3

If you're playing a 3 player game and hand out 1 card to each player, out of a deck of 3 ally and 1 traitor cards, how likely are the players all allies?

I think the math is you have a (3/4)^3 or approximately %32 chance of everybody being an ally. Is this correct? Thanks.

  • 2
    I presume you know this, but in a three player game you would be dealing from a set of 7 cards (2 non-traitor per player, plus 1 traitor) – Omertron Jul 6 '16 at 10:35
  • @Omertron unless you play Betrayer variant: During setup players may choose to add only 1 non-betrayal secret objective per player rather than 2, greatly increasing the odds of a betrayer. – Viktor Mellgren Jul 4 at 16:10
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Deal out the four cards in a line. How likely is it that the one on the far right is the traitor card? 1 in 4 right? Now have the players pick up everything except the far right card. What are the chances that they didn't pick up the traitor card? It's the same question. So your answer is still 1 in 4.

6

The statistic theory presented by the question is best described Hypergeometric Distribution. During the game set up, 2 Non-betrayal objectives per player, and 1 Betrayal objective, are combined in the opening deck. From that 7 card deck, each player will draw 1 objective, 3 total. Your goal is that 3 non-betrayal objectives are drawn.

N = number of items in the population = number of cards in the deck = 7
k: The number of items in the population that are classified as successes. = number of non-betrayal objectives = 6
n: The number of items in the sample = 3
x: The number of items in the sample that are classified as successes. = number of non betrayal objectives drawn = 3

h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]
where [aCb] = a(a - 1)(a - 2) ... (a - b + 1)/b! = a! / b!(a - b)!

There is a 57.1428571428571% probability the betrayal objective isn't drawn, or that 3 non-betrayal objectives are drawn.

Your question stipulates that the starting deck is only 4 objectivess, 3 non-betrayal and 1 betrayal.

N = number of items in the population = number of cards in the deck = 4
k: The number of items in the population that are classified as successes. = number of non-betrayal objectives = 3
n: The number of items in the sample = 3
x: The number of items in the sample that are classified as successes. = number of non betrayal objectives drawn = 3

There is a 25% probability the betrayal objective isn't drawn, or that 3 non-betrayal objectives are drawn. This is equivalent to the prospect of choosing one card from the Deck of 4, and letting the players select what remains.

3

You math is slightly off because you aren't accounting for cards being removed from the deck. In order for all three to be allies:

  1. Player 1 gets the first card, where the ratio is 3:1. This means they have a .75 probability of drawing an ally.
  2. Player 2 gets the second card, but the ratio has been reduced to 2:1 since player 1 removed an ally. Player 2 has a .67 (rounded) probability of drawing an ally.
  3. Player 3 gets the third card, but the ratio is now 1:1, giving a .5 probability of drawing an ally.

If we multiple those together (3/4 * 2/3 * 1/2) we get a total probability of .25 that all three players will be allies, depending on how you want to round the 2:1 ratio.

  • You are doing way too much math to get to the answer backwards – Andrey Oct 6 '16 at 13:51

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