Its an easy question for all you experts.

How do you calculate how many time a given hand will win at any stage of the game when you know all the hands? (pre-flop, flop and turn. I think the river is obvious )

What I mean is, how do they calculate the winning percentages on live TV games when they display it next to the player's hand?

I'm sure there are good links to this, I just need to be pointed to the right direction cause google keeps bringing up programs that do it for you. I need the formula though.

Thanks!

Why do you need the formula? The programs out there will do it for you. Check out Pokerstove. It's the one most used here.

If you really want to know the formulas...

The formula will differ depending on how you want to treat ties. Some will show a tie as a win for all tied players (On TV, they'll both have 100%), some will ignore the ties and only show winning chances (I saw one where player 1 had like 20% to win and player 2 had like 6%, but they would split the pot in all other situations...)

Anyway, I'm going to treat a tie as a win...

The easiest calculation would be on the river. With two players, there are 44 unseen cards (52-2 hole cards P1 -2 hole cards P2 -4 board cards)

Say player 1 is behind, but has a flush draw. There are 9 cards that make him a winner, he loses if any of the others hit. P1 is (9/44)*100 = 20.45% to win.

formula (outs/unseen cards) * 100

The turn is trickier. There are 45 unseen cards, but two to deal. 45 possible on the turn, 44 on the river making 1980 combinations of turn and river. Find out how many combinations make P1 a winner, divide by 1980, multiply by 100 and there's the answer.

For the flop, there are 205,476,480 combination of cards that can become community cards. Good luck figuring this out without the program.

Thanks for the help. The reason I need this is for a report.

I guess I'll have to find how to do it for the flop myself...somewhere!

Originally Posted by triple-t

Why do you need the formula? The programs out there will do it for you. Check out Pokerstove. It's the one most used here.

If you really want to know the formulas...

The formula will differ depending on how you want to treat ties. Some will show a tie as a win for all tied players (On TV, they'll both have 100%), some will ignore the ties and only show winning chances (I saw one where player 1 had like 20% to win and player 2 had like 6%, but they would split the pot in all other situations...)

Anyway, I'm going to treat a tie as a win...

The easiest calculation would be on the river. With two players, there are 44 unseen cards (52-2 hole cards P1 -2 hole cards P2 -4 board cards)

Say player 1 is behind, but has a flush draw. There are 9 cards that make him a winner, he loses if any of the others hit. P1 is (9/44)*100 = 20.45% to win.

formula (outs/unseen cards) * 100

The turn is trickier. There are 45 unseen cards, but two to deal. 45 possible on the turn, 44 on the river making 1980 combinations of turn and river. Find out how many combinations make P1 a winner, divide by 1980, multiply by 100 and there's the answer.

For the flop, there are 205,476,480 combination of cards that can become community cards. Good luck figuring this out without the program.

I'm actually looking for how to calculate the odds knowing ALL the hands in a game, i.e. all cards face up; like they do on TV . Not for figuring out whether to call a big raise, or small raise etc.

I was slightly incorrect earlier, so I'll post all the information you need...

Variables:
C = community cards left to deal
D = Cards left in the deck

Say you are at a 10 player table. There are 20 cards dealt leaving 32 in the deck (D=32). PF there are 5 community cards to deal (C=5) therefore there are:

D!/(C!(D-C)!) = 32!/(5!(32-5)!) = 201,376 combinations of community cards available. A players win% is simply the ratio of combinations that give him the win divided by total combinations of community cards. So if Player 1 will win when any of 100,000 combinations of community cards hits, he is 49.66% to win. You can treat ties however you want. Some count it as a win, some disregard, some count it as a fraction of a win. You would HAVE to use a program to figure these percentages as there are 201,376 possible boards. On a Heads-up table (2 players) there are 1,712,304 possible boards.

