The Problem of the Unfinished Game

@Chase Seibert
Again, if we did Trial A 1000 times, and Trial B 1000 times,
are you saying that the number of times you’re right would
be statistically different?

Okay, let me get this clear. We do this 1000 times.

You flip two coins. You tell me one is heads.

I predict the other is tails. I will be right 66% of the time.

You tell me the right one is heads.

Here’s your problem. This is only going to be the case 66% of the time – in 33% of these trials, the right one is not going to be heads.

So what do you do in this case? Just pick one that’s heads and tell me that one? If the right one is ALWAYS heads, then the game is fixed, so of course my estimates are going to be wrong.

You aren’t giving me any new information in this case.

This is truly getting STUPID. You people need to get a life.

@xooorx

I’ll buy that argument, but I don’t think that that was the intent of the question. (Jeff can tell us for sure.) I also don’t think that’s the same logic that most 50%ers are using – I believe most of them are arriving at the same conclusion through erroneous logic.

@Mike

Shut up already! Who gives a flip about a sheep, anyway? asked the software developer. I’m trying to get some work done on my laptop, and there you are jabbering away. Here’s a real puzzle for you: ‘Let’s say, hypothetically speaking, you met someone who told you they had two children …’

@John Fouhy
I am pretty sure the proportion of men:women stays at roughly 1:1. Half the population has a girl and stops. The other half has a boy. At this point the ratio is 1:1 (ignoring multiples). Of the people who continue to have children, the same occurs again.

@Chris
Why are you even here and reading the comments? It’s stupid, so go do something else.

The 2/3rds answer seems to hinge on treating B-G and G-B as two separate possibilities. I’m still struggling to see why one would do such a thing in a situation where you haven’t been asked to do so. It seems to be creating a discrete possibility out of something in which there isn’t one. Of the genders of children one can have, I only see B-B, B-G, and G-G. Why should I treat G-B as a discrete entity in a problem like this? That’s what I don’t get.

I’m not a mathy person but I haven’t seen anyone explain satisfactorily why this apparently commonsensical observation is incorrect in terms that can similarly appeal to common sense.

50% vs 66% -or- B-G vs G-B

People are confusing Events with Outcomes!!!

An Event is a set of related Outcomes. An Outcome is, simply put, all that one can reasonably measure and/or record

about an (ideally instantaneous) observation.

For example, consider the Events on the roll of one standard six sided die:

- Event Even is Outcome of 2, 4, or 6

- Event Odd is Outcome of 1, 3, or 5

- Event Factor-of-3 is Outcome of 3 or 6

- Event One is Outcome of 1

There is also the special Event called the Sample Space that is the set of all possible Outcomes. The Sample Space

for the one die roll is of course 1, 2, 3, 4, 5, or 6.

Note: An Event CANNOT contain an Outcome that is NOT contained in the Sample Space! This is because, by definition,

the Sample Space must contain all possible Outcomes for a given context.

An Event is said to occur when the Outcome one observes matches one of the Outcomes that defines the Event.

(Ideally, the match should be exact, but in real life the match can be close enough.) And, yes, it is possible for

one to define Events such that more than one occurs within a single (instant) observation. If the roll had an

Outcome of 3, then Events Odd and Factor-of-3 have occurred (simultaneously).

Note also that the Outcomes in the Sample Space are (ideally) distinct. In other words, if one is to work with a six

sided die that is labelled with 1, 2, 2, 3, 3, and 3, your theoretical Sample Space might need to look like 1, 2A,

2B, 3A, 3B, and 3C, and you may need Events like Rolled-a-2 (2A or 2B) and Rolled-a-3 (3A, 3B, or 3C). In real

life, you’re not likely to tell the 2’s and 3’s apart, but your theory (of 2A, 2B, 3A, etc.) will aid you in getting

the odds right.

Each Outcome may have a numerical weight of its own. Ideally all Outcomes should have the same weight, but this is

not always the case. Still, for most problems, giving each Outcome the same weight works well enough.

Each Event, in turn, has a weight equal to the sum of the weights of its Outcomes. Assuming the above Outcomes each

have a weight of 1, then Events Even, Odd, Factor-of-3, and One have weights of 3, 3, 2, and 1.

One can normalize these weights by dividing each with the weight of the Sample Space. This will give the weights,

going from 0 to 1, that is conventionally used in Probability. For the one die roll, this is a weight of 6. Hence,

the Probability of Event Even is 3/6 = 50%. In other words, P(Even) = 50%. Also, P(Odd) = 3/6 = 50%, P(Factor-of-3)

= 2/6 = 33.3%, P(One) = 1/6 = 16.7%.

As many have already pointed out, the possible Outcomes for the one is a Girl, odds on a Boy problem are:

(Boy, Girl), (Girl, Boy), (Girl, Girl)

The Sample Space of this problem has a weight of 3 (keeping it simple, only two kids, gender mix at 50/50, and no

twins). The observation being recorded is the gender of the first born and the gender of the second born. Hence the

Outcomes here are ORDERED sets - (Boy, Girl) and (Girl, Boy) are NOT the same! (Mathematically speaking, an Outcome

is really more like a Vector or a Matrix.)

The Event we would like to track is Girl-and-Boy:

- Event Girl-and-Boy is Outcome of (Boy, Girl) or (Girl, Boy)

Events are NOT ordered sets! {(Boy, Girl), (Girl, Boy)} IS the SAME as {(Girl, Boy), (Boy, Girl)}.

