Multideck Probabilities
08-21-2015
, 03:40 PM
3*C(13,3)*[C(20,3)*20^2 - 4*C(5,3)*5^2] = 390,390,000
Actually, it depends on what is meant by "combinations". The above would be the numerator if you were finding the probability, but if you just want the distinguishable combinations the number is less. For instance, the above counts {7
,7,7
} more than once because it treats each card as distinct even if the two cards have the same rank and suit.
Now I also notice the same thing happens in the 3-deck example (but to a lesser extent). The Wiz, however, did not take it into consideration, so his interest must have been on the probability numerator.
Actually, it depends on what is meant by "combinations". The above would be the numerator if you were finding the probability, but if you just want the distinguishable combinations the number is less. For instance, the above counts {7
,7,7} more than once because it treats each card as distinct even if the two cards have the same rank and suit.Now I also notice the same thing happens in the 3-deck example (but to a lesser extent). The Wiz, however, did not take it into consideration, so his interest must have been on the probability numerator.