Least common PPs (lc/nc)

[Gauss]

The reason the original poster picked this data out is because of the rank-order. If I was to test the rank-order through ,say, a Spearman rank test then the results obviously be biased. This bias is somewhat reduced as I tested the relative frequencies without respect to rank-order (chi-squared p-value is in the order of 1e-6). I grant that there will be some bias, how much is almost impossible to guess.

[kak1154]

Does anyone have any way of calculating how probable this is?

I guess you could say that having AA be the least common is just a 1/13 chance, then having KK be next is 1/12, QQ next is 1/11, etc. This would give you a probability of...

(1/13)(1/12)(1/11)(1/10)(1/9) = 0.000006475 or 0.0006475%

Even taking the rank out of it, the chance that any five will come in order will be...

The first one is given, so that probability is one. Assuming that you can go around the corner, the next probability is 2/12, because it could be the card above or below the first one. Then it continues normally: 1/11, 1/10, etc.

(1)(2/12)(1/11)(1/10)(1/9) = 0.000168 or 0.0168%

Does this seem about right? I realize that it is quite simplistic, since it doesn't take into account the sample size or anything, but is it at least close?

[DcifrThs]

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It's amazing that you set a null hypothesis before getting this data, and then a sample just showed up for you!

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The null hypothesis is that it is equally likely that each rank pair occur. Is there something wrong with this reasoning?

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You should come up with the hypothesis, THEN gather data, rather than the other way around. Otherwise you have a bias problem, since this sample was posted only because it was weird.

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before hypotheses...do you know the research question?

if yes you very likely know in advance the null hypothesis for standard statistical methods.

Barron