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Nice! I had fun solving this problem in my head: > I have two children. At least one of them is a boy born on Tuesday. What is the probability that both childr

Nice! I had fun solving this problem in my head: > I have two children. At least one of them is a boy born on Tuesday. What is the probability that both children are boys? An interesting thing about this problem is the unspoken assumption of what happens in other counterfactual worlds. If the person always answers the question "is one of your kids a boy born on Tuesday?" then the problem is solvable. But if a different family history would've caused the person to answer a different question ("born on a Monday" instead of Tuesday), then the answer would depend on the person's algorithm. Eliezer gave a dramatized explanation here: https://www.lesswrong.com/posts/Ti3Z7eZtud32LhGZT/my-bayesian-enlightenment https://www.lesswrong.com/posts/Ti3Z7eZtud32LhGZT/my-bayesia... Further on this path, there are seemingly basic questions that cause disagreement among actual statisticians. For example, see the voltmeter story in https://en.wikipedia.org/wiki/Likelihood_principle https://en.wikipedia.org/wiki/Likelihood_principle: > An engineer draws a random sample of electron tubes and measures their voltages. The measurements range from 75 to 99 Volts. A statistician computes the sample mean and a confidence interval for the true mean. Later the statistician discovers that the voltmeter reads only as far as 100 Volts, so technically, the population appears to be "censored". If the statistician is orthodox this necessitates a new analysis. However, the engineer says he has another meter reading to 1000 Volts, which he would have used if any voltage had been over 100. This is a relief to the statistician, because it means the population was effectively uncensored after all. But later, the statistician ascertains that the second meter was not working at the time of the measurements. The engineer informs the statistician that he would not have held up the original measurements until the second meter was fixed, and the statistician informs him that new measurements are required. The engineer is astounded: "Next you'll be asking about my oscilloscope!"

benlivengood 5y ago
Maybe I'm missing something about the voltmeter example. My assumption is that the 100-volt-maximum meter can distinguish between 100 volts and more than 100 volts, in which case there's no problem. If the voltmeter doesn't accurately indicate whether or not a measurement is outside of its range then the statistician is correct that everything should be re-measured. Do some people think that the possibility of not being able to take accurate measurements is the same as not having taken accurate measurements? EDIT: Maybe the ambiguity is in what the engineer would have recorded if finding a voltage >100 volts while the other meter was broken? It's like undefined behavior in programming; if you know your software will have undefined behavior when encountering certain data then you can't trust whether the output is valid unless there's independent confirmation that the data won't cause undefined behavior. If the statistician doesn't have certainty that the engineer will have defined behavior (e.g. say "I couldn't complete the measurements" vs. undefined behavior like writing down "99" or exploding) then they of course want to re-measure.
- tinalumfoil 5y ago
  > in which case there's no problem I have a box with 1000 tubes, exactly 1% of which are beyond my capacity to measure. I take a random sample of 20, and call that a roughly 80% chance of being able too measure all of them. Let's say the average is 50V and the ones beyond my measurement average 150V. We can use this info to solve for the average of the tubes we can measure: 25V. So, in 80% of the cases we'll get an average measurement of 25V. In 20% of the cases we're going to realize our mistake, buy a better voltmeter, and redo everything. In this case we can measure everything so we get the correct number: 50V. 25V * 0.8 + 50 * 0.2 = 30V This, of course, won't be statistically significant. But let's get a lot of researchers together to measure a lot of tubes. And let's say 100V limit meters are common so half make the same mistake. Now we have a statistically significant 40V. Wrong, wrong wrong.
vlovich123 5y ago
> In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys. I don’t understand this reasoning. If at least one is a boy, the only configurations I can think of is 1 boy 1 girl or 2 boys. Where does the 1/3 come from?
- tijsvd 5y ago
  With 2 children, there are 4 configurations of equal probability. The one with 1 boy 1 girl occurs twice. Take away the 2 girl case, then 2 boys is 1 in 3.
  - vlovich123 5y ago
    Yeah, the way the problem is formulated though there’s absolutely no indication that order matters so how are there two configurations within which there’s 1 boy and 1 girl?
    - maxov 5y ago
      Another way to think about it is counting the probability of getting k boys out of 2 children. 0 boys - 1/4 1 boy - 1/2 2 boys - 1/4 There's a half chance of getting exactly one boy, and one way to calculate this is by noticing there are two different ways to get one boy if we take order in account. You are right that the orderings don't matter in this case, so we could also e.g. model this with a binomial distribution. Once you know there are >= 1 boys, the chance you have two is 0.25/(0.25+0.5) = 1/3.
    - kgwgk 5y ago
      Because the order exists even if it doesn't matter (at least for two children, maybe not for two quantum particles). With the risk of being accused of binarism, there are four distinct possibilities with (close to) equal a priory probability of 25%: older boy/younger boy, older boy/younger girl, older girl/younger boy, and older girl/younger girl. Discarding the girl/girl case leaves three equally probable cases.

