Game of Snap Probability

Question

If you don't know how snap works - two players get 26 cards each from a standard 52 card deck. They then place their cards in a shared pile of cards where the cards are visible to both of them. If two cards of the same rank are placed right after each other, they call the snap. My question is what the probability is that this does not occur until both players have placed down 3 cards each. If possible, could you use the hypergeomtric distrubition, seeing as the cards are not replaced and kept in a pile in the middle.

Answer 1

I gather from the comments that the two-player structure of the game is in fact irrelevant and you're simply interested in the probability of no two consecutive cards of seven cards drawn from a shuffled standard deck having the same rank.

The probability for a rank match in each particular pair of consecutive cards is

$$ \frac{13\cdot\binom42}{\binom{52}2}=\frac1{17}\;. $$

As a first crude approximation, the probability of a match occurring is roughly the expected number of matches, $\frac6{17}$, which would make the probability for no match to occur

$$ 1-\frac6{17}=\frac{11}{17}\approx0.647\;. $$

As a second, much better approximation, we can treat the match events as roughly independent and calculate the probability of none of them occurring as

$$ \left(1-\frac1{17}\right)^6=\left(\frac{16}{17}\right)^6=\frac{16777216}{24137569}\approx0.6951\;. $$

The exact probability can be calculated using a Markov chain or inclusion–exclusion, but with both approaches it would require quite a bit of casework.

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P.S.: Here's the casework to do it by inclusion–exclusion.

We have $6$ conditions, namely that none of the $6$ possible rank matches occur.

The probability for one particular condition to be violated is $\frac1{17}$, and there are $6$ of them, for a contribution of $-\frac6{17}$.

Two particular conditions can be violated in two different ways. There are $5$ cases where the two matches are adjacent, and in this case the probability for both matches to occur is

$$ \frac{13\cdot\binom43}{\binom{52}3}=\frac1{425}\;, $$

so this yields a contribution $\frac5{425}=\frac1{85}$. And there are $\binom62-5=10$ cases where the two matches aren't adjacent, and in this case the probability for both matches to occur is

$$ \frac{\frac13\binom{13}2\binom42^2+13}{\binom{52}4}=\frac{73}{20825}\;, $$

so this yields a contribution $\frac{10\cdot73}{20825}=\frac{146}{4165}$.

Since it's quite unlikely for $3$ matches to occur, we should already get quite a good approximation if we were to stop at this point:

$$ 1-\frac6{17}+\frac1{85}+\frac{146}{4165}=\frac{34}{49}\approx0.6938776\;, $$

as expected quite close to the already quite good second approximation above.

With diminishing returns in precision, let's do the calculation for three matches. There are $4$ cases in which these are all adjacent, and the probability for three adjacent matches is

$$ \frac{13}{\binom{52}4}=\frac1{20825}\;, $$

for a contribution of $-\frac4{20825}$. There are $4$ cases in which none of the three matches are adjacent, and the probability for such a case to occur is

$$ \frac{\frac1{15}\binom{13}3\binom42^3+\frac15\cdot13\cdot12\cdot\binom42}{\binom{52}6}=\frac{207}{978775}\;, $$

for a contribution of $-\frac{4\cdot207}{978775}=-\frac{828}{978775}$. In the remaining $\binom63-4-4=12$ cases, two matches are adjacent and the third isn't adjacent to either of them. The probability for such a case to occur is

$$ \frac{\frac1{10}13\cdot\binom43\cdot12\cdot\binom42}{\binom{52}5}=\frac3{20825}\;, $$

for a contribution of $-\frac{12\cdot3}{20825}=-\frac{36}{20825}$.

So our next level of approximation, already quite precise, is

$$ \frac{34}{49}-\frac4{20825}-\frac{828}{978775}-\frac{36}{20825}=\frac{676442}{978775}\approx0.6911108\;. $$

Since not too many cases remain, we might as well do the rest, too.

Four matches can't all be adjacent. We can have three of them adjacent and one separate. There are $6$ such cases, and the probability for each is

$$ \frac{\frac1{15}\cdot13\cdot12\cdot\binom42}{\binom{52}6}=\frac3{978775}\;, $$

for a contribution of $\frac{6\cdot3}{978775}=\frac{18}{978775}$.

Or we can have two pairs of them adjacent. There are $3$ such cases, and the probability for each is

$$ \frac{\frac1{10}\binom{13}2\binom43^2}{\binom{52}6}=\frac6{978775}\;, $$

for another contribution of $\frac{3\cdot6}{978775}=\frac{18}{978775}$.

Or we can have two of them adjacent and the other two separate. There are $3$ such cases, and the probability for one of them to occur is

$$ \frac{\frac1{35}\cdot\frac13\cdot13\binom{12}2\binom43\binom42^2+\frac1{35}\cdot13\cdot12\binom43}{\binom{52}7}=\frac{201}{22511825}\;, $$

for a contribution of $\frac{3\cdot201}{22511825}=\frac{603}{22511825}$.

So our next approximation is

$$ \frac{676442}{978775}+\frac{18}{978775}+\frac{18}{978775}+\frac{603}{22511825}=\frac{15559597}{22511825}\approx0.6911744\;. $$

Finally, with $5$ matches, no $4$ of them can be adjacent. That only leaves $2$ cases in which $3$ and $2$ of them are adjacent, and the probability for such a case to occur is

$$ \frac{\frac1{35}\cdot13\cdot12\binom43}{\binom{52}7}=\frac3{22511825}\;, $$

for a contribution of $-\frac{2\cdot3}{22511825}=-\frac6{22511825}$. Since $6$ matches are impossible, that gives us our final exact result:

$$ \frac{15559597}{22511825}-\frac6{22511825}=\frac{15559591}{22511825}\approx0.6911741\;. $$

Here's code that checks these results (both the final tally and the individual contributions).