How often will you end up with the nuts?

RoundTower

An interesting question, little bit mathsy. If you play every hand to the river in a hold'em game, how often will you have the nuts on the river? This was posted on 2+2, I made a reasonable guess at it, but no one has given a conclusive answer yet.

NeVeR

complete guess but 1 out of every 50 hands played.

Norwich Fan Rob

not as often as once in 50

roryc

0.57%

Doctor Fell

RoundTower wrote:

An interesting question, little bit mathsy. If you play every hand to the river in a hold'em game, how often will you have the nuts on the river?

If you mean that you take each one of the possible starting hold 'em hands (269 or something) and compare it to every possible 5 card combination (representing the board) to see if u have the nuts, u would have the nuts very rarely - don't know the % but it would have to be in the order of 0.0001% I guess.
E.g 23o will only make the nuts on a board of A45XY, where XY don't pair the board or cause a higher straight or flush. Or a board any time they make quads. That's only a couple of potential boards, so compared to all possible combinations of 5 card boards that's nothing really.

RoundTower

Doctor Fell wrote:

If you mean that you take each one of the possible starting hold 'em hands (269 or something) and compare it to every possible 5 card combination (representing the board) to see if u have the nuts, u would have the nuts very rarely - don't know the % but it would have to be in the order of 0.0001% I guess.
E.g 23o will only make the nuts on a board of A45XY, where XY don't pair the board or cause a higher straight or flush. Or a board any time they make quads. That's only a couple of potential boards, so compared to all possible combinations of 5 card boards that's nothing really.

Think again.

Doctor Fell

RoundTower wrote:

Think again.

OK......think about what exactly???

Doctor Fell

RoundTower wrote:

Think again.

If uv got more than 2 words to say, let's hear it.
Think about it...

roryc

Doctor Fell wrote:

If you mean that you take each one of the possible starting hold 'em hands (269 or something) and compare it to every possible 5 card combination (representing the board) to see if u have the nuts, u would have the nuts very rarely - don't know the % but it would have to be in the order of 0.0001% I guess.
E.g 23o will only make the nuts on a board of A45XY, where XY don't pair the board or cause a higher straight or flush. Or a board any time they make quads. That's only a couple of potential boards, so compared to all possible combinations of 5 card boards that's nothing really.

I dont think this is what he meant. I think what he is asking is if you dealt out say, 1 million hands and played them down to the river, how often would you have the nuts.

Doctor Fell

roryc wrote:

I dont think this is what he meant. I think what he is asking is if you dealt out say, 1 million hands and played them down to the river, how often would you have the nuts.

Oh right, fair enough. Wasn't sure what exactly he meant.

NeVeR

only 1 way to find out. get a deck of cards and some friends. and start dealing. playing 1 million hands should take you a while but at least you'll have an answer.

marius

Personally I would have the nuts about 98.7% of the time, so if you are going to call me, expect to get beaten:D

TBH I would say it is much less than 1 in 50. I'd go for about 1 in 220

NeVeR

to be honest there is no way of telling,

You could play 1 million hands and get the nuts 10 times and play another million and get the nuts 10,000 times.

Shortstack

Anybody any good at writing programs?

Dealing 10million hands and flops should get you an answer that is near enough accurate.

cardshark202

I could try write a program now to figure that out. If I have any luck I'll let you know.

BigDragon

Find that post from a while back that worked out the worst possible 'nuts' (it was a Qxxxx straight) and start there.

I cant be arsed tho.:p

Norwich Fan Rob

set of Qs = worst nuts.

BigDragon

Norwich Fan Rob wrote:

set of Qs = worst nuts.

Yeah thats right. Crap memory, knew there was a Q involved. Doh!

kincsem

RoundTower wrote:

An interesting question, little bit mathsy. If you play every hand to the river in a hold'em game, how often will you have the nuts on the river? This was posted on 2+2, I made a reasonable guess at it, but no one has given a conclusive answer yet.

Is that if everyone at the table plays to the river?

