How often will you end up with the nuts?

[RoundTower]

Don't know if this has been asked before, couldn't find it in the search. It was originally asked in Mid-High PL/NL, I've now got quite interested in it.

If you are playing hold'em and you play every hand to the river, how often will you end up with the nuts?

[Blackwolf]

Propably depends on your bankroll /images/graemlins/wink.gif

[AaronBrown]

This is an interesting question. I don't know the answer, but I know how you can get it if you have the energy.

Start with a list of all five-card poker hands, along with their probabilites. Consider each one as a board, then figure out what the nut card or cards are. For example if the board is a royal flush, no one (or everyone if you prefer, but you have to decide how to count this) can have the nuts. If there is any other straight flush, the nuts is any hand with the next suited card in sequence. If there are four Aces on the board, any hand with a King is the nuts (again, you have to decide how to count hands with possibly shared nuts); any other four of a kind means an Ace is the nuts.

Once you do this, you figure the chance of getting each five-card hand on the board, and multiply by the chance of having the nuts if it is there. For example, if there is a non-royal straight flush on the board, you need one of 47 remaining cards to have the nuts, which will happen 4.3% of the time. If the board is four Aces, no King, you need one of four cards out of the remaining 47, that will happen 16.5% of the time.

Multiply the chance of having the five card hand on the board times the chance of having the nuts if it is dealt, add it up, and you have your answer.

The only tricky part of this concerns straights. Once you get to hands of two pair and below, you have to consider for each one whether or not there is a straight possibility. It's not too bad, because there will always be two specific ranks of cards you need for the nut straight (except for Ace-high straights); you don't have to consider every possible straight.

[RoundTower]

[ QUOTE ]
This is an interesting question. I don't know the answer, but I know how you can get it if you have the energy.

Start with a list of all five-card poker hands, along with their probabilites. Consider each one as a board, then figure out what the nut card or cards are. For example if the board is a royal flush, no one (or everyone if you prefer, but you have to decide how to count this) can have the nuts. If there is any other straight flush, the nuts is any hand with the next suited card in sequence. If there are four Aces on the board, any hand with a King is the nuts (again, you have to decide how to count hands with possibly shared nuts); any other four of a kind means an Ace is the nuts.


[/ QUOTE ]

Thanks for this, I have done something like this in the nl thread. The way I see the question you can ignore "shared nuts", I want to find the possibility of ending up with an unbeatable hand (which might be tied). So 74o would be the nuts on a KKKKA board, and AK would be the nuts on an AAK62 board. There are still some tricky siutations, e.g. if the board is 2367 of hearts and a blank, A4 of hearts is the nuts but AJ of hearts is not.

AaronBrown, would you like to hazard a guess at the actual answer? You don't have to put your professional reputation at stake.

[CallMeIshmael]

Using homebrew software, I just did the following 50,000 times:

- Deal a hand
- Deal a flop EDIT: I meant to say ENTIRE BOARD
- Find that hands equity vs random hands

That equity was 100% 162 times.

162/50,000 = =0.00324 = 0.324%

[WhiteWolf]

[ QUOTE ]
Using homebrew software, I just did the following 50,000 times:

- Deal a hand
- Deal a flop EDIT: I meant to say ENTIRE BOARD
- Find that hands equity vs random hands

That equity was 100% 162 times.

162/50,000 = =0.00324 = 0.324%

[/ QUOTE ]

Simulation is probably the best way to solve this. I think your method may suffer from a flaw, however, in that it disregards split pots when you have the nuts. For example, a board of TJQ23 (no flush possible), AK is the nuts but will not have 100% equity because it splits with other AK hands.

[RoundTower]

[ QUOTE ]
Using homebrew software, I just did the following 50,000 times:

- Deal a hand
- Deal a flop EDIT: I meant to say ENTIRE BOARD
- Find that hands equity vs random hands

That equity was 100% 162 times.

162/50,000 = =0.00324 = 0.324%

[/ QUOTE ]

This is awesome, great start. Whitewolf is correct in that this should produce an underestimate. Any chance you could change the code so it produces not just a value for equity against a random hand, but win/tie/lose percentages like Pokerstove? This would make it easy to calculate the exact percentage.

Also, a simulated answer would be great, but not as great as an answer demonstrated mathematically. Maybe this is me being a maths nit though.

[elitegimp]

[ QUOTE ]
Start with a list of all five-card poker hands, along with their probabilites. Consider each one as a board, then figure out what the nut card or cards are.