An Outcome should be (in principle, anyway) a perfect record of all that can be observed. Dropping information, such

as birth order, often brings misleading results. This is because, ideally, all Outcomes should have equal weight,

and following the mechanics of the situation is your best hope of achieving this. With birth order, Girl-And-Boy

is an Event with a weight of 2. Without birth order, Girl-And-Boy is an Outcome with a (presumed) weight of 1.
If

you insist on dropping birth order from your Outcome record, you may miss out on whether the Girl-And-Boy Outcome

needs a weight of other than 1.

Event Girl-and-Boy has a Probability of 2/3 = 66.7% (weight of Girl-and-Boy divided by weight of Sample

Space).

Since the problem originally asks for the odds of a boy and a girl, that is simply the weight of
Girl-and-Boy to

the weight of Only-Girls:

{(Boy, Girl), (Girl, Boy)} : {(Girl, Girl)}

2 : 1

What makes this problem Bayesian is in the way information is used to remove Outcomes from the Sample Space and

Events. Once this reduction happens, the weights on Events and the Sample Space can change, which in turn changes

the probability.

Let’s say in addition to one is a girl we are also told and she is the first born. The Sample Space then

becomes:

(Girl, Boy), (Girl, Girl)

because the information she is the first born now makes (Boy, Girl) impossible.

By definition, Sample Space indicates all possible Outcomes FOR THE GIVEN CONTEXT, and so (Boy, Girl) must be

removed from Event Girl-and-Boy too:

- Event Girl-and-Boy is Outcome of (Girl, Boy)

And so, in the context of one is a girl and she is the first born, the weights of Sample Space and Girl-and-Boy

are 2 and 1.

Thus P(Girl-and-Boy) is now 1/2 = 50%, and the odds change to:

{(Girl, Boy)} : {(Girl, Girl)}

1 : 1

Gosh … look at all those replies. Would it make a difference if they told you that the oldest child is a girl?! Hmmmm …

@Therac-25
You aren’t giving me any new information in this case.

Just to clarify what I’m saying here.

Coin A is heads tells me something.

Saying One of them is heads, and then picking one of the ones that are heads after the fact – necessarily changing it on each trial – does not tell me anything new. It’s an obfuscated way of saying One of them is heads.

The point is there are cases where you can say One of them is heads and you CAN’T say Coin A is heads.

That’s why Coin A is heads gives me information, and the process of One of them is heads - Coin [A|B] is one of the the ones that is heads doesn’t. If you then asked me to undergo a SUBSEQUENT guess, then I would wager at 50% the face of the other coin.

Keep posting:

p1 = 50%
p2 = 66%

n1 = posts for 50%
n2 = posts for 66%

answer = (n1p1 + n2p2) / (n1 + n2)

If somebody says in casual conversation exactly I have two children and one of them is a girl obviously they have 1 girl and 1 boy. Believing otherwise is to fail at understanding how normal people talk.

As the problem is stated, one should probably infer that there were two statements: I have two children and Blah blah, my daughter Jane is in kindergarten (or something like that). In this case the answer is 2/3 chance of having a boy and a girl as many have explained.

Now get back to work everybody!

0% they have a boy or girl. You are only hypothetically meeting this person. Hypothetical people do not have real children.

@Rich K

Thank you, that is the first satisfactorily complete answer to why BG and GB should both be considered as unique for the purposes of probability (rather than just squirting that into the explanation as a ‘given’)

66% it is then.

Shmork,

Treating BG and GB as separate possibilities is only useful for calculating the probability of having one of each.

Thinking of it another way, if you have 8 children, is it more likely that you have 4 and 4, or 1 and 7? They are the same outcome, but they do not have the same probability. 4 and 4 is much more likely than 1 and 7.

So the probability of 1 and 1 is 50%, the probability of 2 and 0 is 25%, and the probability of 0 and 2 is 25%.

50%

Happy new year everyone!! Go home and drink beer…

@Jon Ericson: Rather, when I flip two coins and reveal the result of one of the coins, there is a 50% chance the other coin will end up the same as the first and 50% chance it will be different.

If you randomly select whether to reveal heads or tails each time, then I might agree. But that’s not the problem we’re given in this post. We’re given that at least one child (coin) is a girl (heads). It doesn’t matter whether or not the hypothetical person in the problem randomly selected which child to tell us the gender of: we know the outcome.

The New Year’s partying is still hours away, but I’m giddy already. Thank you, Jeff Atwood, for that clever post, and stirring up all the entertaining and educational commentary. Could you make it a December 31 tradition???

All you posters out there, you’re awesome! Happy New Year!

I read through most of the thread, then drew a little diagram to help me think through it.

    B1 50%                 G1 50%
   / \                    / \
  /   \                  /   \
 B 50% G 50%            G 50% B 50%

When a couple has a child, 50% that’s it a boy or a girl. Another child = 50% boy or girl. It doesn’t matter which is born first. When a couple has two children,there’s a 75% chance that one is a girl. When we know that one child is girl, there is a 66.6 (2/3) chance that the other is a boy. This chart does not mean the the B1 and G1 on top are the first child. B1 and G1 could just as easily be the second child. Order does not really matter.

Actually, I think it is not an answerable question and invalid. Who is ‘that person’? Is it the hypothetical person that is telling us or is it the person who we know is a girl? Completely changes the nature of the question. Since it is not clear who ‘that person’ is, we are unable to determine the correct answer because the sentence is not clear.