Texas Hold ‘Em, Statistics Problem Example

The Problem

We are playing Texas Hold ‘Em and my opponent has the 10 of spades and clubs. I have the 6 and 7 of diamonds. The house cards are: 10 of diamonds, 9 of spades, 3 of diamonds and 4 of clubs. So, I need to beat a 3 of a kind to win, but my opponent could still beat me if he gets a card that would help him beat what I built to beat the 3 of a kind. Should I call the bet by tossing the rest of my money into the pot, or should I fold and give the pot to my opponent?

The Approach

One approach is to account for probabilities. It ends up being an analysis of my opponent’s chances of winning against my own. Tabular representation of the situation helps keep count of cards and keep track of probabilities. In order to win, I need to beat 3 of a kind.

The Conclusion

The solution predicates that I would fold. The combined probability that my opponent would beat a 3 of kind on the next draw exceeds my own. So, it is not a good bet for me. The wise move is to fold.

The Solution

The following table represents the cards that have been dealt so far. Cards that will not result in a hand greater better than 3 of a kind for either player have been omitted. The boxes marked “t” are the dealer’s cards – hence, “t” for “on the Table.” My cards are marked “m” for “my cards,” and my opponent’s cards are marked “o” for “opponent’s cards.”

The rest of the labels denominate what we would need to make a particular hand. So, I need one of the cards for the boxes marked “ms” to make a straight – hence “ms” for “my straight.” “O4” is the card my opponent needs to make 4 of a kind. My opponent can also build a full house if he gets one of the cards labeled “of” for “opponent’s full house.”

The following table includes the probabilities that the next card drawn would build each hand. Hence, I have 0 probability of making anything but a straight that would beat a 3 of kind. There are 44 cards left from which to draw; thus, each ratio is represented as a quotient of 44. My opponent could build a 4 of a kind if he gets the last 10, but he would only need one of 9 possible cards to make a full house. The row marked “Totals” is the combined probabilities for each of us to build a hand that beats a 3 of a kind.

His total probability of beating a 3 of a kind exceeds mine. That does not necessarily dictate how I will act though. I still need to take the money into account. If I fold, I will certainly lose $12K of my money. If I call, I will have an 18.18% chance of winning $44K minus my investment of $22K. Thus, it is a $4K possibility for me. For my opponent, it is $5K possibility. It is not a good bet, and I fold.