Sorry I offended you.
Nah, I guess I’m just the square who doesn’t get the joke :P
I said $20 because it’s fair and I don’t want to rip off my friend.
On the other hand, maybe my friend offerred a bet because of inside information. Maybe I have a sneaky friend.
It’s a trick question, computing the odds is not the problem: deciding how much money to bet with is… But if I know my friend will accept the bet anyway, $0 is the easy greedy option. I have nothing to lose, right?
Or to compare it to Martin’s answer to the Monty Hall problem: Any money I put on the line is only going to cost me more money if I lose, and doesn’t increase my earnings if I win. The lowest amount of money is therefore optimal, without knowing the odds of the bet.
More boy/girl problems:
- less than 50%
- exactly 50%
- greater than 50%
- need more information / problem is unclear
“Expected” means, as usual, that you should imagine averaging the female percentage of many such countries.
And one more:
- less than 1
- exactly 1
- greater than 1
- need more information / problem is unclear
link to spoilers
Are we supposed to answer this realistically? Because the answer really depends on whether humans can get infinitely many children…
Wait, it’s not only obvious in the way I thought, but obvious in a different way as well! And doesn’t depend on getting infinitely many children. Really good puzzle!
Also, something can be both obvious and surprising.
Family size limits are clearly not part of the problem as given, but I don’t think it changes anything fundamentally and should make it easier to develop the right intuition, so feel free!
Yeah, I brought a sledgehammer to knock down a house of cards.
It’s surprisingly obvious, how about that?
There may yet be more obvious things waiting to surprise you!
Now I wonder what the most obvious thing is that I’ll never realise…
Answering this question literally (about actual births) is very difficult, requiring consideration of the practical limits of human reproduction and dealing with tricky questions about gender (such as ambiguity, non-binary, rare genetic conditions).
An idealized alternative can be expressed like this:
Make fair, independent coin flips until you get a single “heads” (and then stop). What is the expected number of “tails” that you will get?
Answering this is simply a matter of considering the mean of a specific case of the geometric distribution, a common exercise in a probability 101 class.
Of course, the expected number of heads is clearly one (since every possible sequence of coin flips contains just one “heads” ending the sequence).
In fact, I only realised it was obvious after I changed the setting to coin flips, intuition is weird like that.
It’s more obvious than that, actually.
Try to come up with a strategy of flipping a (very large) number of coins so that you end up with (significantly) more heads than tails
Okay. So in your version, what’s the expected fraction of tails?
I guess the simple, intuitive reasoning is just that on the population level, males and females arrive at roughly the same rate, and that ultimately won’t be biased by how people individually choose to stop having children.
Half of your trials will end on one flip with 0% tails.
For the fraction to average out to 50%, what would you need from the rest of the trials?
The other half of your trials have a tails on the first throw. So 50% of the first throw is heads, 50% is tails.
For those trials needing a second throw, you can use the same logic: 50% of the second throws will be heads, 50% will be tails. And same for the third throws, etc.
Same for any throw, for that matter, there is no strategy to fill a population of random coin flips with a different distribution than the distribution of the coin itself.
Suppose you cut off the experiment after three flips even if you haven’t gotten a “heads”. Try filling in this table (I’ll make it a wiki):
|sequence||probability||fraction of “tails”|
Now, compute the expected fraction of tails: _____
Assume Family A gets either two girls and one boy with probability 2/3, and with probability 1/3 they get three girls and one boy. The expectation of the fraction of boys would be 1/3 * 2/3 + 1/4 * 1/3 = 11/36.
But what if we consider a large number of families with the same pattern? Let’s say 3000 families. Then on average there’ll be 2000 falling in the first category and 1000 in the second. That corresponds to 7000 girls and 3000 boys, yielding a fraction of boys of 3/10, which is different than 11/36.
So I believe that the first way of calculating the average fraction does not behave like one might expect. I believe the problem is that for the average, it doesn’t make a difference whether you plug in 2000/4000 or 1/2. Both are the same fraction, but for the whole population, the two cases (in general) yield different results. But please correct me if I’m wrong.
If half of the families get 1 child, 1/4 get 2 children and 1/4 of the families get 3 children, you have to multiply the fraction of daughters by 1, 2 or 3 respectively according to the number of children in that family.
Suppose we have 24 trials / families
|sequence||probability||fraction of “tails”||total number of coins in trial||number of heads||number of tails|
|TH||1/4||1/2||6 • 2 = 12||6||6|
|TTH||1/8||2/3||3 • 3 = 9||3||6|
|TTT||1/8||1||3 • 3 = 9||0||9|
Hence in total we have 12 + 6 + 3 = 21 heads, and 6 + 6 + 9 = 21 tails.
So 50 / 50, as expected.
In essence the common mistake one would make here is to take the average without taking weight into account. Although families with few children are more common than families with many children, the families with many children deliver, in proportion, a larger number of people to the total population, since they have more children per family.
The easier way to count it, would be by taking this weight into account at the start.
The probability of getting at least 1 child is 1, 50% of these children are daughters, 50% are sons. The probability of getting at least 2 children is 1/2, 50% of these second children are daughters, 50% are sons. The probability of getting at least 3 children is 1/4, 50% of these third children are daughters, 50% are sons.
Hence, if we have 24 families, we get 12 firstborn daughters, 12 firstborn sons, 6 secondborn daughters, 6 secondborn sons and 3 thirdborn daughters, 3 thirdborn sons.
In total, this gives 12 + 6 + 3 = 21 daughters, and clearly the same number of sons.
grouping these together:
TTT - TH - H - H TTT - TH - H - H TTT - TH - H - H TTH - TH - H - H TTH - TH - H - H TTH - TH - H - H
TTTTHHH TTTTHHH TTTTHHH TTTHHHH TTTHHHH TTTHHHH
Oooh, I like this quote :D