# Confusion on Symmetry in probability

This may be a simple problem, but it is something that I still cannot wrap my head around:

Consider a 52-cards deck of cards. A person takes a card randomly without replacement then the second person takes a card. What is the probability that the value of the first person's card is bigger than the second person's?

The argument presented in the solution is that by symetry, the answer should equal to 0.5. However, I still find this a little hard to understand, as I feel that when doing this experiment wihtout replacement, the number of cards left for the second person is less than the first person, so the condition is different.

Why can symmetry exactly be applied here?

Thanks!

• There are 4 different cards of each number. So if both players get a 3, neither player has a higher card. Commented Jul 15 at 1:24
• You say it "feels" different when doing the experiment and draw without replacement, but have you actually done the experiment in a statistically valid way and run a statistically valid analysis of the data? Your intuition is definitely broken here, it should be clear why both players have equal chance of drawing the higher card of different ranks. Trying playing the game with just two cards to start with, then add cards, you'll start to figure it out.
– Nij
Commented Jul 15 at 1:35
• @Lucenaposition - They probably intended to deal with equal valued cards via some standard ranking of the suites. Commented Jul 15 at 1:38
• Consider a deck with only two cards. Commented Jul 15 at 14:29
• Where exactly was this "solution" presented? Commented Jul 15 at 19:32

The symmetry isn't between what happens when the card is dealt to Player 1 and Player 2, but between the situation where Player 1's card is the higher and the situation where Player 2's is higher.

Consider a simpler case where there are only 4 cards numbered 1 through 4. After Player 1 draws their card, we are in one of four possible universes:

• Player 1's card was the 1, and Player 2 has a 100% chance of drawing a higher card.

• Player 1's card was the 2, and Player 2 has a 67% chance of drawing a higher card.

• Player 1's card was the 3, and Player 2 has a 33% chance of drawing a higher card.

• Player 1's card was the 4, and Player 2 has a 0% chance of drawing a higher card.

Notice that in some sense the first and fourth case are mirror images of each other, as are the second and third. In other words, any time there's a scenario where Player 2 has a $$p$$ chance of drawing a higher card than Player 1, there's another scenario where Player 2 has a $$1 - p$$ chance. Because of that, if we write the overall probability that Player 2's card is higher, it will look something like

$$P(P2 > P1) = \frac{1}{n} p_1 + \frac{1}{n} p_2 + \ldots + \frac{1}{n} (1 - p_2) + \frac{1}{n} (1 - p_1)$$

But notice that if we flip the situation around and consider the probability that Player 1's card is higher, then we'll get an expression that looks like

$$P(P1 > P2) = \frac{1}{n} (1 - p_1) + \frac{1}{n} (1 - p_2) + \ldots + \frac{1}{n} p_2 + \frac{1}{n} p_1$$

which has all of the same terms, just written in reverse order. Which means that the two expressions will be equal, i.e. $$P(P1 > P2) = P(P2 > P1)$$. This is where we get the symmetry argument - we get the same overall set of calculations regardless of whether we're looking in one direction or the other, resulting in the two having equal value.

By the rule of total probability we have $$P(P1 > P2) + P(P1 = P2) + P(P1 < P2) = 1$$, which we can easily solve to give $$P(P2 > P1) = \frac{1}{2}(1 - P(P1 = P2))$$ or, if we know that ties are impossible, $$P(P2 > P1) = 0.5$$.

There are $$52 \times 51=2652$$ equally likely ways of Player 1 drawing a card without replacement and then Player 2 drawing a different card.

(There would have been $$52 \times 52= 2704$$ possibilities with replacement.)

Player 2 is as likely to have a card ranking below Player 1's card as having a card ranking above Player 1's card, since for each case that Player 1 has card $$X$$ and Player 2 has card $$Y$$ with $$X>Y$$, there is a corresponding case where Player 1 has card $$Y$$ and Player 2 has card $$X$$ and $$Y, and similarly in reverse.

(The same argument can be made with replacement.)

So the two without replacement probabilities are equal to each other. They are therefore both $$\frac12$$ if there are no equally ranking cards, or both less than $$\frac12$$ if your ranking system allows distinct cards of equal rank.

(What is different with replacement is the higher probability of equal rankings, as the additional $$2704-2652=52$$ with replacement possibilities are those when the identical card is drawn two times.)

We shuffle a deck and give P1 the top card and P2 the second card.

Every permutation of the deck is equally likely. So for every pair of cards A,B, the chance of the top two cards being (A,B) in that order is the same as the chance of (B,A).

So for any particular situation, such as P1 getting card A and P2 getting B, the symmetric situation of (B,A) is equally likely.

So for example, in every case where P1 wins, there is a symmetric, equally-likely case where the cards are reversed and P2 wins.

Here, being symmetric is not about the first player versus the second player, but about winning versus losing. It seems the question assumes that there is no situation where the two players tie. With this assumption, the first player will either win or lose. In your head, you might imagine how to calculate the probability of the first player winning, as well as the probability of the first player losing. To calculate either of these, you would approach it as follows:

sum {
(the probability of the first player picking each card)
* (the probability of the second player picking certain cards for the first player to win or lose)}


Then you will realize that the actual calculation with numbers to determine the probability of winning should be the same as for the probability of losing! It's akin to exchanging the locations of numbers in the calculation, like summing up from 1 to 10 versus summing up from 10 to 1. That's why it's considered symmetric. Therefore, the probability of winning equals the probability of losing, and the sum of both should be 1, as there are no other outcomes. Thus, each probability is 0.5.

I agree that the solution is in error. Assuming by "value", we mean the "rank" in a standard deck 52-cards, there are 4 duplicates of each rank, and therefore a positive chance of tying and no one winning. This is a common issue in games of poker, for example (called a "split").

See Wikipedia: Standard 52-card deck: Composition

Specifically, having chosen the first card, there are 3 cards left in the deck of the same rank, so the chance of a tie is 3/51 ~ 6%. By symmetry, we can then legitimately split the remaining probability between the two players, i.e., about 94%/2 = 47% of either player winning.

Contrary to what some other answer are saying, there is a symmetry between the two players (let's name them Alice and Bob). Even if not immediatly obvious, drawing the first or second card does not change this game!

Let's note X the card drawn by Alice and Y the card drawn by Bob. When we look at the random variable Z = (X, Y), we notice that it is uniformly chosen between all ordered pairs (all ordered pairs have probability $$\frac{1}{52*51}$$ to be chosen). But the random variable Z' = (Y, X) obtained by letting Bob pick first and Alice pick second is also uniformly chosen between all ordered pairs. Z and Z' actually follow the same law and entirely determine the issue of the game, so the game is symmetric between Alice and Bob.

To continue building intuition on this non-apparent symmetry, you can try to convince yourself that picking a card second (or even third or fourth) without replacement still gives you a uniformly chosen random card.

The symmetry between the players shows that Alice and Bob have the same probability P of winning. Also they win if and only if the other loses so they also have probability P of losing. The symmetry also shows that they have the same probability of drawing. Be cautious that if a draw is possible, the probability of winning will not be 0.5!

By the way, this argument holds even if there is no symmetry between the winning and losing cards (for example if you add 2 jokers to your deck).