I know I'm way late to the party here, but what about a draw bag of 100 beads, 99 of one color (e.g. white) and 1 of another (e.g. black)? Each time you pull from the Tower, pull from the bag, discard the white ones; when you pull black you lose.

A Jenga tower usually collapses after about 30-35 pieces are drawn. Here's one way to simulate that.

A d6 dice pool. Every time you use a card you roll 1d6. If you roll a 1 you lose one die from the dice pool. On average it takes 6 rolls of a d6 to get a 1 so to simulate a jenga tower you want a dice pool of 5-6 six sided dice.

Here's a simpler method that only requires one six sided die.

Write down on a sheet of paper the number 5. Roll 1d6 each time you draw a card. If you roll a 1, reduce the number by 1. When you reach 0 it's the same as the jenga tower collapsing.

5 dice will give you 30 rolls on average before you reach 0. 6 dice will give you 36 rolls. You can increase the average number of rolls very easily by increasing the number you start with. Increase by 1 to increase the average number of rolls by 6.

Speak the Sky first suggested 100 d6s, getting rid any dice that show 1s. You can get similar results with a dozen d10s (mean 30.09 rolls, standard deviation 12.13) or 7 d12s (mean 30.37 rolls, standard deviation 14.08).

The larger the die size, the higher standard deviation (i.e. "swingier" results). You *could* try 2 d20s for 30 rolls on average, but you'd still have a lot of games with more than 50 rolls.

I also offer a variant to stevozip's method: Shuffle a deck of cards with a single joker. Draw cards until you hit the joker. It *could* happen the very first draw, but on average it happens around draw 27. The longer you play, the more likely you are to draw that joker. Std dev is 15.20, so you're not far off from the 20-40 draws Speak the Sky mentioned. As a bonus, no game is going to drag on for more than 53 rounds.

Note: I haven't tried all of these yet, just typed up some quick python code.