Positive Expectancy with Random Chance A conversation came up at work about positive expectancy. I am having difficulty getting the same answer as the guys so I am throwing the question out to you folks...  Any help is appreciated.
A game costs €50 to play. If you win you get €100. You get 52 chances to pick the Ace of hearts from a deck of cards. Every time you pick a card it is then returned to the deck and the deck is shuffled again. You pay the €50 once and you then have up to and including 52 tries to get the Ace of Hearts. Is there a positive expectancy on this game?
I worked it that you have a 1/52 chance of winning every time you pick a card. While you do get to pick 52 times your chance of winning is still 1/52 each time you play. 
( 1/52 * 100 ) - ( 51/52 * 50) = (200 - 4750) {I used rounded 2% for 1/52 and rounded 98% for 51/52}
What do you folks think?

To clarify for OP. The expected return of the game is 
$E[R] = 100 \cdot P(\text{win}) - 50$
Where the probability that you lose is
$P(\text{lose}) = (51/52)^{52}$
and the probability that you win is
$P(\text{win}) = 1 - P(\text{lose}) = 1 - (51/52)^{52}$
Hence
$E[R] = 100 \cdot P(\text{win}) - 50$
$E[R] = 100 \cdot (1 - (51/52)^{52}) - 50 \approx \$13.7$
So this is a fantastic game :)
