Question 135683
Since the successful outcome is AT LEAST 1 face card in the 3 cards, you are faced with calculating the probability that there are exactly 3 face cards, then the probability that there are exactly 2, and then the probability that there is exactly 1, and then summing the result.  This is a very convoluted and complex calculation.


Much easier is to realize that if you draw 3 cards the probability that at least one is a face card plus the probability that NONE of them are face cards is certainty -- there are no other possibilities.  So, calculate the probability that NONE of them are face cards -- a fairly straightforward calculation -- and then subtract that value from 1.


In a standard deck of 52 cards, 3 times 4 = 12 of them are face cards and the other 40 are not face cards.


If I pick one card out of the deck, I have a 40/52 probability that it is not a face card.  Presuming I am successful and draw a non-face on the first draw, I will now have a deck with 51 total cards and 39 non-face, so my probability of drawing a second card that is non-face given that the first card was non-face is 39/51.  And the probability for the third card being non-face given that the first two were non-face is 38/50.


The total probability is then the product of these three probabilities:


{{{(40/52)(39/51)(38/50)=59280/132600=.45}}} (rounded to the nearest 0.01)


So the probability that you will draw at least 1 face card in a three card draw is {{{1-0.45=0.55}}}


Give me the name of this guy who told you it was 70%.  I want to play poker with him.