As I have graduated and experienced real life usage of data, I have truly experienced the impact of “assumptions” and their importance. Theory is fun and all, but what happens in reality should be looked at very carefully. This paradox is a great example of that.
Another natural way to understand why the geometric test has such a different result is to consider the other five ways heads could have appeared in that sample. If it was the first flip, it would appear trivially heads-biased. second: no bias. third, fourth, fifth: increasing levels of tails bias. Adding up all six possibilities: G(1)/6 + G(2)/6 + ... = 0.5/6 + (0.5)^2/6 + ... (0.5)^5/6 = 63/(64*6), about a 1/6 chance. This strangely produces a different paradox since now it appears that the geometric interpretation would call this MORE likely than the binomial interpretation would
oh 8:00 i watch just a few more seconds and notice my attempt was flawed. using G(n)/7 instead, the probability is 127/(128*7), but this is still a lot more likely than the binomial interpretation would have suggested
@@MrRyanroberson1 I'm not too sure where your calculation is coming from with the G(n)/6 or the G(n)/7. If, for example, the geometric random variable Y is =1, then the probability of this is P(Y = 1) = 0.5, and similarly P(Y = 2) = (0.5)^2 and so on, but I'm not sure where the division by 6 or 7 has come from? With the geometric interpretation, the probabilities of getting HTTTTT, THTTTT, ..., or TTTTTH should all still be (0.5)^6 each. Because if you look at e.g. HTTTTT and say the probability is 0.5, then you're only really saying that you get H the first time, and it could be e.g. HTHTHT or anything else for the next 5 throws after getting H the first time.
@@DrBarker because the binomial interpretation allows 6 possible things to count as the "same"- those being the 6 positions where H can appear. Applying the geometric interpretation to each of them (only taking the data up to the first H each time) gives G(n). Since all 6 are treated equally by the binomial, there is a 1/6 chance for H to have been in each position. Taking the weighted sum (which here is just the average) of probabilities gives almost 1/6, which is more than the 1/10 given by the binomial interpretation
This all said, i can definitely see why using (0.5)^6 for all six variations would be meaningful, it's just that this interpretation doesn't count the data that "would have been collected" in each possible case.
@@MrRyanroberson1 I see what you've done - that makes more sense now. If you're starting with e.g. TTHTTT, which has probability (0.5)^6, and then changing it into TTH, which has probability (0.5)^3, I think this difference explains why you got a different (higher) probability of the data from the geometric model in your first comment.
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As I have graduated and experienced real life usage of data, I have truly experienced the impact of “assumptions” and their importance.
Theory is fun and all, but what happens in reality should be looked at very carefully.
This paradox is a great example of that.
Very interesting conclusions. Gladly it's only a semi-paradox
This is brilliant❤
Another natural way to understand why the geometric test has such a different result is to consider the other five ways heads could have appeared in that sample. If it was the first flip, it would appear trivially heads-biased. second: no bias. third, fourth, fifth: increasing levels of tails bias. Adding up all six possibilities: G(1)/6 + G(2)/6 + ... = 0.5/6 + (0.5)^2/6 + ... (0.5)^5/6 = 63/(64*6), about a 1/6 chance. This strangely produces a different paradox since now it appears that the geometric interpretation would call this MORE likely than the binomial interpretation would
oh 8:00 i watch just a few more seconds and notice my attempt was flawed. using G(n)/7 instead, the probability is 127/(128*7), but this is still a lot more likely than the binomial interpretation would have suggested
@@MrRyanroberson1 I'm not too sure where your calculation is coming from with the G(n)/6 or the G(n)/7.
If, for example, the geometric random variable Y is =1, then the probability of this is P(Y = 1) = 0.5, and similarly P(Y = 2) = (0.5)^2 and so on, but I'm not sure where the division by 6 or 7 has come from?
With the geometric interpretation, the probabilities of getting HTTTTT, THTTTT, ..., or TTTTTH should all still be (0.5)^6 each. Because if you look at e.g. HTTTTT and say the probability is 0.5, then you're only really saying that you get H the first time, and it could be e.g. HTHTHT or anything else for the next 5 throws after getting H the first time.
@@DrBarker because the binomial interpretation allows 6 possible things to count as the "same"- those being the 6 positions where H can appear. Applying the geometric interpretation to each of them (only taking the data up to the first H each time) gives G(n). Since all 6 are treated equally by the binomial, there is a 1/6 chance for H to have been in each position. Taking the weighted sum (which here is just the average) of probabilities gives almost 1/6, which is more than the 1/10 given by the binomial interpretation
This all said, i can definitely see why using (0.5)^6 for all six variations would be meaningful, it's just that this interpretation doesn't count the data that "would have been collected" in each possible case.
@@MrRyanroberson1 I see what you've done - that makes more sense now.
If you're starting with e.g. TTHTTT, which has probability (0.5)^6, and then changing it into TTH, which has probability (0.5)^3, I think this difference explains why you got a different (higher) probability of the data from the geometric model in your first comment.