What are the odds of guessing higher or lower on 12 consecutive playing cards as they appear from 1 deck?

This is not is an exact calculation...as someone already said it really depends on the cards dealt out, and is probably best dealt with via simulation.





But a rough approximation can be determined if we ignore card removal and calculate the average "guessing probability"





I am going to assume getting the same rank is a LOSS.





A and 2 (2/13)*(12/13) = 24/169


K and 3 (2/13)*(11/13) = 22/169


Q and 4 (2/13)*(10/13) = 20/169


J and 5 (2/13)*(9/13) = 18/169


T and 6 (2/13)*(8/13) = 16/169


9 and 7 (2/13)*(7/13) = 14/169


the dreaded 8 (1/13)*(6/13) = 6/169





Average correct guessing probability (excluding card removal): 120/169 = 71%





To guess 12 times correctly in a row is simply: (120/169)^12 = 0.0164 = 1.64%





Less rare than my first inituition to be honest.

What are the odds of guessing higher or lower on 12 consecutive playing cards as they appear from 1 deck?
The odds would differ depending on each card as it's turned up....
Reply:Eeeerrrrrrr.... good game, good game... rrrrrrrr
Reply:who cares!!!!
Reply:the only thing which can be readily calculated precisely is that there is less than a one in three chance of avoiding an ace i.e. 1 in 2.9623043