One Suit Question
Say we remove all the cards from a deck except a single suit, any suit.
With the remaining 13 cards, there are 13C5 = 1,287 possible 5 card hands.
If you ignore straights and look at card rank only, one hand will beat 643 hands and lose to the other 643 hands. What is that hand?
If we do not ignore straights, what hand will have zero equity?
8 Replies
Combinations of hands are as follows
A high 12C4 = 495
K high 11C4 = 330.
Q high 10C4 = 210
J high 9C4 = 126
T high 8C4 = 70
9 high 7C4 = 35
8 high 6C4 = 15
7 high 5C4 = 5
6 high 4C4 = 1
The desired hand therefore will be a king high hand that loses to 148 of the king high combos and beats 181. Extending the analysis there are 10C3 = 120 KQ hands, and 84 KJ combos. Our desired hand must be KJxxx, where we lose to 28 KJ combos and beat 55. KJT has 8C2 = 28 combos — precisely the number we must lose to, so our desired hand is KJ987, which loses to all KQxxx and KJTxx combos plus all ace high totaling 28 + 120 + 495 = 643.
As for the second question, it seems too simple so maybe I’m missing something, but wouldn’t it just be 76542?
Interesting result for the first question that seems counterintuitive to me. You could potentially exploit someone by setting a hand strength that seems pretty good like QJT98 and betting them that when you deal five cards from your deck, you will get a higher hand. If they would take even money, you would have a very profitable bet.
Combinations of hands are as follows
A high 12C4 = 495
K high 11C4 = 330.
Q high 10C4 = 210
J high 9C4 = 126
T high 8C4 = 70
9 high 7C4 = 35
8 high 6C4 = 15
7 high 5C4 = 5
6 high 4C4 = 1
The desired hand therefore will be a king high hand that loses to 148 of the king high combos and beats 181. Extend
Thanks for your well explained solution!! Makes perfect sense.
As for the second question, it seems too simple so maybe I’m missing something, but wouldn’t it just be 76542?
Sorry if I fumbled the terminology. When I said zero equity, I meant to imply the hand would win 50% and lose 50%, therefore breaking even and earning zero in the long run.
Working from your solution to the first problem, KJ987,...since there are ten straights A-5 thru T-A, the middle of the road hand will now need to beat 5 additional hands. Those hands you must beat would be KJTxx where xx is 32,42,43,52, and 53, so the answer is now KJT54... sound right?
Thanks again!
Thanks for your well explained solution!! Makes perfect sense.
Sorry if I fumbled the terminology. When I said zero equity, I meant to imply the hand would win 50% and lose 50%, therefore breaking even and earning zero in the long run.
Working from your solution to the first problem, KJ987,...since
Right idea, but not quite. There are indeed 10 straights but we already are counting AKQJT, A2345, and KQJT9 among the 643 losing combos. We only add 7 combos to our losers that weren’t there previously, giving 650. That means we must raise our desired in so that we beat 7 additional combos to get back to 643 losing combos. We started with KJ987, so we must add 7 KJTxx combos. By simple brute force and going smallest to largest these would have the xx equal to 32, 42, 43, 52, 53, 54, and 62. Since we must beat all these our new hand must be KJT63.
Right idea, but not quite. There are indeed 10 straights but we already are counting AKQJT, A2345, and KQJT9 among the 643 losing combos. We only add 7 combos to our losers that weren’t there previously, giving 650. That means we must raise our desired in so that we beat 7 additional combos to get b
Yes, I realized my double counting error after a nights sleep. I came back to correct myself but did not realize the other mistake I had made re: dividing new hands we must beat in two. Thank you for correcting me. Much Respect!
Maybe it's a silly point, but in the calculations above the card removal effect is not taken into account. So, if you have KJ987, you can not face any K high hand. And basically, any Ace high is the nuts. This unless different players use different decks to get their hand.
Maybe it's a silly point, but in the calculations above the card removal effect is not taken into account. So, if you have KJ987, you can not face any K high hand. And basically, any Ace high is the nuts. This unless different players use different decks to get their hand.
The original problem was looking for the media hand. That is, which hand beats exactly the same number of hands it loses to. The solutions given are for this problem. As you say, if you were actually comparing to a second hand, this median hand would not have a 50% chance of winning. The hand that gives a 50% winning chance against another hand is a different problem.
To be complete, suppose we deal our opponent a hand using a 13 card deck with all spades and ourselves a hand using a different 13 card deck with all the hearts. The solutions given above would then represent the hand that gives exactly a 50% chance of winning.
