## Sunday, November 27, 2016

### The Odds of Getting Three Consecutive Wars in a Row in the Card Game

What better way to spend the Sunday after Thanksgiving than playing card games with your family and then arguing about the odds?

As pictured, my daughter and I just got three consecutive "wars" in the card game of war. (I lost with a 3 at the end!)

What are the odds of that?

Well, the odds of getting just one war are 3/51, right? Here's why. It doesn't matter whether my or my daughter's card is turned first. That card can be anything. The second card needs to match it. With the first card out of the deck, 51 cards remain. Three of them match the first-turned card. So 3/51 = .058824 = about a 5.9% chance.

Then you each play three face down "soldier" cards. Those could be any cards, and we don't know anything about them, so they can be ignored for purposes of calculation. What's relevant are the next upturned cards, the "generals". Here there are two possibilities. First possibility: The first general is the same value as the original war cards. Since there are 50 unplayed cards and two that match the original two war cards, the odds of that are 2/50 = .040000 = 4.0%. The other possibility is that the value of the first general differs from that of the war cards: 48/50 = .960000 = 96.0%.

(As I write this, my son is sleeping late and my wife and daughter are playing with Musical.ly -- other excellent ways to spend a lazy Sunday!)

In the first case, the odds of the second general matching are only one in 49 (.020408, about 2.0%), since three of the four cards of that value have already been played and there are 49 cards left in the deck (disregarding the soldiers). In the second case, the odds are three in 49 (.061224, about 6.1%).

So the odds of two wars consecutively are: .058824 * .04 * .020408 (first war, followed by matching generals, i.e. all four up cards the same) + .058824 * .96 * .061124 (first war, followed by a different pair of matching generals) = .000048 + .003457 = .003505. In other words, there's about a 0.35% chance, or about one in 300 chance, of two consecutive wars.

If the second war had generals that matched the original war cards, then there's only one way for the third war to happen. Player one draws any new general. The odds of player two's new general matching are 3/47 (.063830).

If the second war had generals that did not match the original war cards, then there are two possibilities.

First possibility: The first new general is the same value as one of the original war cards or previous generals. There's a 4 in 48 (.083333) chance of that happening (two remaining cards of each of those two values). Finally, there's a 1/47 (.021277) chance that the last general matches this one (last remaining card of that value).

Second possibility: The first new general is a different value from either the original war cards or the previous generals. The odds of that are 44/48 (.916667), followed by a 3/47 (.063830) chance of match.

Okay, now we can total up the possibilities. There are three relevantly different ways to get three consecutive wars in a row.

A: First war, followed by second war with same values, followed by third war with different values: .058824 (first war) * .04000 (first general matches war cards) * .020408 (second general matches first general) * .063830 (odds of third war with fresh card values) = .000003 (.0003% or about 1 in 330,000).

B: First war, followed by second war with different values, followed by third war with same values as one of the previous wars: .058824 (first war) * .960000 (first general doesn't match war cards) * .061224 (second general matches first general) * .083333 (first new general matches either war cards or previous generals) * .021277 (second new general matches first new general) = .000006 (.0006% or about 1 in 160,000).

C: First war, followed by second and third wars, each with different values: .058824 (first war) * .960000 (first general doesn't match war cards) * .061224 (second general matches first general) * .916667 (first new general doesn't match either war cards or previous generals) * .063830 (second new general matches first new general) = .000202 (.02% or about 1 in 5000).

Summing up these three paths: .000003 + .000006 + .000202 = .000211. In other words, the chance of three wars in a row is 0.0211% or 1 in 4739.

Now for some leftover turkey.

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As it happens we were playing the variant game Modern War -- which is much less tedious than the traditional card game of war! But since it was only the first campaign the odds are the same. (In later campaigns the odds of war increase, because smaller cards fall disproportionately out of the deck.)

Unknown said...

I agree with your calculations assuming you each have 26 random cards, but I suspect that this assumption is unjustified in the Game of War. There are only a certain number of possible games of War, and thus, only so many possible game states. As you play through the game, because of the way wars work, I'd imagine that players tend to get multiples of cards more than the random distribution would predict (since one player wins both "warring" cards whenever there is a war). But even if that is wrong, I'd be pretty surprised if the average War game state is the same w/r/t distribution of multiples as the random starting case...

Eric Schwitzgebel said...

Yes, that seems right. These calculations probably only make sense in the first run-through of the deck after the shuffle (and even then only with enough cards remaining to play out the triple war without reusing cards). After that, it will get more complicated due to the non-random composition of each player's deck. As it happens, this was our first run-through -- maybe about the eighth trick total.

Unknown said...

Right. I also don't remember: what is the rule for when all your cards are involved in wars, and your last card at the end of the war chain is the same as your opponent's? That might be another complicating factor.

Eric Schwitzgebel said...

In Modern War, we draw randomly from the won-cards pile. If you need to draw and your won-cards pile is empty and your opponent's is not, you lose.

Unknown said...

What are the chances of getting four wars in a row...this happened in a game today. Thanks so much!!

Eric Schwitzgebel said...

Hi Andrew. I can't take the time to work out every path, but starting with 0.0211% for three in a row and then multiplying by something a little bigger than 3/51 (since some pairs will have been removed from the deck, making matching pairs more likely in the remainder), will give in the ballpark of 0.0015%, or about 1 in 70,000!

Chris Barnes said...

I just got 4 wars in a row. What are the odds of that?

Unknown said...

My daughter and I hit 5 consecutive wars in a row tonight. It was unbelievable.