Before the turn on our 10 handed-table (20 cards dealt, 3 flop cards) there are 29 cards left in the deck. Our formula for number of combinations is 29!/(2!(29-2)!) = 406 combinations of turn and river. Winning percentage is calculated once again by dividing combinations that win by total combinations. If player one has 125 combinations that win, he is 30.79% to win the hand.

Before the river, there are now 28 cards left in the deck with only 1 to deal. There are 28 combinations. Winning percentage is simple here.

I hope this clears up any questions you may have. If you aren't sure what the ! means, google factorial. It's the same process every time, it's just that the number of possible community card combinations is so great before the flop, it'd be almost impossible to calculate without a program.

You posted this when I just figured out how to do it myself Thanks anyway, your information is valuable.

I'll post what I've written, its quite long though so I wont expect a reply. I'm just hoping its correct, I'll check with what you've written, thanks.
I'm only interested in post-flop situations by the way.

Here's my example:
--------------------------
For example assume that Alice holds QcQs, Bob holds 5d7d and the flop is 4d6c2d. At this point Alice is winning but Bob has many ‘outs’. This means that there are many cards in the deck that can improve Bob’s hand. The cards are any of the three remaining 3’s or any of the four remaining 8’s to give him a straight, any of the nine remaining diamonds (d) to give him a flush. That’s 16 cards in total that can make Bob’s hand a winning hand. Therefore it can be said that Bob has 16 ‘outs’.

A draw is also possible (and will be treated as a separate player and if he, the draw, wins then the players are tied). There is currently no chance for a draw, as judging from the flop there is no chance that the cards on the board including the upcoming turn and river will be better than both players’ hand.

There is still the turn and river to be dealt amongst 45 unseen cards; 45 possible on the turn and 44 possible on the river making 1980 possible combinations of turn and river (45 cards remaining on the turn multiplied by 44 cards remaining on the river equals 1980). However any of the 16 cards that can improve Bob’s hand can appear at either the turn or the river giving him a winning hand. The probability that any of those cards will appear in the turn is 16/45 (outs / remaining unknown cards), which is 35.56%. The probability that any of those cards will appear in the river (if it hasn’t appeared already) is now 16/44, which is 36.36%.

It is also possible to hit two of the three remaining 5’s, OR two of the three remaining 7’s to make three of a kind and win. The probability of that happening has to take in account all the 1980 possible combinations of turn and river is 6/1980 (6 possible combinations of 5’s and 7’s divided by the 1980 possible turn and river combinations).Thus 6/1980=0.003=0.3%.

Another possibility is making two pair by hitting one of the three remaining 5’s AND one of the three remaining 7’s. The probability is 3/1980 (the 3 is one of the three 5’s) multiplied by 3/1980 (the 3 is one of the three 7’s). Therefore (3/45)×(3/44)=0.0045=0.45%.

Thus the complete probability of increasing Bob’s hand with the turn and/or the river is 36.36% + 0.3% + 0.45% = 37.11%.
------------------------

What puzzles me is the fact that poker odds calculators give QQ 44% and 57s 56%. I did not manage to do any calculations to reach these numbers and I've tried loads!
I did my calculations similar to: http://www.cs.ualberta.ca/~darse/Pap...vidson.msc.pdf
and I tried the results from the above link in a poker odds calculator and the difference is similar to the difference I had with my results that I just posted. Strange...

Originally Posted by nicko82

Here's my example:
--------------------------
For example assume that Alice holds QcQs, Bob holds 5d7d and the flop is 4d6c2d. At this point Alice is winning but Bob has many ‘outs’. This means that there are many cards in the deck that can improve Bob’s hand. The cards are any of the three remaining 3’s or any of the four remaining 8’s to give him a straight, any of the nine remaining diamonds (d) to give him a flush. That’s 16 cards in total that can make Bob’s hand a winning hand. Therefore it can be said that Bob has 16 ‘outs’.

Close. Bob has four 3's, four 8's and seven diamonds (You've already counted the 3 and the 8). Bob has 15 outs.

