Average time to win "Catch (Chase) the Ace!" Catch (Chase) the Ace! is a popular lottery card game that is often played to win large sums of money and to donate funds to worthy charities. See:
https://en.wikipedia.org/wiki/Chase_the_Ace_(lottery)
Gameplay involves a standard 52-card deck. The objective is to select the Ace of Spades.
According to the Wikipedia Article:
The jackpot accumulates from week-to-week until it is won, and the game is then     over. Each week participants buy lottery tickets. The funds from ticket sales are     divided into three parts. Typically the organizers keep 50% (donated to charity), the winner of the weekly lottery takes 20%, and the remaining 30% goes into the jackpot. The lottery winner also then draws a card from a deck of playing cards and wins the accumulated jackpot if the ace of spades is drawn. If not, the reduced deck is kept for the following week's game, and the jackpot rolls over to the next week.
I'm interested in the mathematical/probabilistic aspects of the game and so wrote a quick R simulation.
# Playing Card Alphabet

    cards <- c("Ace of Clubs",
           "Ace of Diamonds",
           "Ace of Hearts",
           "Ace of Spades",
           "2 of Clubs",
           "2 of Diamonds",
           "2 of Hearts",
           "2 of Spades",
           "3 of Clubs",
           "3 of Diamonds",
           "3 of Hearts",
           "3 of Spades",
           "4 of Clubs",
           "4 of Diamonds",
           "4 of Hearts",
           "4 of Spades",
           "5 of Clubs",
           "5 of Diamonds",
           "5 of Hearts",
           "5 of Spades",
           "6 of Clubs",
           "6 of Diamonds",
           "6 of Hearts",
           "6 of Spades",
           "7 of Clubs",
           "7 of Diamonds",
           "7 of Hearts",
           "7 of Spades",
           "8 of Clubs",
           "8 of Diamonds",
           "8 of Hearts",
           "8 of Spades",
           "9 of Clubs",
           "9 of Diamonds",
           "9 of Hearts",
           "9 of Spades",
           "10 of Clubs",
           "10 of Diamonds",
           "10 of Hearts",
           "10 of Spades",
           "Jack of Clubs",
           "Jack of Diamonds",
           "Jack of Hearts",
           "Jack of Spades",
           "King of Clubs",
           "King of Diamonds",
           "King of Hearts",
           "King of Spades",
           "Queen of Clubs",
           "Queen of Diamonds",
           "Queen of Hearts",
           "Queen of Spades")


    ## Simulation

    catch.ace <- function(ticket.price = ticket.price, 
      max.num.tickets = max.num.tickets) {
    week <- 0 # initialize counter
    payoff <- 0 # initialize weekly winnings
    jackpot <- 0 # initialize progressive jackpot
  
    while(week < 52) {
      max.num.tickets <- ceiling(runif(1, 1, 75000)) 
                    # maximum number of tickets sold
      week <- week + 1 # increment counter
      tickets <- sample(max.num.tickets, size = 1) 
                         # random number of tickets sold
      fill.envelopes <- sample(cards, size = length(cards), 
                replace = FALSE) # assign cards to envelopes
      pick.ticket <- sample(tickets, size = 1) # select random ticket
      pick.envelope <- sample(fill.envelopes, size = 1) 
                                # choose random envelope
      cards <- cards[!(cards == pick.envelope)] 
                                # remove selected card
      payoff <- tickets * ticket.price # ticket sales
      jackpot <- jackpot + (0.30 * payoff) 
          # update weekly winnings; 30% of ticket sales 
          # goes into jackpot
    
    cat("\n Week: ", week,
        "\n Ticket number: ", pick.ticket,
        "\n Card selected: ", pick.envelope)
    
    if ("Ace of Spades" %in% pick.envelope) {
      cat("\n Outcome: Congratulations, you've selected the Ace of 
       Spades! You've won the progressive jackpot! \n Jackpot: 
         $",jackpot, "\n \n") 
  break
  } else {
    cat("\n Outcome: Sorry, you didn't select the Ace of Spades! 
     Better luck next time! \n Payoff: $",(0.20 * payoff)) 
       # 20% of ticket sales goes to winning ticket holder each week  
      }
    
    cat("\n Proceeds donated to charity: $",(0.50 * payoff), "\n \n") 
       # 50% of all ticket sales goes to charity
    
    }
    cards <- cards # reset deck
    return(list("week" = week,
                "payoff" = payoff,
                "jackpot" = jackpot,
                "tickets" = tickets,
                "fill.envelopes" = fill.envelopes,
                "pick.ticket" = pick.ticket,
                "pick.envelope" = pick.envelope))
    }

## Let's play!

    ticket.price <- 5 # price per ticket
    max.num.tickets <- ceiling(runif(1, 1, 75000)) 
         # maximum number of tickets sold

    catch.ace(ticket.price = ticket.price, 
               max.num.tickets =  max.num.tickets)

I wondered about the mean time to winning. So I tried it out, simulating 10000 runs.
Mean number of weeks to win (similarly, cards to draw)
    mean(replicate(10000, catch.ace(ticket.price = ticket.price, 
      max.num.tickets =  max.num.tickets)$week))

The answer is around 26 (52/2) weeks (similarly, 26 cards to draw), which is about 6 months.
My question:
Is there an easy way to arrive at a closed form expression for the expectation and other quantities (such as the variance)?

Imagine we number the positions in the originally shuffled deck 1 to 52. Assume that the deck is well-shuffled (well enough that we may treat all possible orders of the cards as equally, likely, say).
We're interested in the position of a specific card, which should be uniformly distributed over those positions. That is, the position of the card in the deck after the initial shuffle has a discrete uniform distribution on the values $\{1, 2, 3,..., 52\}$. If the card it's in position $i$, it is drawn in week $i$.
Consequently, for the specific quantities mentioned, your question is simply, what's the mean and variance of a discrete uniform on $\{1, 2, 3,..., 52\}$?
That's a simple matter of calculation from the definitions, we get a mean[1] of $\mu=(1+52)/2 = 26.5$ and variance[2] of $\sigma^2=51\times 53/12=225.25$, or a standard deviation of just over $15$.
However, "other quantities" is too open-ended to say whether it's easily represented in closed form. Many things are easily computed for the discrete uniform (many things you might care about are in the sidebar there), but it's a bit hard to guess at your ingenuity in coming up with some particularly tricky "other".
However, simulation is a useful strategy in any case, and will yield answers even when there isn't some neat expression. I'd just advise using considerably larger sample sizes -- in R (which is not the fastest way to simulate this, but will do fine), $10^{7}$ simulations of the position takes somewhere between 1 and 2 seconds on my laptop, which is not long to wait.
[1]: https://en.wikipedia.org/wiki/Expected_value#Finite_case
[2]: https://en.wikipedia.org/wiki/Variance#Discrete_random_variable
