# The Problem of the Unfinished Game

Today's post is a simple question.

Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?

This is a companion discussion topic for the original blog entry at: http://www.codinghorror.com/blog/2008/12/the-problem-of-the-unfinished-game.html

Jeff has a lot of reading to do.

@Arnos

Your 1 million coins example is wrong. The question would be 1 million coins were tossed. 999,999 were tails. What’s the chance the other one was heads?

I’m not going to solve this problem but the key is that the single heads result doesn’t have to be the 1,000,000th toss - it could be the 999,998th toss or 57th toss or any single toss as long as the other 999,999 are tails. So the probability will not be 50%.

@sam

The number of trials that would match the conditions you give would be different between the two

How so? There are no conditions for A and are not in B also. I only added the step I tell you the one on the right is heads.

You knew one was heads already, so it tells you nothing about the state of the coins.

*one on the right == where they land on the table, not some kind of order.

@Sam A.
The 999,999 coins that you placed on the table have NO bearing on the one you are actually flipping. The 999,999 are the given, known information. The probability of the coin you flipped (unknown information) is still 50%.

COIN TOSS EXPLAINED

2 coins were tossed. One is heads. What’s the probability that the other is tails?

1. Assume coin on left is heads. Then the other coin has 50% chance of heads and 50% tails.

2. BUT the question didn’t say the coin on the left was heads. So let’s now assume that the coin on the right is heads, then the coin on the left is 50% chance of heads and 50% tails.

3. Combine 1 and 2 (naive solution): (2 * 50%) / (4 * 50%) = 50%

Step 3 is wrong. The problem is that the heads-heads combination was double-counted. The times in 1) when we got heads-heads and the times in 2) when we got heads-heads were the SAME events, so we should only count them once. So to adjust: (250%) / (450% - 50%) = 100 / 150 = 2/3

I’m not sure… I’ll make a wild guess…
100%

I think a person will say I have two girls rather than mentioning one girl…
I’m not sure I’m taking the question in the right direction.

Now, I do the search…

I’m not sure… I’ll make a wild guess…
100%

I think a person will say I have two girls rather than mentioning one girl…
I’m not sure I’m taking the question in the right direction.

Now, I do the search…

I’m not sure… I’ll make a wild guess…
100%

I think a person will say I have two girls rather than mentioning one girl…
I’m not sure I’m taking the question in the right direction.

Now, I do the search…

@Arnos

Ah… But i flipped all of the coins.

For those who are confused if you remove the order than GB and BG are the same thing (because there are is just a B and a G somewhere in the pot, they are not ordered in any way).

@Chase Seibert
Ah I was interpreting you to mean the one on the right was the second coin flipped.

For those who are confused if you remove the order than GB and BG are the same thing (because there are is just a B and a G somewhere in the pot, they are not ordered in any way).

So you cannot call BG and GB separate solutions.

If you call them separate solutions then you are saying the order matters. The question did not say the order mattered, so these are not separate solutions.

I am convinced that the common voodoo beliefs portrayed as statistics and logic are diametrically opposed (IE, Statistics theory is fundamentally flawed).

Sorry, I don’t buy anyone’s reasoning. A firm believer in empirical evidence, I’m going to go out right now and try to make some babies.

For those who are confused if you remove the order than GB and BG are the same thing (because there are is just a B and a G somewhere in the pot, they are not ordered in any way).

So you cannot call BG and GB separate solutions. They would be the SAME SOLUTION.

If you call them separate solutions then you are saying the order matters. The question did not say the order mattered, so these are not separate solutions.

Ok, I was wrong. I didn’t believe that the child’s age makes a difference but it does. There are two working possibilities out of three. Guess I should have though that through more carefully.

Anyway, here is link with a chart that explains it well:

I was thinking of the first question on that page.

So if BG and GB are the same solution then we started with three possibilities (which they are if order does not matter):

BB
GG
BG (which is the same as GB)

and we eliminated BB.

That leaves GG and BG.

@Chase Seibert

Well, asking me to go back in time and revise my previous guess based upon knowledge I didn’t have when I made it is a bit of a tall order

The probability estimates in my head have changed with the new information, but I’m temporally incapable of changing my bet now.

… or do you still think that probability exists anywhere outside of a mind?

you nothing about the state of the coins.

Yes it does.

Given coin A and coin B.

With no information, each combination of A and B, heads or tails has a 25% chance of occurring.

At least one is heads - A heads, B tails OR A tails, B heads OR A heads B heads. 33% chance of each, given our new information.

Probability estimates are the result of a mind’s information about the world. Adding to that information will change the probability estimates.

To those still hanging onto the 50% argument that the first baby doesn’t affect the chances of the second one…

You’re looking at it wrong, and the most concise argument I can see for this is mike’s:

the information is being presented encompasses the sum of the results of both events

We know LIMITED INFORMATION about the SUM of two babies.

We do not know ANYTHING about one particular baby. Nor is the question about a particular baby, but rather about the SUM of the two babies.

So stop thinking about just one child and the resulting 50% chance.

Make the leap.

@Practicality

You are correct. You simply need to note that P(BG) = 2*P(GG).