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Alice and Bob are going to have a poker night. They have invented a variant where there are no secrets, and no randomness, called Face-Up Poker. Here's how it works.

A standard 52 card deck is spread face up on a table.

  1. Alice picks up any 5 cards from the table.
  2. Bob does the same.
  3. Alice discards any number of cards, throwing them on the floor. She then picks up the same number of cards from the table.
  4. Bob does the same (Bob can't pick up cards that Alice discarded).

They then compare hands, and the best hand wins, with Bob winning ties.

Which player can force a win? What is their winning strategy?

This puzzle assumes familiarity with the rankings of poker hands, but no other knowledge of the rules of poker. For a refresher these rankings, see this helpful page (the hands are listed worst to best from top to bottom).

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    $\begingroup$ I would normally read any number as including zero. Which would make the problem trivial. One or more might be a better formulation. $\endgroup$
    – Taemyr
    Jun 3, 2015 at 11:18
  • $\begingroup$ @Taemyr Any number does include zero. Why does this make the problem trivial? $\endgroup$ Jun 3, 2015 at 15:34
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    $\begingroup$ It makes the problem trivial because both players pick a royal flush, and then always select zero cards. Game never ends. $\endgroup$
    – corsiKa
    Jun 3, 2015 at 15:54
  • $\begingroup$ @corsiKa It would eventually end: there are only 52 cards in a deck. $\endgroup$
    – Zibbobz
    Jun 3, 2015 at 17:03
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    $\begingroup$ @Zibbobz If they could pick zero cards, the game never ends because only 10 cards ever gets picked. $\endgroup$
    – corsiKa
    Jun 3, 2015 at 17:28

3 Answers 3

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There is a winning strategy for

Alice. For her starting hand, she will select all four 10's, along with another lower card (let's say the 2 of clubs for concreteness). This will make it impossible for Bob to end up with a royal flush. Bob needs to prevent Alice from making a royal flush when she discards, so he must select a starting hand containing a card of rank higher than 10 in each of the four suits. This means that for at least three of the four suits, Bob's starting hand cannot contain a card of rank lower than 10 in that suit.

When it is Alice's turn to discard, she chooses one of the suits in which Bob did not take a card of rank lower than 10. She keeps the 10 of that suit, discards her other four cards, and takes the 6, 7, 8, and 9 of the chosen suit (Bob did not select a card of rank lower than 10 in the chosen suit). Alice's final hand will be a straight flush to the 10. Since all of the other 10's are now on the floor, Bob cannot possibly make a straight flush whose high card is 10 or above, so Bob's hand cannot possibly tie or beat Alice's.

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    $\begingroup$ We got the same answer! I guess that means we probably got it right. $\endgroup$
    – Bob
    Jun 3, 2015 at 1:23
  • $\begingroup$ @Bob Indeed. Now I'm curious: are there other winning strategies? $\endgroup$ Jun 3, 2015 at 1:35
  • $\begingroup$ I don't think there can be. Alice needs to force Bob's next move in way that leaves her open to make a winning hand. $\endgroup$
    – Bob
    Jun 3, 2015 at 1:41
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    $\begingroup$ It depends on what you mean by a different strategy. I claim >10S 10H 10D JC 9C also works, but is that different? $\endgroup$ Jun 3, 2015 at 1:42
  • $\begingroup$ @JS1: Bob can beat that with the three other tens and the Q9 of the suit you took the 10 in. If Alice prevents a straight flush, Bob can get four aces. $\endgroup$ Jun 3, 2015 at 1:43
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Alice goes first and picks...

4 tens (and any 1 other card) preventing Bob from making the highest ranked hand the royal flush (A,K,Q,J,10 of the same suit)

Bob must...

prevent her making a royal flush so he must pick 4 cards each of a different suit in the range Ace to Jack (eg 4 Aces) (and any 1 other card)

Alice can then...

keep one of the tens and pick 9,8,7,6 of the same suit making a straight flush with 10 as the highest card.

To equal this Bob must...

also make a straight flush with 10 as the highest card but Alice has already taken all the tens. Oh no Bob has lost.

