# Probability of beating an unknown hand

After watching poker online for a while now, I noticed that there were percentages of each player winning the hand - assuming the hand fully plays out - however they only display the percentages when both players hands are known.

Is there a way to calculate the percentage of player A winning the hand, without knowing player B's cards.

Assume poker is only being played two-handed. Assume Texas Hold'em. Assume all hands are equally likely to be played.

• Yes it is possible, if you assume neither hand can be folded. But betting/calling/folding is the major part of poker and so the probability becomes meaningless after the first round of betting – Henry Jun 12 '17 at 16:08
• So if I received Aces, there is a way to calculate my exact percentage of winning the overall hand. I'd quite like to do this as sort of a mini-project while I have free time over summer. Could you point me in the right direction as to what I'd have to calculate? – Sam Tyler Jun 12 '17 at 16:20
• Sure, subject to lots of details. Given that betting strategy can alter the outcome, what's normally done is to run a simulation. Have the computer play out a hundred thousand hands or so with some programmed strategy (possibly involving some randomized moves). – lulu Jun 12 '17 at 16:36
• @Sam to be a little more concrete, suppose you sit down to play some hands of poker. On one particular hand you calculate you have a 60% chance of winning. Suppose the other player raises by all the money you have staked for the whole evening (or whatever time it is that you're playing these hands). Do you take the bet? What about if you calculate 90% chance of winning? – Χpẘ Jun 13 '17 at 3:20

In Texas Holdem, you have two cards given to you.

The other player have $2$ cards too. Possible combinations are $\binom{50}{2}=1225$. Denote player $B$ cards by $B_i$ where $i$ takes value from $\{1, \ldots, 1225 \}.$

By total law of probability \begin{align}Pr(A \text{ wins })&=\sum_{i=1}^{1225}Pr(A \text{ beats } B_i| \text{player } B \text{ has } B_i)Pr(B_i) \\&=\frac1{1225}\sum_{i=1}^{1225}Pr(A \text{ beats } B_i| \text{player } B \text{ has } B_i)\end{align}

I assume that you have access to $Pr(A \text{ beats } B_i| \text{player } B \text{ has } B_i)$.

Remark: in real poker, your hands of cards is less significant than the style that you play.

You just run a simulation on all possible hole cards and boards.

There many programs out there to do this.

More common is to put your opponent on a range as they should not be playing weak hands.

There are only 169 starting hands. 13 pair, 78 unpaired not suited, 78 unpaired suited. There are tables on the web with the hands ranked. table