[OT] mathematical query

[Canada_K]

I'm also in the crowd that says the chance for both A and B is 50%. Here's my reasoning.

To get the probability of two sequential events, you multiply the probability of each signular event. For example, what is the probability of rolling two 6s in a row on a six sided dice?

probability = (1/6) * (1/6) = 1/36

How about, what is the probability of getting two heads in a row when flipping a coin?

probability = (1/2) * (1/2) = 1/4

Now what if I say I'm going to flip a coin twice, and I want to know what is the probability of getting one head and one tail (in whatever order).

Probability = 1 * (1/2) = 1/2

Why? Well the first toss doesn't really matter. All I really care about is that the second toss is different than the first. So whatever I get for the first toss is OK.

Now what if I say I have just tossed a coin and it came up tails. What is the probability that the next toss will be heads? In this case, the first event has already happened, and all we care about is the second toss.

Probability = 1/2

This is the same as the question about the man with the picture of his son. He's already told us one child is a boy. That event has already occured. So the chance that his other child is a girl is 1/2.

Take this to the extreme. Let's say I tell you I have 5 children and all of them are boys. What is the chance my next child will be a boy? It's 1/2 because all the previous children have no bearing on the next one. If instead I ask you, what are the chances of having six boys in a row, the answer is different.

Probability = (1/2) * (1/2)* (1/2) * (1/2) * (1/2) * (1/2) = 1/64

OK, so I went on for an excessively long time. You have to forgive me, math is my career and I finally got a chance to talk about something know.

 

[Blacksad]

It does NOT change the probability to "we know child 1 is male, whether he is older or younger, hence child 2 has a 50-50 chance of being male or female and hence there is a 50% chance of him having one of each" (a common mistake). We're assuming the picture gives us more information than it does... if he had said, "this is my oldest child and is my son" it would have eliminated possibility 3, thus lending a 50-50 chance.

THAT'S why the answer to B is 67%.

--The Sigil

You made a mistake, if you roll a 20, your chance of your next roll to be a 20 is still 1/20, the answer is indeed 50% for each case.

If you want a confusing issue, here is another problem for your players:

An evil king asks his wizard to brew a potion of longevity.
The equally evil wizard decide to brew 3 potion, two of black lotus extract and one of longevity, and propose a little game to the king.

The king chose potion number 1, and the wizard say that the king was right to not chose potion number 3, then the king change his mind and take potion number 2.

Why?

Well Because you have higher odds with the other, the possibilities are as follow:

when the king chose one potion it has 1/3 chances of being the one of longevity.

so we have 1/3 the king chose the potion of longevity, change his mind and die.

and 2/3 the king chose the wrong potion, the wizard show the other wrong, and the king chose the one of longevity.

 

[The Sigil]

As far as I can tell, it's an "order of information question" - and perhaps I was wrong. But I don't think so.

Telling me that one child is male does NOT have zero bearing on the other child in this case - because you haven't told me WHICH child is male. That means that the other child's birth is NOT randomized to 50-50 because I cannot identify it as a discrete event in the sequence. Flipping two coins and saying, "one of them came up heads" is not the same as telling me the "the first one came up heads."

Here is my reasoning...

I am told that you have two children.

I can then create four possibilities:

MM, MF, FM, FF.

Each of these has equal probability.

I think everyone agrees with me to this point.

By showing me a picture, you essentially tell me that the FF possibility is impossible. Now, we must figure out how to "disperse" the 25% probability that was tied up in the FF possibility before. Why should the entirety of the 25% now be assigned to MM instead of split equally among MM, MF, FM? I miss this point.

Here is a more rigorous derivation:

The answer to "I have two children" may be expressed as:
FF (25%)
FM (25%)
MF (25%)
MM (25%)

Now let us add the condition of "I show you a picture of my son" - what is the chance of you showing me a given child.
Let us look at some strict probabilities (unnormalized)...