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    $\begingroup$ This must have been hard for you to type up knowing you would lose $\endgroup$ Jun 3, 2015 at 1:51
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    $\begingroup$ Alice always seem to get the first turn. It not really fair there are so many games where first turn = auto win. $\endgroup$
    – Bob
    Jun 3, 2015 at 1:52
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    $\begingroup$ @Bob That's what you get for agreeing to a purely deterministic game. For "fun", maybe you can determine who will win the most hands if the game keeps going? $\endgroup$
    – Zibbobz
    Jun 3, 2015 at 17:02
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Out of character: I am choosing to interpret that "Bob wins ties" as not using the standard tie breakers. As in, if both hands are straight flushes, normally the higher card wins, as a course of 'breaking the tie'. But in this case no high card comparison is made and Bob wins a straight flush / straight flush matchup. Same with royal flush, straight, and flush. I don't know why, but in my mind it made sense that a four-of-a-kind, three-of-a-kind, pair, full house, etc still compare high card. Alice's 4 Aces still beats Bob's 4 Queens as if there never were a tie to begin with. This might not have been the intention, but custom rules make sense with the following flavor text:

Alice and Bob are playing Face-Up Poker. For the first game,

Alice chooses a Royal Flush and Bob chooses a Royal Flush. They both keep their hands.

According to the Poker Hand Tie Rules, they split the pot. Being Alice and Bob, it takes them 3 more games of exactly repeated cards to learn that neither of them are getting anywhere. They thus agree that high card or high suit shall determine the winner in the case of a tie. Following this new rule and the previous strategy, it takes Bob 5 more games for him to realize he's not getting anywhere, no matter which lower suit ace he chooses. The rules are thus amended, on threat of Bob playing "never ever again", whereupon high card and high suit are totally ignored and all ties yield Bob's winning. Play thus proceeds, with Alice realizing from past experience she must prevent a Royal Flush:

Alice takes 4 aces and a king, to prevent Bob's Royal Flush.
Bob must settle for a straight flush.
Using one of her aces, Alice is free to take a royal flush, and she wins.
4AK, SF -> RF, SF = Alice

Next game, Bob also realizes he must

thwart Alice's royal flush attempt.
When Alice takes 4 aces and a king, Bob takes a straight of a king, queen, jack, ten and nine, all of different suits.
Alice cannot go "straight" (pun intended) for a royal flush, but she naively thinks her 4 of a kind to be winning and so she keeps it.
Bob beats the four aces with a straight flush.
4AK, KQJ109 -> 4AK, SF = Bob

Realizing what she did, in the next game Alice

takes 4 aces and a king again.
Bob takes king through nine of different suits.
Then Alice goes "straight" for the straight flush with the highest card possible, a queen (even though this was irrelevant under the rules, she likes queens).
Bob also takes a straight flush, and wins the tie without high card comparison.
4AK, KQJ109 -> SF_Q, SF = Bob

On a losing streak, Alice cannot continue with her past method of preventing a royal flush, so for the next game Alice decides to use Bob's own tactic against him.

Alice takes ace through ten, all of different suits!
Bob cannot make a royal flush, so Bob takes a straight flush.
Alice is able to take a royal flush and wins, because Bob, having been unable to obtain a royal flush the first round, is still not magically able to make a royal flush.
AKQJ10, SF -> RF, SF = Alice

Just when Alice thinks she's finally turned the tables, she

takes ace through ten, all of different suits.
Bob cannot make a royal flush, but this time he does the same to Alice, taking a different set of ace through ten, all of different different suits.
Without the possibility of her own royal flush, Alice upgrades her straight to a straight flush using lower cards.
Bob also has a straight flush laid out in front of him somewhere on the table, and he wins the tie without high card comparison.
AKQJ10, AKQJ10 -> SF, SF = Bob

In the end


Alice makes several more attempts.
Try as Alice might, she cannot both block a royal flush and beat a straight flush from Bob.
Bob always wins either flat out or by virtue of the special rules established.

Having determined the answer, the Alice, Bob and Charlie (yes, Charlie) personalities retreat back into the subconscious, allowing Daniel to emerge and write this down.

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  • $\begingroup$ +1 - i don't think this is what OP intended, but it's a very interesting twist and a well written analysis. $\endgroup$
    – IanF1
    Jun 5, 2015 at 6:12

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