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I like probability questions, and it has been a while since tttpp's excellent analysis of losing in the first turn of Forbidden Island. I figured it might be interesting to calculate how likely you are to lose on the first turn in Pandemic. This was inspired by JoeGolton's question about impossible setups, and tttppp's answer referencing a BBG session report. This question is only interested in figuring the odds of losing the game during the first players turn. To figure out the odds of losing, you will need to the following:

The rules for Pandemic.

  • Calculate how likely you are to lose by running out of 24 cubes of a single color.

  • Calculate how likely you are to lose by Outbreaking 8 times.

  • Calculate how likely you are to draw an Epidemic cards on the first turn.

  • Calculate the odds of being able to remove 0 or more cubes from possible chain outbreak cities.

  • Calculate the odds of no player receiving a Special Event (linked all 5) card that will prevent a loss.

  • Determine the effect of each difficulty level (4-6 Epidemic cards in evenly divided piles), and for 2-4 Players on the result.

Assume the following:

  • The First Player character is randomly determined.

  • The other players out of the 5 roles are randomly determined.

  • Assume players do not know which Infection cards will be drawn (unless a Special Action card allows it)

  • Standard Rules and Characters (none from On the Brink)
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This happened to us once. It was epic! The sheer carnage was fun to watch. –  Pat Ludwig Sep 28 '12 at 23:34
    
You can't draw two epidemics the first turn. –  Matt Oct 1 '12 at 11:38
    
@Matt, quite right. I was thinking about two epidemics being back to back, but that only.occurs when the top epidemic is at the bottom if It's stack, and the other epidemic is at the top of its stack. –  user1873 Oct 1 '12 at 12:42
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1 Answer 1

The hardest part of computing the probability is determining the best strategy for the players. I have used a script to simulate games where players use a few simple tactics, and estimated the probability based on these strategies. Given that the strategies I implemented are not the optimal strategy, these figures provide an upper bound on the probability of a first turn loss.

Most of this answer assumes that the players are playing on "Normal" difficulty, that is with five epidemic cards. To find equivalent probabilities for other difficulties, see the section below entitled "The player deck".

To summarise the results, here is a table showing my estimate of the probability of a first turn loss for different strategies.

Strategy        | Probability of first turn loss at Normal difficulty
Do Nothing      | 0.0018
Drive/Treat     | 0.0013
Drive/Fly/Treat | 0.0011-0.0012

The probability of losing in the first turn, given that the players do their best to prevent it, is therefore less than 1/900.

The Player Deck

To set up the player deck, some cards are dealt to the players and then epidemic cards are distributed throughout the deck. The number of cards dealt to the players, d, varies with the number of players. We either have d = 8 in 2 or 4 player games, or d = 9 in 3 player games. The remaining (53-d) cards are split into e roughly equal piles, and an epidemic is shuffled into each, before they are restacked (with any larger piles on top).

We observe that to lose in the first turn an epidemic must be drawn. The probability of drawing an epidemic as one of the top two cards of the deck is:

P(Epidemic drawn) = 2 / (ceil((53-d)/e)+1)

where ceil is the ceiling function. This results in the following table:

d | e | P(Epidemic drawn)
8 | 4 | 2/13 = 0.154
8 | 5 | 2/10 = 0.200
8 | 6 | 2/9  = 0.222
9 | 4 | 2/12 = 0.167
9 | 5 | 2/10 = 0.200
9 | 6 | 2/9  = 0.222

During the rest of the discussion we assume e is 5 ("Normal" difficulty), but the above table can be used to compute equivalent probabilities for other values of e and d.

The "Do Nothing" Strategy

By far the simplest strategy is for the players to do nothing. This is also a strategy that maximises the chance of losing in the first turn.

For the calculation I employed a python script to simulate a million games, assuming that an epidemic card is drawn. It picks nine cities to infect for the set up, and then one more to infect for the epidemic. It repicks two cities to re-infect, and counts the outbreaks and cubes used.

This gave the probability of losing, given that an epidemic is drawn is about 0.009245.

Combining this with the probability of drawing an epidemic card we get:

P(Lose first turn on Normal | Players do nothing) ~= 0.001849

The "Drive/Treat" Strategy

As a simple improvement on doing nothing, I assumed the first player would ignore their cards, and drive to the nearest city with three cubes, removing as many as possible. If they couldn't remove any from three cube cities, they would try for the nearest two cube city, and finally they would try for the nearest one cube city.

This gave the following probability:

P(Lose first turn on Normal | Using "Drive/Treat" Strategy) ~= 0.001343

The "Drive/Fly/Treat" Strategy

Finally, as an improvement on the "Drive/Treat" strategy I considered trying to fly to a city infected with three cubes. If anyone has the "Airlift" or "Government Grant" special event cards, then they can be used to get the first player to any city in 0 or 1 actions respectively, meaning that three cubes can be treated. Similarly if any other player has the Atlanta card it can be passed, and the first player can charter a flight, leaving two actions to remove cubes.

I also considered the case where the player has the card of a 3-cube city, or Atlanta, and they use that to fly and remove three cubes from a city.

There are further variations on this that I did not consider in this tactic, such as moving to Washington, chartering a flight and treating two cubes.

Using this strategy gave the following probabilities:

P(Lose first turn on Normal | 4 players using "Drive/Fly/Treat" Strategy) ~= 0.001208
P(Lose first turn on Normal | 3 players using "Drive/Fly/Treat" Strategy) ~= 0.001145
P(Lose first turn on Normal | 2 players using "Drive/Fly/Treat" Strategy) ~= 0.001167

I don't think my simulation is evidence enough to say that the deviation between these three results is significant, but I think they're good enough to give a first-turn-lose probability between 0.0011 and 0.0012.

Potential Improvements

If you have played the game then I'm sure you can think of situations in which these strategies are not optimal. Here are a few things that could be included, that I think would make a significant improvement to the strategies:

  • Treat more than one city if possible.
  • Target infected cities that are next to each other in preference to isolated ones.
  • Use "One Quiet Night" or "Forecast" to prevent the first turn loss (this is a big one - it's possible about one game in three).
  • Use "Resilient Population" to lower the chance of a first turn loss.
  • Take the researcher into account for passing cards.
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Very cool! One nit: I don't think you meant the researcher - they can only give cards on their turn, so it's not useful to the first player. The epidemiologist is the one (from the expansion) who can take a card from someone else. There are also other event cards which would really help (again from the expansion) - Remote Treatment and Mobile Hospital come to mind. I also wonder: surely in these losing cases there are chains of outbreaks; might the best strategy be to treat a cube in each of two adjacent cities, rather than fully treating one? –  Jefromi Oct 4 '12 at 20:47
    
Jefromi: The researcher's ability works on other players turns too - see the bottom of page 5 in the rules (linked in the question). –  Timothy Jones Oct 5 '12 at 1:36
    
I think targeting cities with 2 or 3 cubes that are separated by 0 or 1 cities would dramatically reduce the probability of a first turn loss, since a first turn loss requires a large number of cascading outbreaks. –  bwarner Oct 5 '12 at 13:35
    
@bwarner Jefromi - I agree - trying to stop the chain reactions would definitely be an improvement on the above strategies. I'll have a look at simulating it when I get a chance. –  tttppp Oct 5 '12 at 16:36
    
@TimothyJones Oops, you're right. I couldn't remember for sure, so I checked the card, which says "you may give a player cards from your hand for 1 action per card" and interpreted that as meaning it had to be your turn, because that's the only time you have actions. (Odd that the epidemiologist doesn't parallel that.) The full rules are way clearer. –  Jefromi Oct 5 '12 at 23:36
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