benlivengood 5y ago

Maybe I'm missing something about the voltmeter example. My assumption is that the 100-volt-maximum meter can distinguish between 100 volts and more than 100 volts, in which case there's no problem. If the voltmeter doesn't accurately indicate whether or not a measurement is outside of its range then the statistician is correct that everything should be re-measured. Do some people think that the possibility of not being able to take accurate measurements is the same as not having taken accurate measurements? EDIT: Maybe the ambiguity is in what the engineer would have recorded if finding a voltage >100 volts while the other meter was broken? It's like undefined behavior in programming; if you know your software will have undefined behavior when encountering certain data then you can't trust whether the output is valid unless there's independent confirmation that the data won't cause undefined behavior. If the statistician doesn't have certainty that the engineer will have defined behavior (e.g. say "I couldn't complete the measurements" vs. undefined behavior like writing down "99" or exploding) then they of course want to re-measure.

tinalumfoil 5y ago
> in which case there's no problem I have a box with 1000 tubes, exactly 1% of which are beyond my capacity to measure. I take a random sample of 20, and call that a roughly 80% chance of being able too measure all of them. Let's say the average is 50V and the ones beyond my measurement average 150V. We can use this info to solve for the average of the tubes we can measure: 25V. So, in 80% of the cases we'll get an average measurement of 25V. In 20% of the cases we're going to realize our mistake, buy a better voltmeter, and redo everything. In this case we can measure everything so we get the correct number: 50V. 25V * 0.8 + 50 * 0.2 = 30V This, of course, won't be statistically significant. But let's get a lot of researchers together to measure a lot of tubes. And let's say 100V limit meters are common so half make the same mistake. Now we have a statistically significant 40V. Wrong, wrong wrong.

vlovich123 5y ago

> In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys. I don’t understand this reasoning. If at least one is a boy, the only configurations I can think of is 1 boy 1 girl or 2 boys. Where does the 1/3 come from?

tijsvd 5y ago
With 2 children, there are 4 configurations of equal probability. The one with 1 boy 1 girl occurs twice. Take away the 2 girl case, then 2 boys is 1 in 3.
- vlovich123 5y ago
  Yeah, the way the problem is formulated though there’s absolutely no indication that order matters so how are there two configurations within which there’s 1 boy and 1 girl?
  - maxov 5y ago
    Another way to think about it is counting the probability of getting k boys out of 2 children. 0 boys - 1/4 1 boy - 1/2 2 boys - 1/4 There's a half chance of getting exactly one boy, and one way to calculate this is by noticing there are two different ways to get one boy if we take order in account. You are right that the orderings don't matter in this case, so we could also e.g. model this with a binomial distribution. Once you know there are >= 1 boys, the chance you have two is 0.25/(0.25+0.5) = 1/3.
  - kgwgk 5y ago
    Because the order exists even if it doesn't matter (at least for two children, maybe not for two quantum particles). With the risk of being accused of binarism, there are four distinct possibilities with (close to) equal a priory probability of 25%: older boy/younger boy, older boy/younger girl, older girl/younger boy, and older girl/younger girl. Discarding the girl/girl case leaves three equally probable cases.