Optimal Pass the Pigs Strategy (Part One)

Since I'm out of town for a bit, I'm migrating over a few relevant posts from an old blog of mine that I'm planning to shut down. Enjoy!

Pass the Pigs is a simple yet addictive dice game that uses cute little plastic pigs as dice. If you've never played, the rules are very straightforward. On each turn, a player rolls two pigs. The pigs will land in different positions, which will determine how many points the player has in their hand for that turn. The player may then decide to "pass the pigs" to the next player. If they do this, all of the points in their hand will be added to their official score. They may also decide to roll the pigs again to try to add more points to their hand before passing the pigs. But they must be careful! If the pigs both land on their sides with one showing a dot and the other showing a blank side, they "pig out" and lose all of the points they've accumulated in their hand! It's risky business. The first player to accumulate a score of 100 or higher wins.

Pass the Pigs

Credit: Larry Moore

Like anything with dice (even pig-shaped dice), Pass the Pigs is a game of chance. That means, with a little effort, we should be able to figure out the probabilities of certain things happening in the game, and develop some optimal strategies. So how do you win at Pass the Pigs? Read on to find out.

The Math

There's only one simple question at the heart of Pass the Pigs: "Do I hold or do I roll?" So, let's start thinking about that. When is it a good idea to "pass the pigs" and keep the points in your hand, and when is it a good idea to risk it and go for more? Simple math will give us a good answer.

There are 6 different ways that a pig may land in Pass the Pigs:

  • Sider (Dot): Laying on its side with a dot showing
  • Sider (No Dot): Laying on the opposite side without a dot showing
  • Razorback: Laying on its back, belly up
  • Trotter: Standing on all four feet
  • Snouter: Balanced on its front two feet and its nose
  • Leaning Jowler: Balanced on one foot, its nose, and one ear

If a player rolls two identical siders, they get a point. If they roll opposite siders, they pig out and lose all the points in their hands. If they roll a sider in combination with anything else, the sider is worth 0. Point values for the rest of the rolls are as follows:

  • Razorback: 5 (20 for 2)
  • Trotter: 5 (20 for 2)
  • Snouter: 10 (40 for 2)
  • Leaning Jowler: 15 (60 for 2)

This lets us make a very simple table of the possible point outcomes of every roll of 2 pigs:

Dot No Dot Razorback Trotter Snouter Leaning Jowler
Dot 1 Pig Out! 5 5 10 15
No Dot Pig Out! 1 5 5 10 15
Razorback 5 5 20 10 15 20
Trotter 5 5 10 20 15 20
Snouter 10 10 15 15 40 25
Leaning Jowler 15 15 20 20 25 60

Now that we know all of the point outcomes, we'll also need to know the probabilities of each outcome. Luckily, other people have done all the dirty work of rolling the pigs thousands of times and calculating the probabilities. The probabilities of each roll work out about like this:

Dot 34.90%
No Dot 30.20%
Razorback 22.40%
Trotter 8.80%
Snouter 3.00%
Leaning Jowler 0.70%

With some simple multiplication, we can use these values to figure out the probabilities of each specific combination of 2 pigs. The following table is a helpful summary (with small decimal values truncated in the interest of readability):

Dot No Dot Razorback Trotter Snouter Leaning Jowler
Dot 12.2% 10.5% 7.8% 3.1% 1.0% 0.2%
No Dot 10.5% 9.1% 6.8% 2.7% 0.9% 0.2%
Razorback 7.8% 6.8% 5.0% 2.0% 0.7% 0.2%
Trotter 3.1% 2.7% 2.0% 0.8% 0.3% 0.1%
Snouter 1.0% 0.9% 0.7% 0.3% 0.1% 0.0%
Leaning Jowler 0.2% 0.2% 0.2% 0.1% 0.0% 0.0%

So, we know all of the probabilities of each possible outcome... and we know all of the points we would get for each possible outcome. But there's an awful lot of data here... We can make it a bit more manageable by adding together all the probabilities for each potential point value. For example, there's a 10.5% chance of pigging out with the first pig having a dot... and a 10.5% chance of pigging out with the second pig having a dot... Let's just summarize and say there's a 21% chance of pigging out... Doing that math looks like this:

Pig Out! 21.080%
1 21.301%
5 40.622%
10 7.848%
15 2.783%
20 6.229%
25 0.042%
40 0.090%
60 0.005%

We're getting close to having our answer. We know the probabilities of every potential score every time we roll... The only missing piece is recognizing the point value of "pigging out." When you pig out, you lose the points in your hand. So, the point value of pigging out is just the negative of however many points you have in your hand. This lets us make a simple equation for the expected value of each roll:

Expected Value =

.21080 * (Current Hand * -1)
+ .21301 * 1
+ .40622 * 5
+ .07848 * 10
+ .02783 * 15
+ .06229 * 20
+.00042 * 25
+ .00090 * 40
+.00005 * 60

When the expected value of the roll is greater than 0, we should roll. When the expected value is less than 0, we shouldn't roll. So, we just need to solve the equation for the value of Current Hand that will set Expected Value to 0... And Wolfram Alpha tells us that the answer is 22.5.

The Strategy

It's simple, then. If our current hand has less than 22.5 points in it, we should roll - on average, we'll gain points. If our current hand has more than 22.5 points in it, we should "pass the pigs" - the dangers of rolling outweigh the potential benefits. Now we know how to maximize our score on any given turn!

But, ultimately, this really isn't an ideal strategy. Consider this scenario: It's the beginning of your turn... you have 0 points in your hand, and 0 points in your score. Your opponent has 99 points in his score. You roll a few times and get your hand up to 25 points. The "stop after 22.5" rule says that you should stop rolling now! The expected benefit of rolling has now become negative! But should you really stop? Of course not... if you stop at 25, your opponent will almost certainly (79% chance) get a point of some kind and win the game on his next turn!

This means that the ideal strategy for maximizing your score on a particular turn is to stop at 22.5... but we're not interested in maximizing our score on a particular turn. We're interested in winning the game! So, what do we do? More on that later.

2 Responses

  1. Kayla J McLaughlin April 3, 2016 / 11:15 am

    Hi Dayne, why do you subtract one from the first equation in the WolframAlpha?

    • daynebatten April 4, 2016 / 10:15 am

      Good question. I’m actually not subtracting one, I’m multiplying by -1. The reason for this is that if you pig out you lose all the points currently in your hand (in other words, your current hand * -1).

      Hope that helps!

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