In the game Who Goes There, you are fighting for survival with other players (up to 6 can play) and fending off The Thing, an alien who infects and attacks your camp.

To be infected, you take an action, resulting in you drawing 1 Vulnerable card from a deck of 12 Vulnerable cards. Only 1 of these cards is infected. The other 11 are clean. So that's

   12 cards
 1/12 Infected
11/12 Clean

What I'd like to know are the various probabilities as more players draw Vulnerable cards. And I'd like to learn the best methodology for figuring these out.

So assuming a 4 player game, with players A, B, C, and D taking turns in that order, there are a few scenarios I'm interested in

  1. Not a scenario, but does turn order affect probability here?
  2. Each player takes 1 card in order until all are gone. Who is most likely to be infected?
  3. Only A and B take 1 card in order until all are gone. Who is most likely to be infected?
  4. A and B take 1 card in order until 2 are left. C takes 1, then D takes 1. Who is most likely to be infected?
  5. Each player takes 1 card in order until 4 are left. Those 4 remain. Who is most likely to be infected or will no one be infected?

For those wondering, Who Goes There is a game based on the book, which was turned into movie The Thing. See more on the game here


This kind of probability theory amounts to careful counting. Here's a key idea that will help you answer several of these kinds of questions: note that you could do all the dealing you're ever going to do immediately, but you could deal everything face down. Think of it like everyone has their cards, and then one of them will be "chosen" randomly after the fact to be the infected card.

The symmetry of this should convince you that the answer to (1) is no; the order of dealing is irrelevant, and all that matters is the total number of cards each player receives. The key idea for the rest is that infection rates are proportional to the number of cards in each player's possession.

  1. Every player ends up with 3 cards in this scenario. All 12 of the cards are equally likely to be infected. Thus, everyone has the same probability, 1/4, of infection.
  2. Similar to #2; I'll leave the answer as an exercise.
  3. Players A and B hold five of the 12 cards, so the chance that one of their cards is selected to be the infected card is $\frac{5}{12}$. Similarly, for players C and D the infection chance if $\frac{1}{12}$.
  4. Construct a fifth pile of cards in the center of the table for the ones that don't get drawn; four will be left here. Hence, the probability that nobody is infected is $\frac{4}{12}$; I'll leave the others as an exercise.

Alternatively to Aaron's answer, you can explicitly calculate the probabilities and check if there's a pattern. Let $P_n(L)$ be the probability that on the $n$-th turn, the player drawing loses. Then
$$P_1(L)=1/12\ \text{and}\ P_2(L)=\frac{11}{12}\cdot\frac{1}{11}=1/12$$ where $11/12$ is by the fact that on the 2nd turn, one of the 11 noninfected cards out of 12 must have been drawn on the first turn.

You can confirm that $P_n(L)=1/12$ for $n=1,\dotsc,12$, i.e. the probability that on any of the 12 turns, the player drawing loses with 1/12 chance. Then the probability of any player $i$ losing is $(\text{# of Player $i$'s turns})\frac{1}{12}$.

For the fifth scenario, the probability that the infected card is amongst the first eight cards is 8/12. And similar reasoning shows $P_n(L)=\frac{1}{8}$ for $n=1,\dotsc,8$. The probability of any player $i$ losing is now $\frac{8}{12}(\text{# of Player $i$'s turns})\frac{1}{8}$.

  • $\begingroup$ So this makes sense, but this assumes no knowledge, right? If I know I the first card was clean, then the next player to draw ends up with a 1/11 chance. So if you assume each card was clean previously, then the next player to go should always have a higher chance of being interested, right? $\endgroup$ – merlinpatt Nov 30 '18 at 23:52
  • $\begingroup$ You are confusing the game and its "subgames". If it is revealed that the first card is clean and we are analyzing the "subgame" consisting of the remaining 11 cards, then sure the next player has an 1/11 chance of losing. But this 1/11 chance is interpreted in the "subgame" and not the original game with all 12 cards. $\endgroup$ – palmpo Dec 1 '18 at 4:10
  • $\begingroup$ If we go with your thinking, assume there are 2 draws left (the previous draws are clean). Suppose the 2nd to last draw is clean. Then the unfortunate last player to draw would have a 100% chance of losing the game! Clearly this 100% chance is interpreted in the subgame of only a single infected card, and not the entire game. $\endgroup$ – palmpo Dec 1 '18 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.