A Statistical Paradox? Different conclusions from the same data

[@cach_dies]

4:50 why is p(Y>= 6) = p{Ts on first 5 in a row)? Shouldn't it be p(Y>= 6) = p(Ts on first 5 in a row and heads on the 6th throw) + p(Y>=7) = (1/2) ^ 6 + p(Y>=7) = 1 - p(Y=1) -p(Y=2) - p(Y=3) - p(Y=4) - p(Y=5) =0.03125 ? I know its the same answer but I think the reasoning could be clearer. If anyone can explain why Dr Barker decided to do it like this I would appreciate your reply.

[@ygalel]

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.

[@maklovitz]

Very interesting conclusions. Gladly it's only a semi-paradox

[@karthikrajeshwaran1997]

This is brilliant❤

[@MrRyanroberson1]

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

[@MrRyanroberson1]

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

[@DrBarker]

@@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.

[@MrRyanroberson1]

@@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

[@MrRyanroberson1]

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.

[@DrBarker]

@@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.