Two people (call them E and F), decide to play a special card game. They use a special deck of playing cards which is a standard $52$ card deck, but with $12$ extra cards added in. You can think of them as $3$ extra ranks, each having $4$ suits. So the new ranks (in ascending order) are $123456789TJQKABC$ (T = ten). Ranks $2$ thru $A$ are the normal ranks found in a standard deck of cards. $A$ is considered the rank just above king and never the same as $1$ for this game. We added a new low $1$ rank which comes before $2$, and we added $2$ new high ranks ($B$ and $C$) after the $A$ (Ace). So for each hand, we shuffle well and then deal out only half ($32$ out of $64$) of the cards. They are community cards shared between E and F. That is, both E and F use the same $32$ dealt cards to determine winners. Both E and F can win in the same hand multiple times.

E gets a win for each occurrence of a quad dealt in order, including multiple wins per hand such as $77772439999$ which would count as $2$ wins for E. Best case for E would be something like $11112222333344445555666677778888$ which would be $8$ wins.

F gets a win if any $5$ card straight is dealt in order (such as $23456$). Wraparound straights such as $ABC12$ are not a win for F. Only ascending straights are wins for F. For something like $234567$, that counts as a double win for F since there are $2$ overlapping straights ($23456$ and $34567$). Best case for F then would be $24$ wins in a single hand ($123456789TJQKABC123456789TJQKABC$). To make it even easier for F to win, we will assign a point multiplier. Each win (as described above) will have points awarded. E gets $8$ points for each win but F gets $15$ points for each win.

So best case for E would be $64$ points in a single hand but F can get up to $360$ points in a single hand.

If there is no winner in a hand, the cards are simply returned to the deck, reshuffled well, and a new hand is dealt.

So who has the point advantage long term and by how much?

  • $\begingroup$ With a small sample of $10^7$ games I got that E got $57528$ points and F got $57390$ points. @David I was calculating multiple wins for both players per hand (but the code didn't record how often that happened). $\endgroup$ – Ian Miller Aug 27 '16 at 14:35
  • $\begingroup$ Using techniques like in the first answer to this I get $9648226241311964275300295224260000000\approx10^{37}$ possible draws of 32 cards out of the deck. $\endgroup$ – Ian Miller Aug 27 '16 at 15:12
  • $\begingroup$ @Ian Miller - I am not even sure if that answer is right. The suits are not important for either player. Just the arrangement of the ranks. $\endgroup$ – David Aug 27 '16 at 15:15
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    $\begingroup$ @IanMiller How are you counting? The technique you linked to advises us to take the coefficient of $x^{32}$ from $(1+x+x^2/2+x^3/6+x^4/24)^{16}$ which is $86320499673449/165112971264$ and multiply with $32!$, giving $137563906395290913582482779401060000000\approx 10^{38}$ possible ways to deal 32 cards from the special deck. $\endgroup$ – Christian Sievers Sep 18 '16 at 12:37
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    $\begingroup$ Looking back at my code/calculations I took the coefficient of $x^{31}$ and multiplied it by $31!$ by mistake. $\endgroup$ – Ian Miller Sep 18 '16 at 15:24

Let's start with E. We number the cards in the hand from 1 to 32. Let $$X_i = \begin{cases} 1 &\text{if there are four of a kind starting at card i}\\ 0 &\text{otherwise} \end{cases}$$ for $i = 1,2, \dots ,29$. Then $$\Pr(X_i = 1) = \frac{3}{63} \cdot \frac{2}{62} \cdot \frac{1}{61}$$ and the expected number of "wins" per hand for E is $$E(\sum_{i=1}^{29} X_i) = \sum_{i=1}^{29} E(X_i) = 29 \cdot \frac{3}{63} \cdot \frac{2}{62} \cdot \frac{1}{61} \approx 0.0007303$$ So E's expected score is about $8 \times 0.0007303 \approx 0.005842$.

Now for F. Let $$Y_i = \begin{cases} 1 &\text{if there is a 5-card straight starting at card i} \\ 0 &\text{otherwise} \end{cases}$$ for $i = 1, 2, \dots ,28$. First, the card at i must be in the range from 1 to 12 in order for a straight to be possible; then the next 4 cards must line up appropriately. So $$\Pr(Y_i = 1) = \frac{12}{16} \cdot \frac{4}{63} \cdot \frac{4}{62} \cdot \frac{4}{61} \cdot \frac{4}{60}$$ and the expected number of "wins" per hand for F is $$E(\sum_{i=1}^{28} Y_i) = \sum_{i=1}^{28} E(Y_i) = 28 \cdot \frac{12}{16} \cdot \frac{4}{63} \cdot \frac{4}{62} \cdot \frac{4}{61} \cdot \frac{4}{60} \approx 0.0003760$$ So F's expected score is about $15 \times 0.0003760 \approx 0.005641$.

Therefore E's expected score is a little higher than F's (0.005842 vs. 0.005641); but the game must be boring to play, because most of the time no-one scores.

  • $\begingroup$ I have a problem with this answer for F. What about cases like 234567 counting as 2 wins? Where in your formula does it handle the case that after seeing something like 23456 then we only need to see an immediate 7? Also for F, how can you use 4 in the numerator 4 times when some (or all) of the cards that make up a 5 card straight could have appeared previously such as 65432A23456? Also, what about cases where the same straight appears multiple times in the hand such as 23456A23456? As the 2nd straight forms, we don't have 4 of each of those ranks remaining in the deck @awkward. $\endgroup$ – David Aug 28 '16 at 12:29
  • $\begingroup$ Also regarding this game being boring cuz wins are so infrequent, that can easily be solved by programming a computer to play out 1 hand at a time and stop when there is a winner. That would be instantaneous on a computer. The stopping could be so that they have time to make a wager (bet). I am running a simulation now of 100 million hands and then I will run 1 billion to see what I get. @awkward $\endgroup$ – David Aug 28 '16 at 13:31
  • $\begingroup$ Out of 100 million simulate hands I got $583,600$ points for E and $565,410$ points for F. Seems very close to the answer by @awkward. $\endgroup$ – David Aug 28 '16 at 15:06
  • $\begingroup$ I guess when you have a 2 player game like this where wins don't interact with each other and they can occur anywhere in the hand equally likely, then you just compute the probability of the win happening as early as possible in the hand and then multiply by the number of possible positions in the hand it can occur. $\endgroup$ – David Aug 28 '16 at 15:36
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    $\begingroup$ @David, in answer to your questions about the analysis for F, cases like 234567 are handled correctly, i.e. as two wins. For example, if the 2 is in position 1, then we would have $Y_1 = 1$ and $Y_2 = 1$, which contributes 2 toward the total. The theorem that makes all this work is that E(X+Y) = E(X) + E(Y) even if X and Y are not independent. $\endgroup$ – awkward Aug 28 '16 at 16:05

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