FF (25%) * no picture (0%) = 0.00
FM (25%) * picture of M2 (100%) = 0.25
MF (25%) * picture of M1 (100%) = 0.25
MM (25%) * picture of M1 (50%) = 0.125
MM (25%) * picture of M2 (50%) = 0.125

Hence, we now have four solutions that still fit our original question (2 children, showed a picture of a male) with the following (unnormalized) probabilities:

FM/M2 - 0.25
MF/M1 - 0.25
MM/M1 - 0.125
MM/M2 - 0.125

The sum of probabilities where there is one male and one female is 0.50. The sum of probabilities where there are two males is 0.25.

Hence, we have 0.50 vs. 0.25.

That looks like 2:1 odds (67%) to me.

If I have made a mistake please explain it to me.

--The Sigil

P.S. - I know some of you are saying the probability tree should look like this:

Picture (100%) * Older (50%) * Other is Male (50%)
Picture (100%) * Older (50%) * Other is Female (50%)
Picture (100%) * Younger (50%) * Other is Male (50%)
Picture (100%) * Younger (50%) * Other is Female (50%)

but I'm not sure/convinced that this is the correct order to use.

 

[The Sigil]

if you roll a 20, your chance of your next roll to be a 20 is still 1/20

I agree with you. But the information given is not equivalent to this statement (even if you had said a d2 instead of d20 - I'm not going to quibble on that). The key word in your sentence is "next." You have been able to assign my roll of 20 to a discrete event (die roll one), thus creating order.

The question is not, "if I have a boy, what is the chance of my NEXT child being a boy." IOW, you can ascribe "boy" to a discrete event (birth one).

The question being asked is, "if I have two children, and one of them is a boy, what is the chance of the other being a boy as well." This IS a different question because I cannot ascribe "boy" to a discrete event - I don't know whether it is "birth one" or "birth two".

--The Sigil

 

[ichabod]

Time for the statistician in training to weigh in again. The answers are 50% in the first case, 67% (2 in 3) in the second case.

As explained before, you have four possibilities:

MM, MF, FM, FF

You have two male/female possibilities, because the female or the male could be born first. This is two different events we're talking about.

If we know one of the children is male, that leaves us with three possibilities:

MM, MF, FM

Since we can't have one male and both female at the same time. Thus the answer to the second question is 2 in 3. Sorry to repeat that, but it seemed to bear repeating.

As to independent probabilities, the birth of the first child is independent of the birth of the second child. However, that is not relevant to the second question. We are not asked the sex of the second child given the sex of the first child, we are asked the sex of two children given the sex of one of them. The latter case is a dependent probability, as obviously the sex make-up of the pair depends on the sex make up of the individuals within the pair. For example, if we know one of the children is male, the chance of both of them being female is 0%, not 50%.

 

[The Sigil]

As to independent probabilities, the birth of the first child is independent of the birth of the second child. However, that is not relevant to the second question. We are not asked the sex of the second child given the sex of the first child, we are asked the sex of two children given the sex of one of them. The latter case is a dependent probability, as obviously the sex make-up of the pair depends on the sex make up of the individuals within the pair. For example, if we know one of the children is male, the chance of both of them being female is 0%, not 50%. [/B]

Beautifully done... better than I could have done.

Thanks, Ichabod.

I think most people are adding "next" to "birth" without realizing it's a faulty assumption.

--The Sigil

 

[Zappo]

The answer is 50% and 67%, for the reasons already explained.

Blacksad... when the king gets told by the wizard that potion #3 was indeed poison, he has a 50% chance of having already chosen the right one, NOT a 33% chance. That's because one of the possibilities (choosing potion 3, one of the wrong ones) has already been discarded. Therefore, at that point, choosing potion 1 or 2 is the same.

BTW, I'd just call a cleric, have the wizard shoved into a zone of truth after having been cursed to have his Will saving throw lowered to nothingness, and then call the torturer.

 

[The Sigil]

Blacksad... when the king gets told by the wizard that potion #3 was indeed poison, he has a 50% chance of having already chosen the right one, NOT a 33% chance. That's because one of the possibilities (choosing potion 3, one of the wrong ones) has already been discarded. Therefore, at that point, choosing potion 1 or 2 is the same.

