# Expected number of cards turned over until a spade is drawn. [duplicate]

• I have given a very detailed answer to a very similar problem yesterday with full justification. You will just have to replace the $4$ (for $4$ Aces) by $13$, and the $48$ by $39$. – André Nicolas Nov 27 '12 at 18:48
• Thank you! It took me a few minutes to work through your solution and understand it, but it makes sense. Pretty much we're summing up the probability of one card being before any spades plus two cards being before any spades and so on. I was thinking of it that way, but was getting the probability wrong as $\frac{13}{52}$. Could this be boiled down in terms of a Bernoulli trial where $n=39$ and $p=\frac{1}{14}$ in the formula $E[X]=np$? – kamikazekent Nov 27 '12 at 19:07
• @kamikazekent the actual value is $53/14$. So you maybe bit off with a Bernoulli assumption. – jay-sun Nov 27 '12 at 19:35