The chances of you having a full hand from seven cards (quads 0.2%, full house 2.6%, straight 4.6%, flush 3.0% = 10.4%).

I guess your straight or flush could be beaten .....

8 players against you would have equal chances of full hands. So I would give you one chance in 9 of the nuts with a 10.4% chance of a nut type hand = 1.115% or 85/1.

I've changed my mind on the odds a number of times. This is tricky. Does anyone have the answer? I've ignored QQQ, KKK, AAA as the nuts although we know these are the lowest possible nuts.

marius

If you are playing to the river it does not matter what cards you have - AA is as likely to be the nuts as 23o (ish....remember it is not the best hand at the table, it is about the 'nuts' hand). i.e any 2 cards are almost as likely to be the nuts as any other 2 cards.
Therefore the likelyhood of having the nuts would be = (the total number of possible hands)269/1.

You can say AK sooted is more likely to be the nuts than 23o but that is missing the point. 'We are asking how ikely is it you will have the nuts?' the answer - before seeing your cards is - as likely as any other possible hand you might have.

Now I am quite hungover so feel free to point out my idiocy.

RoundTower

Here's a summary of the replies so far, and what I think.

NeVer guessed 1 in 50. Fine for a guess but he must be running really good. He also suggested it was impossible to tell, some people just get lucky. However I am looking for the answer in the long run, not over one person's million hands.
Norwich Fan Rob said less than 1 in 50. This is a correct answer.
Doctor Fell said 1 in a million, and tried to justify this mathematically, but it would be kinder to call it a guess. Not even a good guess.
Marius guessd 1/220. said that if you hadn't seen your cards, the answer would be the same as if you did look at your cards every time. This is a useful way to think of it. He then guessed 1/269 based on the fact that there are 269 possible hold'em hands all equally likely to be the nuts. Not so, I think his first guess is closer.

Roryc guessed (?) 0.57% = 1 in 175. This is excellent.
Sean used some unusual mathematical reasoning based on the number of players at the table (which doesn't matter at all) and got 1 in 175, which I also think is very close. Then he changed his mind, 1 in 85 seems too much.

I don't know the answer, but I think 1 in 175 is very close. My analysis on 2+2 is here
http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=4436532&an=0&page=0&gonew=1#UNREAD
There is another thread on it in the "Probability" forum. There is an explanation of how to calculate the answer, but no answers yet:
http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=4436532&an=0&page=0&gonew=1#UNREAD

cardshark202

i was thinking about writing a program to solve this yesterday but then thought the time and effort and complexity of the problem meant it wasnt worth while. However I just read the thread on 2+2 and am now trying to write something to solve this. I doubt I'll have much luck though because my programming skills are fairly atrocius.

roryc

So that proves that I am a legend I was thinking it would be around 1 in 175

kincsem

The chances of you having a full hand from seven cards

Quads 0.2%
Usually the nuts, except when it on the board then you need an ace (had it once). It’s possible to have two sets of quads (Layne Flack quad tens v quad fours). Quads likely if a full house on board.

Full house 2.6%
Can't be the nuts if you hold a pocket pair. Someone else could have quads (board pair and their pocket pair). With a pair on board and you hold two different cards you may have the nuts (or a share). I don't know how many of your full houses are pocket pair; pocket two different cards. Let's say 50/50. Of the three unpaired cards on board do you have a pair to match the high, mid, or low card. So your nuts chances are 2.6% x 1/2 x 1/3 = 0.43%.

Straight 4.6%
Can be the nuts except where there is a possible flush. If there is a pair, then quads are nuts.

Flush 3.0%
Can be the nuts except where there is a pair, then quads are nuts.

The chances of a pair in five cards (the board cards) is I think 42.25% and that means quads are nuts.

Your chances of nuts are quads 0.2% + full house 0.43% + straight 4.6% x (100-42.25) + flush 3.0% x (100-42.25)

How good is your straight? You can have the top end, middle, or bottom ends Multiply your straight by 1/3. There are 1, 2, or 3 possible ways to fill in 3 board cards but there could be 3, 4, or 5 board cards that could be used in straights. I'll ignore the chance of being beaten by a flush.

