Probability question: would you play this card game?

Today I want to share a fun card game that is a twist on drawing to the high card.

The game is played with a deck of cards numbered 1 to 100, and its rules are as follows.

First, you and I each draw one card from the deck and show them face up.

Then, I then draw another card from the deck. In the end, I have two cards and you have one.

I win the game if either of the following happens:

–The first card I drew is higher than your card

–The second card I drew is higher than the first card I drew

Because I have two different ways to win, I agree to payout more money to compensate for the odds.

I will pay you $3 when you win, and you only have to pay me $1 if I win.

Are you willing to play this game?

Give it a thought before reading the answer below.

Naive reasoning

At first glance the game appears to be fair.

A basic analysis might go as follows.

My opponent basically has two chances to win this game. On the first draw, he wins if he gets a higher card than me, which happens 1/2 of the time.

The other half of the time he needs his second card to get him a victory. In that case, he wins only if his second draw is higher than his first draw. This should also happen 1/2 of the time.

Putting this together, this means the total probability my opponent wins is:

Pr(opponent wins) = Pr(first card higher) + Pr(first card lower) * Pr(second card is higher than first)

Pr(opponent wins) = 1/2 + 1/2 * 1/2 = 3/4

I win if my opponent does not, so that means my chances of winning are 1/4.

Because I get $3 when I win, and he only gets $1, the game is fair:

E(game) = Pr(I wins)*payout + Pr(I lose)*loss

E(game) = (1/4)(3) + (3/4)(-1) = 0

So it appears the game is fair–the payout of $3 appears to compensate for the advantage of drawing two cards.

The problem is this reasoning is flawed.

Conditional probability is tricky – the solution

The above reasoning accounted the outcomes incorrectly.

The game involves three cards: the two cards that I draw (call them a₁ and a₂) and the one card you draw (label it b).

The sample space of the game is the number of ways the cards can be ordered. We can write out all 6 possible outcomes:

a₁ > a₂ > b
a₁ > b > a₂
a₂ > a₁ > b
a₂ > b > a₁
b > a₁ > a₂
b > a₂ > a₁

Now I win the game if EITHER my first card is higher than yours (a₁ > b) OR if my second card is higher than my first (a₂ > a₁)

From inspection we can find out this happens in 5 of the 6 cases:

a₁ > a₂ > b
a₁ > b > a₂
a₂ > a₁ > b
a₂ > b > a₁
b > a₁ > a₂
b > a₂ > a₁

So the game is actually more favorable to the person drawing two cards–he wins with probability 5/6. The only time I lose is when your card is the highest, and my second draw is lower than my first draw.

To be fair, I would have to pay out $5 for every time I lose versus $1 for when he wins.

What was wrong with the naive reasoning above?

The incorrect part was calculating the probability of the second card being larger than the first draw, conditional on the first draw being smaller than the other persons.

In other words, we need to consider the times a₁ is less than b and see how many times that a₂ is greater than a₁

These are the relevant events.

a₂ > b > a₁
b > a₁ > a₂
b > a₂ > a₁

As you can see, there are 3 possible events and 2 of them correspond to a₂ being greater than a_1. This makes the conditional probability of 2/3 and not 1/2.

The proper calculation is therefore:

Pr(opponent wins) = Pr(first card higher) + Pr(first card lower) * Pr(second card is higher | first card being lower)

Pr(opponent wins) = 1/2 + 1/2 * 2/3 = 5/6

The bottom line: always be careful with conditional probability calculations, especially when faced with a bet!

Reference: Card game from Game Theory Evolving, Problem 1.21