Probability of beating an unknown hand

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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.

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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.

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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