How good is your flush. You need the ace. Assume you hold two flush cards with three on board. Chance you hold the ace is 2/10.

So multiply the above by the straight and flush nut odds.

Quads 0.2% + full house 0.43% + straight 4.6% x (100-42.25) x 1/3 + flush 3.0% x (100-42.25) x 2/10


Nut percentages - Quads 0.2% + full house 0.43% + straight 0.89% + flush 0.35% = 1.87% or 53/1.

Mr. Flibble

Full house 2.6%
Can never be the nuts as there must be a pair on board and that means quads are the nuts.

If I have 89 in my hand and the board is 88992 I have a nut fullhouse.

kincsem

Mr. Flibble wrote:

If I have 89 in my hand and the board is 88992 I have a nut fullhouse.

You are right, I missed that.

The chance of two pair on board is 4.75% It's 2/47*1/46 = 0.09% that you have both of them, or about 1 in a thousand. That case does not change the odds much.

And of course if you hold KJ to a KKJ flop, turn 4, river 7 you have the nuts, which can only be equalled.

I'm going to revise my earlier post.

mewso

So am I right in thinking the question is how often will you have the winning hand as opposed to the nuts. The nuts is the best possible hand someone can have with the board is it not.

BigDragon

musician wrote:

So am I right in thinking the question is how often will you have the winning hand as opposed to the nuts. The nuts is the best possible hand someone can have with the board is it not.

The 'nuts on the river' not 'winning hand' is my understanding of the question

mewso

Yes but what I'm saying is that the nuts is a possible hand that someone may have. On a board of 8c 8d kh 2d 3c a A8 may be the winning hand on the river but 88 is the nuts whether a player has it or not. Or are you saying that on the river the nuts reduces to the actual best hand out there?

kincsem

My reading of the problem is what are the chances of you ending up with a hand where you can say (not out loud) "I have the nuts, my hand can not be beaten" and bet all or call any bet knowing this.

Anyone out there like to try to figure it out and put their reasoning on boards. I feel like my efforts need to be either improved or dismissed by insightful comment :rolleyes:

If you don't I'll try to prove it's only 5/1 against having the nuts.

Mr. Flibble

The nuts is any hand which you know cannot be beaten (but can possible be equalled). This is not necessarily the best possible hand.

eg. on a board of 88992 the nuts are 89 or 99.

eg. board 6c 7c 9c kd jd
the nuts are
8c Tc or
5c 8c
Ac 8c

Doctor Fell

RoundTower wrote:

Doctor Fell said 1 in a million, and tried to justify this mathematically, but it would be kinder to call it a guess. Not even a good guess.

Norwich Fan Rob said less than 1 in 50. This is a correct answer.

I don't know the answer, but I think 1 in 175 is very close.

Hmmm, u must of missed the part where I said it was a guess!
U don't seem to know what u are talking about really, perhaps a little confused?? For example u seem to think 50/1 and 175/1 are equivalent answers.
Also, there's no way playing any random holding will result in getting the nuts i.e the best possible hand (quads etc) as opposed to merely a winning hand as often as once in 175 times. E.g 23o is going to flop quads or straight flush approx 1 in time in 175 (or whatever odds ur guessing urself).

I'd like to know the answer for sure, u seem to know a lot about it so when u get an answer I'd love to hear it. As opposed to some random guessing on your behalf.

I like Marius' way of thinking about it, but I think he answers a different question: his answer corresponds to if u consider any 5 card board, and u say what are the odds of a particular player having the nuts? E.g board of 10KAh XX -nuts is QJh obviously. Odds of someone having QJh is 220/1 but this is different than taking QJh and comparing it to random 5 card boards and determining when the nuts are hit. There's a huge amount of 5 card boards, and QJh will only hit the nuts in a very small % of them.