[/ QUOTE ]

I haven't thought too much about this, but it seems like it would be easier to enumerate if you start with every possible starting hand and count how many boards give you the nuts.

Example: If you have AA, you have the nuts on a KQJTx board (no flush), an AAxyz board (no straight flush), and any time you have a flush (no straight flush).

If you have 72o, you only have the nuts on a 777xy (no over pair, no straight flush) board, 222xy (no over pair, no straight flush), or if you have 3456x / 4568x / 5689x (no T) all suited w/ your 7.

Computer-wise, a small extension on the method outlined CallMeIshmael: For each possible hand, deal every board and rather than looking for "equity = 1", look for "percent of hands loss = 0"

[CallMeIshmael]

[ QUOTE ]
[ QUOTE ]
Using homebrew software, I just did the following 50,000 times:

- Deal a hand
- Deal a flop EDIT: I meant to say ENTIRE BOARD
- Find that hands equity vs random hands

That equity was 100% 162 times.

162/50,000 = =0.00324 = 0.324%

[/ QUOTE ]

This is awesome, great start. Whitewolf is correct in that this should produce an underestimate. Any chance you could change the code so it produces not just a value for equity against a random hand, but win/tie/lose percentages like Pokerstove? This would make it easy to calculate the exact percentage.

Also, a simulated answer would be great, but not as great as an answer demonstrated mathematically. Maybe this is me being a maths nit though.

[/ QUOTE ]


yeah, I will rework the code tomorrow. Cant believe I made such an obvious error /images/graemlins/frown.gif /images/graemlins/frown.gif


I agree that a 'true' answer is better, but for the life of me I cant think of a truly good way of doing it in an appropriate amount of time

[AaronBrown]

[ QUOTE ]
AaronBrown, would you like to hazard a guess at the actual answer? You don't have to put your professional reputation at stake.

[/ QUOTE ]

I'm not sure there's any reputation to lose, but here's my guess.

8% of boards are three suited, in which case the high outstanding suited card combined with anything is usually the nuts. That's 46 different hands.

I'm guessing about 8% of the time the board can make a straight, in which case 16 possible hands can complete it.

I think half the remaining hands, 42% total, have no pair and no straight or flush possibility, in which case a pair of the top card is the nuts, there are 3 possible hands.

In the other 42%, most of the time you will need one of the pair (or top pair) to have the nuts, that's 92 different hands.

That means there are an average of about 45 out of the 1,081 possible hands that are the nuts, so about 4%. I know there's a lot of sloppiness in the estimate, but the answer feels about right to me. It says on the average hand you need one specific card to have the nuts, the hands on which you need two cards for nuts are balanced by the hands in which you can have one of two or more cards.

That may contradict experience, we know in a 10-player game it's not true that someone holds the nuts on one hand in three. But that's because people fold before the river. If everyone stayed in to showdown, the one in three wouldn't surprise me.

[KJ o]

That sounds way too high to me.

This is my reasoning. Is this completely flawed?

There are 47*46/2=1081 possible hands for each given board. For the vast majority of boards, only one or a few hands are the nuts. While there are certainly boards with loads of possible nut hands, these are very uncommon.

My gut feeling would be that on average the number of possible nut hands for a board is somewhere around 3-5, indicating that the probability should be 0.3%-0.5%. This was also what the simulation suggested, no?

I think this is the flaw in your reasoning:
[ QUOTE ]
In the other 42%, most of the time you will need one of the pair (or top pair) to have the nuts, that's 92 different hands.


[/ QUOTE ]
If the board has paired and the pair is not the highest rank, only a pocket pair of that rank is the nuts. That's only one hand instead of 92.

[RoundTower]

[ QUOTE ]

I think half the remaining hands, 42% total, have no pair and no straight or flush possibility, in which case a pair of the top card is the nuts, there are 3 possible hands.


[/ QUOTE ]

I think this is way off, more like 10% of total boards make top set the nuts. There is quite a long discussion of the question here
http://www.boards.ie/vbulletin/showthread.php?t=2054874589
including a program to find the answer (not my work), which is now on this web page
http://www.absolutebluff.com/personal/holdem/index.html

There were a few problems with this, not sure if they were all fixed, but I made a few changes and got .95% (about once every 105 hands).

[AaronBrown]

[ QUOTE ]
If the board has paired and the pair is not the highest rank, only a pocket pair of that rank is the nuts. That's only one hand instead of 92.

[/ QUOTE ]

You're right. I was just hazarding a guess, and guessed too high.

I now think my answer was much too high, but I wish someone would do the full calculation so we could know the correct answer.