Suppose there is some situation where Player A moves all-in and ALL Players call such that all players are all-in (they have equal amounts of chips, so no side-pots). Is there a possible distribution of cards such that Player A is drawing dead BEFORE THE FLOP? And if so, how likely is it, and what is the lowest number of players for which this is still possible?

Stipulations:

No possible chance for Player A to win the pot or split it

All players who are dealt a hand will call (No cards can be folded)

No other player may have a hand identical to Player A (disregarding suit, of course); for example, if PLayer A has pocket twos, no player may have pocket twos except for Player A

The number of players may not be so high that there aren't enough cards to deal the flop, turn, and river (including burn cards)

One deck only

Texas Hold-em (Two cards per hand)

Obviously, the cards to come are not known (You can't just say "A misses his flush draw and loses to Aces; therefore, he's drawing dead")

Thanks!

• Define how many 'All players' is equaled to?
– Grinch91
Sep 9, 2018 at 22:17
• Progressive hints For this interesting puzzle. Use rot13.com to read it. 1. Rules clarification: (Ignoring suit,) Player A can't have the same hand as any other player, but B can have the same hand as C, e.g.. 2. Importantly, we require that Player A rot13(pnaabg rira gvr). 3. rot13(ubj pna bgure cynlref oybpx fgenvtugf naq syhfurf)? 4. rot13(pna lbh qb guvf zber rssvpvragyl? gur pbeerpg nafjre vf va gur fvatyr qvtvgf) 5. rot13(vs lbh trg rvtug (v.r. frira bguref), pna lbh oybpx fgenvtugf orggre?) 6. rot13(pna nabgure cynlre jva rira vs gur obneq vf n fgenvtug?) Oct 27, 2021 at 0:00

This is an interesting question. The key consideration to me is "Player A cannot win the pot via a split pot", meaning that there should be no scenario whereby the board contains the nuts.

This is essentially the error with the 2 answers above. IF the board comes A♠2♠3♠4♠5♠, then someone other than player A needs to be holding the 6♠, otherwise it is a split pot. same for straights on the board (i.e. 7♠8♣9♠ T♥J♥). In that scenario, someone other than player A needs to hold a Q. (or KQ if player A holds a Q).

edit: I thought it required 20, but I think I have a solution with 8.

Player 1: 4♠4♣

Player 2: 5♠5♣

Player 3: 5♦5♥

Player 4: T♥4♥

Player 5: T♦4♦

Player 6: T♠T♣

Player 7: A♦A♥

Player 8: A♠A♣

All TTTT5555 are in villians' hands, therefore there are no possible straight/straightflush options for the board (every straight will have to hold either a 5 or a T).

All Aces are in villians' hands, therefore it is not possible to splitpot with a 4 of a kind nuts on the board (i.e. KKKKA or any XXXXA combination).

Player A cannot draw to straight (no 5s), flush, set, or hope to play the board and split. Player A is drawing dead.

• This is the first correct answer so far; maybe not optimal, but it does avoid the 5-way splits. Sep 10, 2018 at 16:36
• It is actually the best answer. I cannot come up with a better one myself. As this one already contain the minimum number of players needed to prevent any potential ties that involved the first player (that player can't win nor tie). Sep 10, 2018 at 20:53
• You can do it with 7. I think 6 may be possible, maybe but haven't gotten anything for 6 yet. I can post the answer with 7 if people want, but I'll preface this with it's not my work, found it online to this very interesting situation.
– Grinch91
Sep 13, 2018 at 17:15
• @Grinch91 Add an answer and give the reference? Sep 14, 2018 at 6:13
• Interestingly, with ten people it's possible to have all but two players drawing dead if hands aren't required to be unique, and all but three drawing dead if they are (55/55/66/66/TT/TT/JJ/JJ/AA/AA, e.g. would work; at least three hands must be live if duplication isn't allowed because at least three hands would need to contain an ace to guard against a four-of-a-kind with an ace kicker). Oct 12, 2018 at 23:09

As I mentioned in a comment above, this was not my answer, found it here. Although I did find you could also substitute the 7s with 8s, and then the Queens with Jacks, 10s or 9s. Suits matter in the solution.

K♣K♠

A♣A♠

A♦K♥

A♥K♦

Q♣Q♠

7♦7♥

7♣7♠

Can check it out here.

• what a great answer. (I'm hoping this was done through thought and not computer trial-and-error). The 2x 7s are integral to the answer though (they block the 6-high straight flush on the board), which 8s do not block. Sep 18, 2018 at 6:25
• The logic could have been: while you need to block 4-board-card SFs every 5 cards (i.e. you need the Q♣Q♠ (or J/T/9) in between hero’s K♣K♠ and the 7♣7♠), you only need to block the 5-board-card SFs every 6 cards (i.e. nothing needed between the K♦, K♥ and the 7♦7♥). The main improvement of this solution over @sakon 's is that it uses the two cards K♥ and K♦ to block both the King-high 5-board-card SFs and the outs to three-of-a-kind. Although not relevant to the question, this solution has the added appeal that these hands might actually go all-in in a real-life game (no 10-4 suited). Nov 26, 2022 at 10:43