RoundTower

Doctor Fell wrote:

Hmmm, u must of missed the part where I said it was a guess!
U don't seem to know what u are talking about really, perhaps a little confused?? For example u seem to think 50/1 and 175/1 are equivalent answers.
Also, there's no way playing any random holding will result in getting the nuts i.e the best possible hand (quads etc) as opposed to merely a winning hand as often as once in 175 times. E.g 23o is going to flop quads or straight flush approx 1 in time in 175 (or whatever odds ur guessing urself).

I'd like to know the answer for sure, u seem to know a lot about it so when u get an answer I'd love to hear it. As opposed to some random guessing on your behalf.

I like Marius' way of thinking about it, but I think he answers a different question: his answer corresponds to if u consider any 5 card board, and u say what are the odds of a particular player having the nuts? E.g board of 10KAh XX -nuts is QJh obviously. Odds of someone having QJh is 220/1 but this is different than taking QJh and comparing it to random 5 card boards and determining when the nuts are hit. There's a huge amount of 5 card boards, and QJh will only hit the nuts in a very small % of them.

I don't understand what you are saying. I didn't understand what you were saying the first time, when you guessed one in a million. No, I still don't know the right answer. No, I don't think 50/1 is equivalent to 175/1 -- if you do, this may shed some light on how you came up with your answer. Yes, I do think 175/1 is close. As usual, I don't mind putting up a little bit of money on it, if you're interested. I'll only do this in the next day or so, or before a correct answer is posted with justification here or on 2+2.

padser

Lads there is a much simpler way to at least get part way towards the answer.

Assume the board pairs.

Now the nuts is (almost ) always a specific two cards. Its either another pair to make quads, or its going to be two cards to make a straight flush. (since im only trying to get very close to the answer im going to ignore the times when one card can make a straight flush).

the chances of having the two cards needed for the nuts in this case are
1/47 x 1/46 x 2 = 1 / 1031

Now this answer is also correct if the board is not paired but there are 3 or 4 or the same suit showing. THis is because on a board

Ad Qd 9d 5h 4h

the nuts strictly speaking is Kd Jd

not Kd Xd

Therefore if the board has a possible flush, or possible full house the chances of someone having the nuts is
1 / 1031

If you think about it thats a pretty big proportion of the time.


Also there is something absolute that can be deduced from this. The chances of you having the nuts MUST be greater then 1/1031 since this is the chance of having the nuts when its most unlikely you could have it.


Sorry if that didnt make a lot of sense, im happy that the reasoning is correct even if the english is a little shakey

RoundTower

padser wrote:

Now this answer is also correct if the board is not paired but there are 3 or 4 or the same suit showing. THis is because on a board

Ad Qd 9d 5h 4h

the nuts strictly speaking is Kd Jd

not Kd Xd

Therefore if the board has a possible flush, or possible full house the chances of someone having the nuts is
1 / 1031

No, Kd xd is the nuts here. In the same way, AK is the nuts on a AAK26 board, 74o is the nuts on a KKKKA board, and Ah5h (but not AhJh) is the nuts on a 3h4h7h8hQd board. In all these cases you could make a better 5-card hand if you had different cards. But you have the nuts if your hand cannot be beaten regardless of what cards your opponents hold.

If you don't like this definition of "the nuts", use it anyway, because that's what I'd like to figure out. Your maths is good, I have already done similar calculations if you follow the first link I gave.

ocallagh

When the board shows:

1 pair.......................: 2/47 X 1/46 (either st8flush or quads - 2 of 2 cards)
three of a kind, no ace........: 1/47 X 4/46
three of a kind, with an ace: 1/47 + 1/46
4 of a kind, no ace.......: 4/47 + 4/46
no str8 or flush possible....: 3/47 X 2/46 (top trips - 2 of 3 cards)
3 to str8, no flush............: 8/47 X 7/46 (top straight - 2 of 8 cards)
4 to str8, no flush............: 4/47 + 4/46 (1 of 4 cards twice)
3 to flush, no str8flush......: 1/47 X 9/46
4 or 5 to flush, no str8flush......: 2/47 + 1/46
3 to str8 flush..................: 2/47 X 1/46
4 to str8 flush..................: 1/47 + 1/46

Now you have the odds of holding the nuts for each type of board, you need to work out the chances for each of the possible boards above. (that is the hard bit!)

