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Wow that is...quite the edit lol. I'm all for small corrections and addendums, but completely changing the meaning of your comment from an attempted callout is something else.
Glad to have impressed you. My edit was intended to bring the comment in line with HN guidelines.
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Not really. In the son/daughter case, the calculations are: expected daughters: 1 expected sons: 1/2 0 + 1/4 1 + 1/8 2 + 1/16 3 + 1/32 4 + … So number of expected daughters = 1, number of expected sons = 1. In practice since women can't have an infinite number of children, then this wouldn't be an infinite series, so the real number of expected boys would be lower than one, but there you go… Now, for the bus case, yo…
I guess it would be balanced if the rule was, > 1 when your bus turns up, -1 for every bus going the other way.
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Wow that is...quite the edit lol. I'm all for small corrections and addendums, but completely changing the meaning of your comment from an attempted callout is something else.
Glad to have impressed you. My edit was intended to bring the comment in line with HN guidelines.
This reminds me the bet in the bitcoin community [1]. If on average bitcoin blocks are produced every 10 minutes, and you learn that 5 minutes ago someone found a block, what is the average time you will wait for the next block? It turns out it's 10 minutes, not 5 minutes as you would intuitively think. (it's a memoryless process, so average expected time till block is always the same - 10 minutes - no matter how man…
Suppose this is your first flip, one would intuitively think that there is 50% chance of H and 50% T.
Suppose you flipped once and got H. For your second flip, one might intuitively think that since the number of times getting H over the long run is 50% of the total flips, and we already have a flip of H, to "balance it out" the next flip should have a smaller probability of getting H. Instead that is wrong. The next flip still has 50% chance of H.
Suppose further that one has performed N flips, all of them H. One might even think that because of the way the geometric distribution works, it is very unlikely for the next flip to be H again. Instead that is wrong. The next flip still has 50% chance of H.
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Send all buses at once
If something like that is an option, just don't ever send any.
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The expected number of sons is 1, and the expected number of daughters is 1 (by the framing of the problem, in every possible scenario, there is exactly one daughter), but the expected value of the ratio is not 1:1. E[X]/E[Y] = E[X/Y] is not a valid identity. http://www.thebigquestions.com/2010/12/22/a-big-answer-2/
I read that and it seems wrong. The question asked "what fraction of the pop is female" but his argument is that 3 families of 4 girls and 1 family of 12 boys make the fraction of girls in the average family 75% (the average of 100% x3 and 0% x1) which is non-sensical to me. >E[X]/E[Y] = E[X/Y] is not a valid identity. is completely irrelevant here because it is being used to point out that a non-answer is wrong.
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Not really. In the son/daughter case, the calculations are: expected daughters: 1 expected sons: 1/2 0 + 1/4 1 + 1/8 2 + 1/16 3 + 1/32 4 + … So number of expected daughters = 1, number of expected sons = 1. In practice since women can't have an infinite number of children, then this wouldn't be an infinite series, so the real number of expected boys would be lower than one, but there you go… Now, for the bus case, yo…
I guess it would be balanced if the rule was, > 1 when your bus turns up, -1 for every bus going the other way.