I was playing Texas Hold'em at a local cardroom last night keeping a meticulous record of the hands I was dealt. Perhaps I am totally wrong but I thought the occurrences of certain events in this poker session were considerably outside the realm of statistical expectancy but, although I took statistics in college, I am having trouble computing some probabilities. Can someone please help me?
This poker session lasted 4 hours 10 minutes and consisted of 142 hands.
In the course of these 142 hands, there were 38 instances of two (2) consecutively dealt hands in which the latter hand contained at least one card of the same rank as one of the cards in the prior hand. This seemed to me to be a high percentage (26.76%) of such events in a session of this length and consisting of this number of hands.
Could someone please tell me the probability of two consecutively dealt hands containing at least one card of identical rank (not rank and suit) and how this is calculated?
Additionally, there were three (3) instances of two (2) consecutively dealt hands in which both cards in the latter hand contained were the exact same rank (not rank and suit) as the first hand.
Same question as for the 38 instances?
I know that, relatively speaking, this is a very small sample of poker hands. However, is it probable that this could happen if the player is receiving "random hands"?
To anyone who chooses to help me with this I really appreciate your time.
The number of ways of choosing $2$ cards out of $52$ twice is $\binom{52}2^2=1758276$. The number of ways of doing it such that the cards don't share any ranks is
$$ \binom{13}{4}\cdot\frac{4!}{2!^2}\cdot4^4+2\cdot13\cdot\binom42\cdot\binom{12}2\cdot4^2+13\cdot12\cdot\binom42^2=1268592\;. $$
The three terms represent the three possibilities that $0$, $1$ or $2$ of the hands are pairs. If there are no pairs, we can choose $4$ of $13$ ranks; their $4!$ different permutations are distinct up to two factors of $2!$ for permuting within each hand; and each rank can independently be from one of $4$ suits. If there is one pair, it can be in one of two hands and have one of $13$ ranks, and we can choose $2$ of $4$ suits for it, which leaves $12$ ranks to choose the two ranks in the other hand from, and again $4$ suits for each of those two cards. If there are two pairs, we can choose one of $13$ ranks for the first and one of $12$ for the second, and for each pair we can again choose $2$ of $4$ suits.
Thus the probability for successive deals not to share a rank is
$$ \frac{1268592}{1758276}=\frac{8132}{11271}\approx72\%\;, $$
so the probability that they do share a rank is about $28\%$, so your observed value of $38$ is very slightly below the expected value of $141\cdot\frac{8132}{11271}\approx39.3$.
The second question is a bit easier. Of the $\binom{52}2$ deals, there are $\binom{13}2$ rank doublets that each occur $4^2=16$ times and $13$ rank singlets (pairs) that each occur $\binom42=6$ times. Thus the probability for the rank signatures to coincide is
$$ \frac{\binom{13}2\cdot16^2+13\cdot6^2}{\binom{52}2^2}=\frac{20436}{1758276}=\frac{131}{11271}\approx1.2\%\;, $$
so in this case you did see a substantially above-average number of occurrences, $3$ compared to the expected value $141\cdot\frac{131}{11271}\approx1.6$. The probability of observing $3$ successes in $141$ Bernoulli trials with $p=\frac{131}{11271}$ is $\binom{141}3p^3(1-p)^{138}\approx14\%$. This isn't quite the probability for your observation, a) because you would have been at least as surprised if there had been even more such coincidences, and b) because the $141$ trials aren't fully independent because pair hands differ from unpaired hands. Both effects should be quite small, though. (Note that this affects only the last result, the probability; the expectation values are exact since the linearity of expectation doesn't require independence.)