RoundTower

ocallagh wrote:

When the board shows:

1 pair.......................: 2/47 X 1/46 (either st8flush or quads - 2 of 2 cards)
three of a kind, no ace........: 1/47 X 4/46
three of a kind, with an ace: 1/47 + 1/46
4 of a kind, no ace.......: 4/47 + 4/46
no str8 or flush possible....: 3/47 X 2/46 (top trips - 2 of 3 cards)
3 to str8, no flush............: 8/47 X 7/46 (top straight - 2 of 8 cards)
4 to str8, no flush............: 4/47 + 4/46 (1 of 4 cards twice)
3 to flush, no str8flush......: 1/47 X 9/46
4 or 5 to flush, no str8flush......: 2/47 + 1/46
3 to str8 flush..................: 2/47 X 1/46
4 to str8 flush..................: 1/47 + 1/46

Now you have the odds of holding the nuts for each type of board, you need to work out the chances for each of the possible boards above. (that is the hard bit!)

How do you do the calculation here for 3 of a kind on the board, no ace? Are you assuming you need an ace to have the nuts?

ocallagh

RoundTower wrote:

How do you do the calculation here for 3 of a kind on the board, no ace? Are you assuming you need an ace to have the nuts?

yes, do you not need an ace for the best 5 card hand?

333K2 - u need the one remaining 3 (1/47) plus 1 of the 4 remaining aces (4/46)

5starpool

ocallagh wrote:

yes, do you not need an ace for the best 5 card hand?

333K2 - u need the one remaining 3 (1/47) plus 1 of the 4 remaining aces (4/46)

If you have the 3 with that board, then you do not need the ace for the nuts, any kicker will do since you cannot be beaten which I believe is Dave's definition of the nuts, not the technically highest hand. You are playing with the knowledge of your own cards and don't care about the other players hands at this point.

ocallagh

ah ok, i got this one wrong.. i'm calculating the chances of holding the highest possible hand

would this be closer?:

when board shows:

1 pair.......................: 2/47 X 1/46 (either st8flush or quads - 2 of 2 cards)
three of a kind........: 1/47 + 1/46
4 of a kind, no ace.......: 4/47 + 4/46
no str8 or flush possible....: 3/47 X 2/46 (top trips - 2 of 3 cards)
3 to str8, no flush............: 8/47 X 7/46 (top straight - 2 of 8 cards)
4 to str8, no flush............: 4/47 + 4/46 (1 of 4 cards twice)
3 to flush, no str8flush......: 1/47 X 9/46
4 or 5 to flush, no str8flush......: 2/47 + 1/46
3 to str8 flush..................: 2/47 X 1/46
4 to str8 flush..................: 1/47 + 1/46

RoundTower

ocallagh wrote:

ah ok, i got this one wrong.. i'm calculating the chances of holding the highest possible hand

Yeah I thought you must be, but where you calculate the odds of having the nut flush it seems you allow 9 ways of having it (any Ax suited) instead of only one (AK suited) which would give you the highest possible hand.

ocallagh

RoundTower wrote:

Yeah I thought you must be, but where you calculate the odds of having the nut flush it seems you allow 9 ways of having it (any Ax suited) instead of only one (AK suited) which would give you the highest possible hand.

yep.. that too.. thought i'd give writing a prog a quick bash.. but don't know if i could be arsed. maybe if i have nothing to do in work tomorrow

Shortstack

Would a better way not be something to do with how many possibe boards that each given hand can be the nuts on. For example

7 2o is only the nuts when the board is:

777xx
222xx
where the xx does not allow for str flush or higher quads.

If you can work out these and use it against the number of different boards possible and then calculate the odds of getting each hand you get a pretty accurate idea. No idea how to automate or show the calculation though. /requesting maths boffins.