No, the probabilities are NOT discarded.

Here goes...

Step 1: The right answer is:
A (1/3)
B (1/3)
C (1/3)

King chooses:
A (1/3)
B (1/3)
C (1/3)

Let the first one in the pair be the right answer, the second one in the pair be the King's choice and we get:
AA (1/9) - King has the right one
AB (1/9)
AC (1/9)
BA (1/9)
BB (1/9) - King has the right one
BC (1/9)
CA (1/9)
CB (1/9)
CC (1/9) - King has the right one

Next step: Eliminate one potion:
1.) AA (1/9) - eliminate B (50%) - King has the right one
2.) AA (1/9) - eliminate C (50%) - King has the right one
3.) AB (1/9) - eliminate C (100%)
4.) AC (1/9) - eliminate B (100%)
5.) BA (1/9) - eliminate C (100%)
6.) BB (1/9) - eliminate A (50%) - King has the right one
7.) BB (1/9) - eliminate C (50%) - King has the right one
8.) BC (1/9) - eliminate A (100%)
9.) CA (1/9) - eliminate B (100%)
10.) CB (1/9) - eliminate A (100%)
11.) CC (1/9) - eliminate A (50%) - King has the right one
12.) CC (1/9) - eliminate B (50%) - King has the right one

Now the question is - what is the sum of the probabilities that the King has the correct potion?

Well, we have 1/9 * 50% * 6 (possiblities 1, 2, 6, 7, 11, and 12) or 1/9*1/2*6 or 1/18*6 or 1/3.

The King STILL has (only) a 1/3 chance of having the correct potion, even after we eliminate one of the potions from contention.

Odds are now 2:1 that the "other" potion is the right one.

Go through things step-by-step and watch out for thinking a dependent probability is an independent one.

--The Sigil

 

[CRGreathouse]

No, the probabilities are NOT discarded.

[...]

Go through things step-by-step and watch out for thinking a dependent probability is an independent one.

You're the one making assumptions here.

Your reasoning is correct, but only if the wizard chooses to never tell you that your choice is wrong. Since D&D wizards have Int as their primary stat, I'd assume they'd figure this one out.

If the wizard is biased, you can double your probability of success by switching, but you can't gain anything if he's unbiased. (What's more, all of these assume he's telling the truth and knows which potions are which. There's no reason, really, for us to believe that!)

 

[2WS-Steve]

The Ugly Answer

Okay, the formal solution to this gets kind of ugly since it relies on Bayes’ Theorem but if you’re interested here it goes:

Bayes’ Theorem tells us the proper way to update our degrees of belief when presented with new evidence. Everything in this theorem is derivable from the standard probability axioms.

Pr(H/E & c) = [Pr(H/c)*Pr(E/H & c)]/Pr(E/c)

Pr(x) stands for the probability of x
Pr(x/y) stands for the probability of x given y

H represents the hypothesis (i.e. that both children are the same sex)
E represents the evidence (i.e. the photo of the male child)
c represents the prior conditions (i.e. that probability space prior to the evidence); the prior conditions are just the state space described previously, i.e. the following options:

1) M, F
2) M, M
3) F, F
4) F, M.

We’re solving for Pr(H/E & c) in this case…

Pr(H/c) = .5; this is just the prior probability we assign before seeing the photograph.

Pr(E/H & c) = .5; this is the probability that we’ll actually see a photo of a male child given that the hypothesis is true and the background conditions. If the hypothesis is true we eliminate options 1 and 4 leaving us with just 2 and 3; 50% of those options shows a photo of a male.

Pr(E/c) = .5; this is the probability of the evidence given the prior conditions. In other words, the straight probability of seeing a photo of a male child. Since there are 8 possible “photos” we can see (2 for each option) and 4 of those possibilities are male we end up with .5 here as well.

That gives us:

Pr(H/E & c) = (.5 * .5)/.5 = .5


---------------------
Cool puzzle! The intuitive answer is the right one but figuring out why it’s right is the real puzzle.