Poker probability (Texas hold 'em)

Any fellow RP player can suggest a 100% sure Poker probability list what is should accepted by the whole known universe?

Not sure I can post this here, but at web site, learn-texas-holdem. com. there is a chart with relevant easy to memorize stats and oddsl

Hugo X

I like this one:'em)

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BRB, learning to poker. Thanks for the links!

Wiki not allways accurate ( albeit my headline is copyed/pasted from them too)

The expanded question is , all who have any doubt, would be satisfied by it(?), also, how much % different would be accepted by them as for a 100% need a lots hand . ( how much lots and how much the % closer to 100% by higher number draws)
Also variables have to be in count and minimum hand played as explaned …

I like to see different opinions then wiki, if exist different.

The simple rule of thumb goes something like this…

If there were 50 cards in the deck, each would be worth 2%. This isn’t perfectly accurate, since their are 52 cards, but it’s close enough for simple, on the fly calculations.

So, say you have 2 spades in the hole and flop 2 spades. There are 13 spades total, you know where 4 of them are, so there are 9 left. You have roughly an 18% chance of hitting a spade on the next card. (the actual odds would be 9 in 47, or 19%)

So you have about an 8% chance of hitting that inside straight on the next card, about 16% if you are open-ended, and about an 18% chance of making the flush on the next card if you are 4 to a flush.

The essential skill of poker is knowing what to do with this information. Learn about pot odds, implied odds, fold equity, and, most important, expected value, or EV.


One simple (or tough I don’t know) probability question:

If the probability of winning against one player is 70%, what is the probability of winning against two players? Is this computable?

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depends, if it’s poker related there are almost always more variables that alter the probablility dependant on how the hands are sharing the same outs, blockers, etc.
but when you take all the variables out and only use the raw 70% to both then your winning odds become 49%


If you want to start making decisions based on your hand vs the ranges of 1 or 2+ opponents, I’d suggest looking at Equilab and/or Flopzilla. You can also start examining range v range(s) with these tools. Not sure if this is what you were looking for so I’m taking my best guess. If you had another application in mind, please post.


well, if you have aces or kings preflop, you are probably going to win…

Actually my question is one part of two questions. Supposing I have open ended straight on the flop, my outs are eight, which has the probability of making the straight is 32%, which can be hand calculated (as mentioned in one of your previous responses) or from published statistics as this one. Now with this number how to find the probability of winning against one opponent holding any card. Against two, against three and so on.

I tried the apps you have mentioned. They are so confusing. I gave up. May be they are giving such stat. If they do what kind of calculation they make.

I posted the second part of my question because it looked simple enough of a probability question. May be its not.

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Yes, leaving out other outs, blockers, etc and working with raw probability of 70%, like open ended straight has raw probability of 32%. What’s my odds against one opponent, two three etc.

49% may not be. Like if against two players is (0.7x0.7) 49%, against six players it will be 12%, which doesn’t seem to be.

Your profile is very nice.

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in general if you use the rule of 4&2 you only use the outs which are next to certiain they make the winning hand anyway. just take the example of the str8, lets say you hold 67o on a rainbow board. on the flop lies 58A. if you hit your 4 or 9 (8 outs) you are almost sure you will gonna win the pot. of course with more opponents something like hitting 4(turn) and 8(river) with one opponent holding 55 might happen earlier, but in general when you hit your str8 on such a dry board you might almost always be good.
lets take an slightly more difficult example here. you hold the exact same hand but instead of a rainbow board there lies 2 hearts. in this example it’s getting much more important to see the amount of opponents since many opponents are quite likely to have at the very least one of them with 2 hearts. so here there are a few things you can do. first of all you obviously have 6 clean outs and 2 filthy outs. depending on how threatening the filthy outs are i use a few things. first of all i try to imagine how bad they are and make an adjustment in my calculation based on that. here a few examples: lets say you have 1 opponent and no reads. simply said only if (s)he has 2 hearts you are in trouble with your outs, since he is the one who bet (because you are caclulating your odds means you consider a call) it’s much more likely he has something like an ace then he has 2 hearts. but he might also be semi bluffing with 2 hearts. here i just assume the chances low but possible he has these hearts, so i just take a half out of those 2, making it 7,5 outs and base my calclation on that. also i only consider my implied odds based on only these 6 pure outs and decrease my implied odds when i hit one of my wrong outs .
now lets say the same thing happens but with 6 opponets. to make the example easy lets forget about the fact that there are people after you to act (don’t forget about this with a real game however). since it’s very likely at least one of them has hearts i don’t count those 2 outs at all, and only take the 6 outs i have and fully erase my implied odds on the filthy outs. besides of that i decrease my actual odds with 18% ((because if you might hit your str8 the other card might be a heart anyway (18 because of the 9 outs) )).

all things considered know that this are just examples on how to alter your outs based on filthy outs and your amount of opponents. since all situations are different it’s more like calclulation based on feel (like the same way fold equity works).
so the point i try to make is try altering your outs and implied odds based on how much you deem likely hitting your hand won’t be enough and add the reverse implied odds to your wrong outs.


I did some thinking, made some guesses.

Player one has the probability of winning 70%. Player two 30%. Another one arrives with the same probability of 30%, with the total probability of 1.3. Bringing it back to 1, probabilities will be 0.54, 0.23 and 0.23. If a fourth player arrives, it will be 0.44,0.18,0.18 and 0.18. So, if the probability of winning against one player is 70%, winning against two players is 54%, against three players 44%.

Another one: If one has, say 32% with open straight, this remains almost same irrespective of the number of players, probably with slight reduction with every player as the another player may beat with the higher straight. The rest 68% is distributed and changes with number of players.

Just an empirical guess.

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Good word empirical !!!

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