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Two players (call them C and D), decide to play a card game using a well shuffled standard 52 card deck. Player C draws 7 cards from the deck and D draws 4 cards from the deck. Of the remaining cards in the deck, 1 card at a time is drawn and "shared" by both C and D. A winner (for the hand) is declared if either C matches 3 of his original 7 cards or D matches 2 of his original 4 cards. A match is defined as a same rank card that has not already been matched for that same hand and for that same player. Once an original card is matched, it cannot be "rematched" in that same hand. For example, if C has cards 3 4 5 6 7 8 9 and draws a 3 that is a match but if C draws another 3 in that same hand, it is NOT a match (but the first matching 3 remains a match). Also, a single card can be a "double" match (for both C and D) such as if C and D both hold an unmatched rank 3 card and a "community" rank 3 card is drawn. That counts as a match for both C and D.

Some things to note are some original hands can never be winners. Such as if C gets 333 7777. There are only 2 unique ranks to match so 3 ranks cannot be matched. Also if D gets 4444 for example, that is only 1 rank so 2 ranks cannot be matched. Pairs, triples, and even quads in the original hand make it more difficult (or impossible) to be a winner such as if C draws 3 5 7 3 5 7 3. In that case, the only way for C to win would be to draw another 3 5 and 7 before D wins the hand but that is "hard" cuz there are at most 2 of each of those ranks remaining in the pile of undrawn cards. Also, in the rare cases where C and D draw identical "overlapping" ranks, some interesting things can happen. Suppose C draws 4 5 7 4 5 7 7 and D draws 4 4 5 6. Here, all the 4s are used up already so they cannot be matched in either hand, but C cannot win cuz there are only 2 more unique ranks to match and C must match 3 to win. This case will be a win for D as soon as a 5 and 6 appear in the drawn community (shared) cards.

Lastly, ties are possible and end the hand just like a win would except no score is awarded for that hand since it is not a win for either player.

Note I changed this problem slightly from the original post because I realized I had an "error" since there were 41 shared cards and alternating one card each would give one player an "extra" chance to match so I changed it to community (shared) cards instead for the remaining 41.

So the question is who has the winning advantage in the longrun and by how much?

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  • $\begingroup$ So far, all I have tried is a few hands using a real deck of cards but eventually I will try a computer simulation. This problem seems to have some "compound" probability in that it may be first required to determine what the chances are of getting unique ranks vs. repeated ranks in the original hand (such as 3 4 5 6 7 8 9 vs 3 3 5 6 7 7 9). Then from there, the probability of matching those ranks, then the probability of doing so before the opponent wins. It seems rather difficult to do this "on paper" mathematically but perhaps someone can shed some insight on how to set this up. Thanks. $\endgroup$
    – David
    Commented Apr 17, 2017 at 13:24
  • $\begingroup$ Just to be clear, if say C's hand contains a pair of 3's, and the other two 3's are somewhere in the remaining deck, is C allowed to match both 3's when they're drawn (one with the first and the other with the second), or just the first drawn 3? $\endgroup$
    – Barry Cipra
    Commented Apr 18, 2017 at 13:20
  • $\begingroup$ @Barry Cipra - C and D try to match their original unique ranks so if C gets 3 3 4 5 6 7 8 as the original 7 cards, C is then trying to match the 6 unique ranks (3,4,5,6,7,8) and a match can only occur once per rank per player per hand so if another 3 is drawn from the deck, it will match C once but if the 4th 3 is drawn, that is NOT a match for C because rank 3 was already matched in that hand. C can have anywhere from 2 to 7 unique ranks to match per hand with the average being somewhere around 6. $\endgroup$
    – David
    Commented Apr 18, 2017 at 13:23
  • $\begingroup$ Thank you for the clarification. In other words, each player should simply disregard (e.g., discard) any duplicates in his hand. I'm picturing play proceeding with each player then "laying out" a card from his (remaining) hand in a given round if it matches the card that's drawn from the deck, with the game ending as soon as C has laid out three cards or D has laid out two cards. Interesting that your simulation indicates it's close to a fair game. $\endgroup$
    – Barry Cipra
    Commented Apr 18, 2017 at 13:44
  • $\begingroup$ @BarryCipra - When I first tried this game with a real deck of cards, I would first lay out the original cards and if I got any multiples of the same rank (such as a pair of 7s), I would just stack them as if they were only one card. So suppose player C got original cards 3 5 7 7 A K 10. What I would see would be 3 5 7 A K 10 (only 6 cards to match). Then as I drew community cards, if there was a match for C or D, I would just flip over the matching rank and then I could easily see if C had 3 matches or D had 2 matches. It is interesting how even it is. $\endgroup$
    – David
    Commented Apr 18, 2017 at 13:49

