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Poker Probabilities Inquiry -- November 15, 2018

Recently, someone inquired on my Wizard of Vegas forum about experiencing the same pocket pair three times consecutively in Texas Hold 'Em and sought clarification on the odds of this occurring.

For an accurate answer, we would need to know the overall number of hands dealt, which wasn't provided. If we assume an average of 30 hands played each hour, typical for Texas Hold 'Em, we could make a rough estimate based on how long they played. Since that information was also missing, let’s assume they played for four hours, totaling 120 hands. Within 120 hands, there exist 118 distinct sequences of three hands.

The mathematics become intricate as there are four potential scenarios for the player during the game:

  1. State 1: The previous hand wasn't a pocket pair or it was the very first hand dealt.
  2. State 2: Last hand was a pocket pair.
  3. State 3: The last two hands were the same pocket pair.
  4. State 4: The player has successfully seen the same pocket pair at least three times during the playing session.

The subsequent step involves determining how each state transitions to others. I won’t delve into the detailed math behind this. After completing these calculations, the transition matrix is structured as follows:

0.941176 0.058824 0.000000 0.000000
0.941176 0.054299 0.004525 0.000000
0.941176 0.054299 0.000000 0.004525
0.000000 0.000000 0.000000 1.000000

Each row corresponds to the current state, beginning with 1 at the top and descending to 4 at the bottom. Each column depicts the next hand's state, starting with 1 on the left and going to 4 on the right.

Next, you'll need to perform matrix calculations, specifically raising this matrix to the 118th power. Thankfully, this is quite manageable in Excel. I suggest calculating T^64*T^32*T^16*T^4*T^2 to arrive at T^118, which is:

0.941047 0.058549 0.000265 0.000139
0.941028 0.058548 0.000265 0.000159
0.936789 0.058284 0.000264 0.004663
0.000000 0.000000 0.000000 1.000000

The figure located in the upper right corner provides the solution to our query: 0.000139, which translates to approximately 1 in 7,190.

I apologize if this explanation felt a bit rushed. I intend to explore this topic further in my upcoming Ask the Wizard column.