Wizard Of Odds Weekly Update November 15, 2018
This week somebody wrote on my forum at Wizard of Vegas that he got the same pocket pair three hands in a row in Texas Hold 'Em and asked about the probability of that.
To give an exact answer, you would need to know the total number of hands played, which he didn't state. Assuming 30 hands per hour, which is the benchmark for Texas Hold 'Em, we could estimate if we knew the time played. He didn't state that either. So, if we assume four hours of play at 30 hands per hour, that would be 120 total hands. There are 118 different three-hand sequences in 120 hands.
Where the math gets complicated is there are four states the player could be in at any one time:
- State 1: Last hand was not a pocket pair or the first hand played.
- State 2: Last hand was a pocket pair.
- State 3: Last two hands were the same pocket pair.
- State 4: The player has had the same pocket pair at least three times successfully in the session.
The next step is to calculate the probability of each state leading to each other state. I will spare you the math on that. After all those calculations, the transition matrix is as follows:
Each row is for the current state, starting with 1 on top and going down to 4 on the bottom. Each column is for the state of the next hand, starting with 1 on the left and going to 4 on the right.
Then you have to do some matrix, taking this matrix to the 118th power. Fortunately, that is not difficult to do in Excel. I would recommend doing T^64*T^32*T^16*T^4*T^2. That gives us T^118, which is:
The number in the upper right corner is the answer to our problem, 0.000139, or 1 in 7,190.
Sorry if this went rather fast. I plan to address this problem in more depth in my next Ask the Wizard column.