- 0.25 / (0.25+0.5) = 0.33
- 1 1 --- 0.5 / (0.25+0.5) = 0.66

Thus, the probability of both coins being heads is 0.33, or 33%.

Sorry Cheeze, you can argue all you want, but you're still going to be wrong.

 

[FrcknFrckn]

But the probability of flipping one of each is the sum of the probability of heads THEN tails and the probability of tails THEN heads.

Sorry Cheeze, you can argue all you want, but you're still going to be wrong.

You've flipped twice trying to get the possibility of One of Each, while only flipping once for Two Heads or Two Tails.

That's the problem--people are mistaking permutations for actual flips/puppies/etc.

Nope - the actual probability of getting one head and one tail is twice that of getting 2 heads.

Try it - flip a pair of coins 100 times. You'll end up with approx. 25 H-H pairs, 25 T-T pairs, and 50 mixed pairs.

Heck, I could write you a program that would do thousands of tests if you'd like, it's pretty simple...

 

[FrcknFrckn]

Yeah, we know one is heads. BUT WE DON'T KNOW WHICH ONE. That, believe it or not, is the key factor. Either coin, or both, could be heads. Lets say the coins are different - say, a dime and a penny. There are 3 *equally likely* situations: DH PH, DH PT, and DT PH. Only one of the 3 has 2 heads. 33% chance.

So I went ahead and wrote the program, here's the output of 5 runs of 100000 flips each:


-----------------------------------

2 Heads: 24762 (24.762)%
2 Tails: 25135 (25.135)%
Mixed: 50103 (50.103)%

1+ Heads: 74865 (74.865)%
2 Heads: 24762 (24.762)%
Chance of 2 heads: 33.07553%

-----------------------------------

2 Heads: 24465 (24.465)%
2 Tails: 25270 (25.27)%
Mixed: 50265 (50.265)%

1+ Heads: 74730 (74.73)%
2 Heads: 24465 (24.465)%
Chance of 2 heads: 32.73786%

-----------------------------------

2 Heads: 24428 (24.428)%
2 Tails: 25591 (25.591)%
Mixed: 49981 (49.981)%

1+ Heads: 74409 (74.409)%
2 Heads: 24428 (24.428)%
Chance of 2 heads: 32.82936%

-----------------------------------

2 Heads: 24724 (24.724)%
2 Tails: 25337 (25.337)%
Mixed: 49939 (49.939)%

1+ Heads: 74663 (74.663)%
2 Heads: 24724 (24.724)%
Chance of 2 heads: 33.11412%

-----------------------------------

2 Heads: 24779 (24.779)%
2 Tails: 25257 (25.257)%
Mixed: 49964 (49.964)%

1+ Heads: 74743 (74.743)%
2 Heads: 24779 (24.779)%
Chance of 2 heads: 33.15227%

-----------------------------------

Should be pretty self-explanatory. Let me know if you want the source code.

 

[Alex_P]

Jeez, go on a business trip for 24 hours and another hundred posts flow in...

Scenario #1: Toss a fair coin ten times in a row. If the first nine tosses come up heads, the tenth toss still has a 50-50 chain of being heads. Each coin flip is independent. This is sequential probability.

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Your description of #2 sound fishy.

These scenarios are equivalent:
A. I flip a coin and look at it, then I flip another coin and look at it, &c., up to 10 coins.
B. I flip ten coins and then look at them in some specific order (it can be different from the order in which I flipped them).

The probability only changes when you start cherry-picking: searching until you find a male.

-- Alex

Not at all. Remember that the question is not "Is the tenth coin heads?" but rather "Is the tenth coin heads given that the first nine are heads"?

In the first scenario, I'm looking at the possibility of the tenth coin coming up heads KNOWING that the first nine came up heads. At this point, nothing I can do can change the outcome that at least nine of ten are heads. I have selected my set of ten with knowledge that eliminates the vast majority of probability distributions. In fact, only two possible probability distributions remain - nine heads followed by a tails, OR ten heads in a row. In selecting nine consecutive "heads" coin tosses, I have done the lion's share of obtaining ten straight "heads" coin tosses.

