Thoroughly shuffled

The “natural” way to verify if all suits (D, S, H, C) of a 52- cards deck are not ill distributed among the 4 hands, 13 cards each, is to use the Chi-squared test.
We know that every player expects to have 13 / 4 = 3.25 cards of each suit. Therefore one can evaluate the Chi-squared easily

The following examples are real samples I got through a Program , one immediately after the other, any choose.

Sample I
_______North______East______South______West
__D____2__________5_________3__________3___
__S____3__________2_________4__________4___
__H____4__________3_________3__________4___
__C____5__________3_________3__________2___


Sample II
_______North______East______South______West
__D____6__________3_________1__________3___
__S____1__________4_________4__________4___
__H____1__________4_________4__________4___
__C____5__________2_________4__________2___

Chi-2 respectively Q(I) = 4.15, Q(II) =10.15.

The problem is HOW TO FIND the critical values?

I think that it was not too much inappropriate to consider that the number of degrees of freedom will be k=16 – 1 – 4, because 16 is the number of cells (4 players, 4 suits) and 4 are the constraints: each suit have13 cards. If so Crit = 19.68.
But, honestly I ask: is this realistic? This is not to rely abusively on the test robustness? (The population is multinomial with restrains, and the cells have too little expectations - less than 5.
What to do? Simulation, of course.

Results:

Cumulative probability equal to 0.956 for Q=16.92.
and p(Q>4.15) = 0.901, p(Q>10.15)=0.397.
Conclusion: for both samples do not reject H0 (the packs were thoroughly, or sufficiently, shuffled in respect to the suits).

Note that the critical value for 11 d.f. (19.68) is rather greater than the REAL one, 16.92, even though the d.f. were adjusted for the constraints.


__________________licas
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