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.