Originally Posted by nicko82

A draw is also possible (and will be treated as a separate player and if he, the draw, wins then the players are tied). There is currently no chance for a draw, as judging from the flop there is no chance that the cards on the board including the upcoming turn and river will be better than both players’ hand.

There is still the turn and river to be dealt amongst 45 unseen cards; 45 possible on the turn and 44 possible on the river making 1980 possible combinations of turn and river (45 cards remaining on the turn multiplied by 44 cards remaining on the river equals 1980). However any of the 16 cards that can improve Bob’s hand can appear at either the turn or the river giving him a winning hand. The probability that any of those cards will appear in the turn is 16/45 (outs / remaining unknown cards), which is 35.56%. The probability that any of those cards will appear in the river (if it hasn’t appeared already) is now 16/44, which is 36.36%.

It is also possible to hit two of the three remaining 5’s, OR two of the three remaining 7’s to make three of a kind and win. The probability of that happening has to take in account all the 1980 possible combinations of turn and river is 6/1980 (6 possible combinations of 5’s and 7’s divided by the 1980 possible turn and river combinations).Thus 6/1980=0.003=0.3%.

Another possibility is making two pair by hitting one of the three remaining 5’s AND one of the three remaining 7’s. The probability is 3/1980 (the 3 is one of the three 5’s) multiplied by 3/1980 (the 3 is one of the three 7’s). Therefore (3/45)×(3/44)=0.0045=0.45%.

Thus the complete probability of increasing Bob’s hand with the turn and/or the river is 36.36% + 0.3% + 0.45% = 37.11%.
------------------------

What puzzles me is the fact that poker odds calculators give QQ 44% and 57s 56%. I did not manage to do any calculations to reach these numbers and I've tried loads!
I did my calculations similar to: http://www.cs.ualberta.ca/~darse/Pap...vidson.msc.pdf
and I tried the results from the above link in a poker odds calculator and the difference is similar to the difference I had with my results that I just posted. Strange...

OK, this will get into some serious math here.
We have 45 cards left. Bob has 15 outs. 2 can hit the turn which gives Alice a redraw to win.

If the turn is any 3, any 8, or any diamond (except Q or 6) then Bob wins the hand no matter what the river is. (13/45) * (44/44)

If the turn is Qd then Alice has 10 outs to win (last Q plus 9 board pairs). (1/45)*(34/44)

Turn is 6d, Alice now has 4 outs to win (6h,6s,qh,qd) (1/45)*(40/44)

If the turn is a non-diamond A, K, J, T, 9 then Bob still has 15 outs (15/45)*(15/44)

If the turn is the Qh, Bob has 13 outs (he loses the qd and 6d) (1/45)*(13/44)

If the turn is a 5 or 7 (6 cards), Bob now has 20 outs (6/45)*(20/45)

Finally, if the turn pairs the board (except the 6d, 8 cards) Bob loses the Qd as an out and has 14 outs (8/45)*(14/44)

Add up all of Bob's chances
[(13/45) * (44/44)] + [(1/45)*(34/44)] + [(1/45)*(40/44)] + [(15/45)*(15/44)] + [(1/45)*(13/44)] + [(6/45)*(20/45)] + [(8/45)*(14/44)] and you get => 56.22% as Bob's chances to win this hand. You could do the same with Alice, but hers will be 100-56.22 = 43.78%. I believe these numbers match what your program gave you.

Looks like I chose quite a complicated hand, I got the point though, great explanation thanks, it makes perfect sense!

My calculation of outs was far too simplistic I guess, there were a lot more possibilities than I saw (maybe thats why I'm a fish ). Also some foolish mistakes from my part.

It looks quite computationally intensive!

Thanks again!

You are right, it is computationally intense. That's why I use Pokerstove to do all the math for me. Granted, it doesn't get that detailed, but it works. Good luck with your project.