The Progressive Ace Game

As a brief break from my usual, more rigorous content, I thought it’d be fun to explore some games. Of course, to stay consistent with the themes of my blogs, these games will still be rooted in statistics.

As we’ll soon see, the puzzles revolve around how uncertainty behaves and how small changes can dramatically change what good strategies look like. This is, in fact, exactly the kind of thinking that appears in probability theory: We’re given a system governed by randomness, and the challenge is to reason our way through it!

I thought card decks make particularly good playgrounds for this kind of reasoning. So, in this short sequence of puzzles, we’ll take a shot on two related games using a shuffled deck of cards. I hope you have fun brainstorming for each.

I also present the solutions in detail after all the problems.

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In this first part of the game, you take a standard 52-card deck and shuffle it thoroughly. You will now draw cards from the top of the deck one at a time without replacement. The game ends as soon as the first Ace appears.

If the first Ace appears on the $k$-th draw, you receive a payout that depends on how long you managed to avoid drawing one. For each non-Ace you draw, you add $1 to your winnings. When you draw an Ace, your winnings are tripled.

For example:

Draw Sequence	First Ace Position	Payout ($)
A	1	3
N, A	2	6
N, N, A	3	9
N, N, N, A	4	12

What’s the expected payout of the game?

***

Now, let’s modify the rules a little. Instead of ending the game when an Ace appears, you’re allowed to stop drawing cards whenever you like. Of course, there’s got to be a catch—else you could simply draw all 52 cards and claim the largest payout possible. The catch is that drawing an Ace immediately ends the game and wipes out your winnings.

Once again, start with a well-shuffled standard deck of 52 cards. You will still draw cards from the deck one at a time, where after each reveal, you must choose between two options:

1. Stop the game and keep a payout equal to the number of cards you have successfully drawn.
2. Continue drawing.

However, again, as I’ve mentioned, your payout becomes $0 upon drawing an Ace. What’s your optimal strategy, or more specifically, at what point should you stop drawing cards in order to maximize your expected payout?

Solutions

Considering the rules of the payout, we define

where the r.v. $X$ represents our winning amount when the $K$-th card drawn is an Ace ($K$ is also an r.v.). Our goal is to calculate the expectation $\mathbb{E}\left[X\right]$.

My favorite way to go about this is by marginalizing conditional expectations. The problem with $X$ is that it’s a function of another r.v. $K$. So, it’d be nice if we could condition on values $K=k$. Since the deck contains 4 Aces and 48 non-Aces, the first Ace could appear anywhere from the 1st draw to the 49th draw. Therefore, $k \in \{1,2,…,49\}$, and we can define, using the law of total expectation,

Since $(X \mid K=k) = 3k$,

where

Using combinatorics, we obtain the equation

where the left term can be further simplified, giving us

We then simply substitute this back into the expectation function

Calculating the summation yields an expected payout of $32.80.

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Let’s say that at this point we have drawn $k$ non-Ace cards, meaning there are still $52-k$ cards left in the deck, 4 of which are Aces. If we tap out now, then our payout is $$k$. If we instead draw another card, our expected payout is

$P\left(\text{Ace}\right)$ is irrelevant because the value associated with it is zero. On the other hand,

Now, we need to consider: Is the extra expected gain from drawing one more card worth the risk of losing everything? In other words, we should call it quits if the expected payout after drawing the next card is less than the current payout, that is,

Substituting $\mathbb{E}\left[Y\right]$,

Solving for $k$, we find that

In other words, after the $k=10$-th draw, the risk of ruin brought from drawing the 11-th card outweighs the potential reward.