RoundTower

Shortstack wrote:

Would a better way not be something to do with how many possibe boards that each given hand can be the nuts on. For example

7 2o is only the nuts when the board is:

777xx
222xx
where the xx does not allow for str flush or higher quads.

If you can work out these and use it against the number of different boards possible and then calculate the odds of getting each hand you get a pretty accurate idea. No idea how to automate or show the calculation though. /requesting maths boffins.

Also when the board is 4568x, etc., of the same suit as your 7. I don't know if this would be easier to do than the other way. I thought the other way. Maybe if me, Niall and cardshark all have nothing to do tomorrow one of us will produce a working program.

Shortstack

RoundTower wrote:

Also when the board is 4568x, etc., of the same suit as your 7. I don't know if this would be easier to do than the other way. I thought the other way. Maybe if me, Niall and cardshark all have nothing to do tomorrow one of us will produce a working program.

Oh yeah it was 3am.

Doctor Fell

Shortstack wrote:

If you can work out these and use it against the number of different boards possible and then calculate the odds of getting each hand you get a pretty accurate idea.

Ye, that's what I meant aswell. If you just calculate the total number of 5 card boards possible, and then work out on how many of these boards each particluar starting hand is the nuts.
Or a quick way of getting an approx answer: get the total number of boards, and pick an average hand (e.g Q7o). For this average hand work out on how many borads this is the nuts and you will have a close enough answer!

I'll give it a lash today when I get a chance, publish the maths and u can check my figures! Hopefully get it done today anyways...:v:

Doctor Fell

This is the way I did it.

There are 52 cards in the deck, so the number of 5 card boards is
52 C 5 (52 combination 5 e.g. nCr where n and r are numbers = n!/(r!(n-r)!).
This works out at 2,598,960 different boards.

Now considering an average hand like Q7o, this will be the nuts as follows:

1) qqqxx (where no straight flush obviously)
2) qq77x (x smaller than 7)
3) qq7xx (7 being 2nd highest card, no straight flush etc)
4) 777xx (as above)
5) 3456s, 4568s, 10JKAs, 910JKs

The number of these boards is as follows:
1) qqqxx occurs 49C2 times = 1,176. U have to take away xx when it is a higher pair, so it occurs 1,174 times.
2) Approx. 40 times (6 ways qq can appear, 6 ways 77)
3) 18 ways qq7, multiplied by the number of combo’s of 2->6 = 10, therefore 180 in total.
4) 1176 – any higher pair (roughly!) = 1169.
5) 48 times for each straight flush = 192 boards.

In total this gives us 2,755 times that Q7o will be the nuts out of a possible 2,598,960, or in other words its 943.4/1 (approximately!!!) that an average starting hand will be the nuts on the river.

If I’ve left anything out, I’m sure someone will point it out, but I think as a rough guide to the answer it’s good. Basically 1000/1 – I thought it would be higher because I guessed there would be a lot more potential boards.

bohsman

1) qqqxx (where no straight flush obviously)
2) qq77x (x smaller than 7)
3) qq7xx (7 being 2nd highest card, no straight flush etc)
4) 777xx (as above)
5) 3456s, 4568s, 10JKAs, 910JKs

5689s
10JKA os
AKxxs
AAAAK KKKKA 10JQKA s/os etc

Doctor Fell

bohsman wrote:

5689s
10JKA os
AKxxs
AAAAK KKKKA 10JQKA s/os etc

Oh yeah, thx. Crap, that'll make a big difference I think!! I'll try it now...

Shortstack

Doctor Fell wrote:

There are 52 cards in the deck, so the number of 5 card boards is
52 C 5 (52 combination 5 e.g. nCr where n and r are numbers = n!/(r!(n-r)!).
This works out at 2,598,960 different boards.

I may be wrong but is the number of possible boards not this:

52*51*50*49*48 = 311,875,200

Seems to obvious to be correct though...

roryc

Im telling you lads, its 0.57%...

it came to me in a dream.... and I forgot it in another dream