At a gambling joint, you are invited to be a player in a game with identical unfair but consistent coins flipped simultaneously in each round, with no cost to enter. Each player gets a different number of coins to flip.
In the first set of games, there are 7 players, having 2, 3, 4, 5, 6, 7, or 8 coins each. In each round, one player (the highest-scoring one) gets a payoff equal to their score, and others get nothing. The score of a player with n coins who gets k heads is k/(n+1-p), where p = 5/17 is the probability of a head appearing. There are many rounds per set of games.
Example (with Cn referring to player n with n coins): Suppose C3 flips 3 coins and gets 1 head, scoring 1/(3+1-p) ~ 0.2698; that C4 flips 4 coins and gets 3 heads, scoring 3/(4+1-p) = 0.6375; and C8 flips 8 coins, gets 5 heads, and scores 5/(8+1-p) ~ 0.5743. If everyone else scores less, C4 receives $0.6375.
Given a choice in this first set of games, how many coins will you use? If that first choice is gone, what's your second choice?
In the second set of games, instead of 7 players with 2, 3, 4, 5, 6, 7, or 8 coins each, there are only 5 players, having 3, 4, 5, 6, or 7 coins each. Again, if given the choice, how many coins will you use, and if that choice is gone, what's your second choice?
Addendum: Now that I've accepted Florian F's correct answer to the question as I asked it, here's the question I meant to ask (but unfortunately mixed up head and tail probabilities in original post). This has a more interesting result, that can be separately answered as seen below. Question: With p = 12/17 as the probability of a head appearing, and score being k/(n+p) for k heads on n coins, what are your first and second choices in the 7-player and 5-player cases?