1 Answer 1

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I have a simulation program. It gives the first 7 cards to C and the next 4 cards to D then it draws community (shared) cards until there is a winner (or tie), or all of the cards in the deck are drawn. For $250$ million hands I got the following results:

$113,477,236$ : C wins requiring on average $7.3062$ additional cards (counting C wins only).
$116,893,384$ : D wins requiring on average $6.1375$ additional cards (counting D wins only).
$~~19,626,698$ : CD ties requiring on average $9.2320$ additional cards (counting ties only).
$~~~~~~~~~~~~2,682$ : no wins (ran out of cards before either player could win or tie).

So it appears D has a slight advantage in the longrun. I am a little surprised at how often ties occur, nearly 8% of the time. That would require each player having a same rank card (such as 9), and then that shared card showing up as the 3rd match for C and the 2nd match for D at the same time.

Also, a real example my computer caught is the following:

C: (sorted) $2,6,6,8,8,8,9$
D: (sorted): $8,9,9,9$

Here, neither player can win since all the 8s and all the 9s are used up. No wins are rare (about $1$ out of every $100,000$ hands).

One thing that puzzles me somewhat is if D needs 1 less card to win on average, why then does that player appear to only have a slight advantage overall? Also, f.y.i, when I modded the simulation program to instead use alternating additional cards starting with player C, instead of community (shared cards), I was hoping to make the game more even steven since player C would be about half a card "ahead" of D (on average). However, when I did this, C is a favorite $52.1$% to D's $47.9$% (excluding ties and no wins).

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  • $\begingroup$ How many of the ties correspond to games that neither player is capable of winning (in my notation, with $c\le2$ and $d\le1$)? $\endgroup$
    – Barry Cipra
    Commented Apr 18, 2017 at 14:30
  • $\begingroup$ I call a game a "tie" if neither player wins, as happens often in tic-tac-toe and occasionally in chess. $\endgroup$
    – Barry Cipra
    Commented Apr 18, 2017 at 15:00
  • $\begingroup$ I will be running a simulation overnight to get the results from 1/4 billion hands and then I will post them. I was hoping someone else could also do a simulation to help confirm my numbers for correctness. $\endgroup$
    – David
    Commented Apr 19, 2017 at 0:46
  • $\begingroup$ Just as a "side" comment, I wonder if instead of using community cards so that ties are possible, if (after the original 11 cards are dealt) I alternated giving cards to C then D but always starting with C so that C has the first chance to win. Since the average # of additional cards needed to win is about 1 more for C, this should (in theory), make the game even more "even Steven". It seems like the slight advantage D currently has will be lessened. I could mod my simulation to check this but for now, I will leave the problem as is since someone may be close to solving it. $\endgroup$
    – David
    Commented Apr 20, 2017 at 4:18
  • $\begingroup$ I had an error in the # of additional cards. I was counting the original 11 cards dealt (7 for C and 4 for D). Also, for clarification, the # of additional cards applies only when that player actually wins. So for example, when C wins, on average, about 7.31 additional cards need to be dealt, in addition to the 11 original cards, meaning the win occurs at 18.31 cards dealt from the deck total. This is somewhat surprising and even more surprising is D only needs about 6.14 additional cards to win on average. In other words, the win appears quite quickly after the original 11 cards are dealt. $\endgroup$
    – David
    Commented Apr 20, 2017 at 13:27

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