# What is the worst possible hand that wins at showdown?

We all know that the worst (least likely to win) starting hand in Texas Hold'em is 7/2 offsuit. But suppose that all the players have made it through all rounds of betting and revealed their cards. What is the worst hand that can actually win?

With a little effort, I can devise the following hand. If the community cards are

``````2 4 6 8 10
``````

, player 1 has

``````3 7
``````

, and player 2 has

``````5 9
``````

with suits such that no one has a flush, then player 2 wins with 10 high. They win because of the kicker 9, of course, but the name of the hand's winning cards is 10 high.

Is this improvable? Is it possible to win a showdown with 9 high or even less? And what happens if there are three players at showdown, 4 players, n players — what is the worst hand that can win?

• As a starting point, the worst 5-card hand you can actually have would be 2,3,4,5,7; but it's impossible for that to be your best 5-card hand out of 7 cards. Jan 6 at 6:21

9-high is the worst possible hand that can still win, as long as you have 3 or fewer opponents to beat at showdown.

Throughout this answer I'm just assuming that flushes are not considered; it is very easy to come up with a set of 9 cards where you don't have any 5 of the same suit.

• Opponent hand: 2,5

• Community cards: 3,4,7,8,9

• Your best 5 for the showdown: 9,8,7,6,4

• Opponent's best 5 for the showdown: 9,8,7,5,4

You win with the 9-high, 6 kicker. This situation applies up to there being 4 people at showdown, since all 3 opponents could have the same hand.

Could your hand be any worse? No.

First off, 8-high is automatically an impossible hand to have. You have 7 available cards to make your hand, and if you cannot have any duplicates, then you'd need 7 distinct cards that are 8 or lower, and those only exist if the cards are 2-8. This means you'd actually have 4,5,6,7,8 for a straight.

So to look at possible worse hands of 9-high: Of the 7 cards available to you, there are only 8 unique numbers from 2-9 to pick. If you leave 2 of those numbers out, then you only have 6 unique numbers to fill 7 slots, so by the pigeonhole principle you'd have at least a pair. Thus there is a single number left out. If that number were 2-4, then you'd have a 5-9 straight. If that number were 7-9, then you'd have a 2-5 straight. Thus the missing number must be either 5 or 6. This only leaves you with 2 possible sets of 7 cards:

• 2,3,4,6,7,8,9
• 2,3,4,5,7,8,9

If you have the first set, that is the solution proposed above. If you have the second set, that is what your opponent had. There is no other worse hand that can be made for 9,8,7,5,4 to beat.

There are other variants where you can switch the 2 with any of the community cards and still end up with the same result. All that matters is that you have a 6 while your opponent has a 5, and that the other card you have is the same as your opponent's other card.

• It works with 4 opponents as well, if you have 5,6 and two opponents have 2,5 while two opponents have 2,6. Jan 6 at 21:29
• @Squala You can’t have both 5 and 6 without having a straight. Jan 7 at 0:20
• ah shoot, missed that. Jan 7 at 0:57
• How does the answer change as the number of players increases? Ideally I'd like to see a table of lowest winning hand for all possible numbers of players up to 23, the max possible players in a hand (23*2 + 5 = 51). Jan 7 at 2:36
• @TheWreckersCompanion the worst winning hand with 23 players is at least a pair because all shown cards will form a pair with someone. Mar 8 at 9:52