In the second scenario, I'm looking at the possibility that I have randomly selected a set of ten fair coin tosses which are all heads. There's a 50% chance the first toss will be tails; therefore half my random sets fail already even if I don't yet know it. I toss the second coin. My chance of being all heads is now 0.50 x 0.50 or 0.25, because both throws have to be heads to meet my criteria of ten heads. I can fail on the first throw, or on the second. It makes no difference whether I look at the values individually or all together at the end; I've selected them as a set. Now I toss the third coin; my chance of remaining all heads is now 0.50 x 0.50 x 0.50 or 0.125 (or 12.5% expressed as a percentage.) Now I toss the fourth coin: my chances of remaining all heads drop to 0.0625. The fifth, sixth, seventh, eighth, and ninth coin tosses take me to 0.3125, 0.15625, 0.0078125, 0.00390625, and 0.001953125 respectively. This number (0.195%) is the chance of a randomly selected set of ten fair coin tosses containing nine heads in any combination. If the first nine tosses were heads, then the tenth has the same base 50% chance to be heads. However, I selected the set randomly. To get the chance that I randomly selected ten coin tosses and they are all heads, I have to multiply 0.195% by 0.50 again - at which point I get 0.00098 or 0.098% - roughly one in a thousand. In this case, I have selected my set with a huge amount of possible permutations and combinations.

I picked ten coin tosses rather than four to emphasize the difference between sequential probability and set probability. If I group items together into a set AFTER I know their value, then the odds of the last item in the set are independent of the other items because I have negated all the possible probability distributions that don't fit that set's values. If I group items into a set BEFORE I know their values, then all possible probability distributions are still valid, and I can only eliminate them as I learn things about the set OR about individual items in the set.

To belabor a point, if I throw nine heads in a row with a coin that I know to be fair, the tenth heads is not remarkable. It's the nine in a row that are remarkable.

EDIT: Forgot to add, the chances are variables in set mathematics. If you alter the percentage chance of a particular permutation occurring, you alter the value distribution of the set. Thus the value of some set items must be changed. Since grouping the items into a set is merely a matter of convenience, it can't change the value of any particular item in the set; set mathematics can only describe probabilities of each particular combination or permutation of values of the items.

Here's the big distinction:

Given N coins,

"At least N-1 of these coins are heads" => 1/(N+1) probability of getting all heads

"At least these N-1 coins are heads" => 1/2 probability of getting all heads

I think your language is obscuring this.

If I say "I looked at three coins out of this set and they are all heads" (nine is too much typing for me), then my only options are:
HHHH
HHHT

I can't have something like HTHH because I already looked at the second coin and know it is not T.

If, however, I say "at least three coins out of this set are heads," then all of these are still kosher:
HHHH
THHH
HTHH
HHTH
HHHT

See what I mean?

-- Alex

 

[Alex_P]

No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The subtext of the problem is that puppy nards are fair coins.

-- Alex

 

[Alex_P]

No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The subtext of the problem is that puppy nards are fair coins.

I know, and that is what is screwing everyone up, even me when I wrote that entry.

Just because puppy nads are as fair as fair coins, that doesn't mean a permutation matrix is a probability matrix.

I mean, if the Law of Large Numbers is what predicts that we'll have an equal amount of HH, HT, TH, and HH, well, if it applicable to a number so small as the bare minimum number of coin flips to even have two results, why is it called the Law of Large Numbers? Why isn't it just called, like, the Law of Numbers, Great and Small.

Wise and Wonderful...


Bright and Beautiful...

okay, I'll stop.

Rephrase. Right now you're kinda just saying "Why is probability valid at all?"

-- Alex

 

[gerrymander61]

It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.

EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.

 

[Alex_P]

It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.

EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.

Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.

-- Alex

 

[Samirat]

Incorrect - we aren't assuming they choose from a random list of people with at least one male.

They selected from a random list of people, INCLUDING people with FF pairings.

Putting a sample population to it of 1000 people with dog pups, we get:

2xMale, 250 people
Answered: YES
1xMale 1xFemale 500 people
Answered: YES
2xFemale 250 people
Answered: NO

Hence, the resulting population is bias towards MF's

250 MM's vs 500 FM's

Good man. Nice to see that you've changed your mind when faced with evidence. Refreshing.

 

[mnimmny]

After this, I think The Escapist would explode if the cs peeps out there put those brain teasers interviewers are so fond of up.

I didn't realize this place was so fuzzy. Feels like peeps are arguing the way they'd argue about religion, politics, or polarizing video games rather than something techie like statistics and probability.