If you keep drawing cards from a shuffled deck, what is the expected number of draws to complete at least one card from each suit?
The expected value is the average number of draws that it would take if you repeated the experiment many times.
If we have a deck that's formed by (a+b+c) cards of three suits, a>0, b>0, c>0), then a card of the fourth suit, and then the (51-a-b-c) rest of cards, the number of draws will be (a+b+c+1). Since the 4th suit can be picked in four ways, the number of ways to form such a deck is D(a,b,c)=4. C(13,a). C(13,b). C(13,c). (a+b+c)!. 13. (51-a-b-c)!
To answer this problem we have to calculate
ΣD(a,b,c)x(a+b+c+1)/52!, where the sum is taken for all 1≤a,b,c≤13.... too much work!!
Too hard by hand
