# GeekDad Puzzle of the Week – Black Eye Odds

Whenever my wife’s parents come down from Canada, or when we go up there to visit, we usually wind up playing a card game called Black Eye. A game consists of fifteen hands, where players gets 7 cards the first hand, 6 cards the next, then 5, 4, etc., until the seventh hand where players get 1 card. For the eighth hand, players get 2 cards, then 3, 4, 5, etc., until the fifteenth and final hand where players are back up to 7 cards. After the cards are dealt, the very next card is turned up and its suit is considered “trump.”

To play each hand, players look at the hand they were dealt and try to properly guess how many of the “tricks” they can win that hand. Each player declares their bid simultaneously in a manner similar to “paper-scissors-rock” and all bids are recorded. The hand is played according to typical Pitch rules (you have to follow the suit initially lead, the trump suit beats all others, etc.), and when the hand is complete, each player counts the number of tricks they won. If that number matches their declared bid, they win 10 points on the scoresheet for making their bid, plus the number they bid. On the score sheet, you simply place a “1” in front of their recorded bid, so a bid of 4 goes to 14 points, 2 to 12 points, and a bid of 0 goes to 10 points. If they get additional or fewer tricks than hoped, they not only don’t win the 10 points, but their bid as previously recorded gets circled out, giving that player a “black eye” (and zero points) for the hand.

This week’s puzzle is to determine the rates or odds of each player winning a game of Black Eye, given their particular playing styles as defined below. Additionally, determine the odds that Allison will beat me, beat her Dad Wilf, or beat both of us during a given game.

Here’s each player’s strategy, based upon the review of several past game’s scoresheets:

Wilf: Wilf is a bit of a sandbagger. A full 2/3 of the time, he bids “zero” and makes these hands 4/5 of the time. He bids “one” 1/6 of the time (making these 1/2 of the time) and two less than the number of cards dealt the remainder of the time (subject to a floor value of zero, obviously.) In this last set of hands he only succeeds 1/4 of the time.

Linda: Linda is an aggressive player. Half of the time, she bids half of the number of cards dealt (rounding down) and has a 60% chance of making it. One quarter of the time she bids zero, and succeeds 1/2 of the time. The remainder of the time she bids 2 on odd numbered hands and 1 on even numbered hands, making these 40% of the time.

Judd: I tend to play in a balanced manner. I bid 1 on odd numbered hands and 2 on even numbered hands, adding an additional point on hands with more than 5 cards (again, all subject to the maximum number of cards dealt.) I make these 3/4 of the time.

Allison: Allison plays to have fun. For the first half of the game, she bids either one (even hands) or zero (odd hands) and makes these bids 1/2 of the time. For the second half of the game, she bids either two (odd hands) or three (even hands) and makes these 45% of the time. As with all other players, she can only bid between zero and the actual number of cards dealt in the hand.

As always, please submit your responses to GeekDad Central. A complete entry will list the odds/probabilities of Allison beating me, beating her Dad Wilf, beating both of us, and the odds of each player winning outright. All correct (or at least reasonably supported) responses will be entered into a random drawing for this week’s prize, a \$50 ThinkGeek Gift Certificate.

Good luck, and happy puzzling.