* * * There are 4*3*2* 1=24 grouping methods (it is conceivable that four A players choose their opponents in turn). We consider the opposite of the problem. How many grouping methods will happen? For the convenience of writing, abcd and ABCD are used respectively.
AbBaCdDc AbBcCdDa AbBdCaDc
Writing here, you find that A is not a group with A, but a group with B. You can write in all three situations (you can't write anymore). It is conceivable that A and C or D can also write three other situations. At this point, we have written all the negative situations. We classify all the situations according to who A's opponent is, and list them one by one.
9/24 is the probability of the opposite, and the answer to this question is 1-9/24=5/8.
If you want to ask the probability that "each group of players has the same corner code", you can easily get the answer by using the above idea: 1/24.
Thank you.