![]() ![]() Public static boolean containsFourOfAKind(int hand)įor(int index = 1 index < hand. If(pairCheck = 1 & startThreeKindCheck = 3) ![]() Public static boolean containsFullHouse(int hand) Public static boolean containsStraight(int hand) In a game of Seven Card Stud Hi/Lo, the value of 8 or higher is a requirement to win for the low hand. However, in the event that neither hand meets the requirements for low, the best hand for high will win the pot. Public static boolean containsThreeOfAKind(int hand) The outcome of a 7 Card Stud poker game ends with a 50 split of the pot for both players. Public static boolean containsTwoPair(int hand) Public static boolean containsPair(int hand)įor(int index = 1 index < hand.length index++) Int startScan, index, minIndex, minValue įor(startScan = 0 startScan < (hand.length-1) startScan++)įor(index = startScan + 1 index add extra damage to these multi-damage events, and being able to trigger both at once can lead to some incredibly powerful attacks. ("Enter five numeric cards, 2-9, no face cards please") įor(int index = 0 index < hand.length index++) The next, and obvious, cards to add are Warrior Skill and Honed Technique. SortHand(hand) //Sort hand in ascending order Same for entering 1, 1,1, 1, 4, it identifies as three of kind instead of 4.Īny suggestions to my code? public static void main(String args) I am having issues though where its not identifying three or four of a kind.Įxample, if I enter 1, 3, 2, 1, 1, it identifies it as TWO PAIRS instead of Three of a Kind. I set up my program to prompt the user and then go through several if else statements calling methods. ![]() I currently have my program set as follows, first prompting the user for 5 cards, 2-9, then sorting the cards in ascending order. Poker hands are categorized according to the following labels: Straight flush, four of a kind, full house, flush, straight, three of a kind, two pairs, pair, high card. I tried testing this out by comparing the probabilities to a Mississippi Stud game (5 card poker) and getting the EV of this scenario, but the EV did not match up.Prompt: Write a program that reads five cards from the user, then analyzes the cards and prints out the category of hand that they represent. Any feedback would be greatly appreciated. I am not sure if this is correct, or if I need to subtract the Full House occurrences from this. I add both binomial sets together to get 792 out of the 3 card flop with 50 remaining cards (52 minus 7 and 2) to get 4.0408%. 2 Choose 1 for one of the two existing ranks (7 or 2) and any of the 3 remaining suits of this rank therefore 3 Choose 1. 11 Choose 1 for any card not 7 or 2, 4 Choose 2 because it must be two suits of that same rank.
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