The example in this post (post #91):

http://forumserver.twoplustwo.com/sh...1&postcount=91

is an example that illustrates that there is a larger gap in pre-flop equities in nil poker. Player B will end up with more chips playing nil poker than regular poker because in nil poker there is a larger gap in pre-flop equities. (Eg. in nil holdem AA has more equity vs. 22 than it does in regular holdem.) Every hand that has a pre-flop equity advantage against another hand in regular holdem has a larger pre-flop equity advantage against the same hand in nil poker. This is why in that example Player B will end up with more chips playing nil poker than regular poker. (And of course we are all assuming that Player B will only call with hands that have greater than 50% equity vs. a random hand.)

So, I think that the following is a good question:

in that why does the fact that there is a larger equity gap between hands when playing nil poker necessarily mean that nil poker is more "skill-based"?

What people are you talking to OP? No offense but they are intellectually broke.

You have not proven the bold section. How do you not get this? You don't get to just say it's true.

There are a lot of assertions without any support in this thread that may or may not be true, but that particular assertion is actually true, assuming that in both scenarios Player B only calls with hands that have greater than 50% equity vs. a random hand*, which is obviously how he should be playing, and assuming that we're really talking about the expected amount of chips and not the actual amount (it may be possible that the actual amounts won't come out that way, although I would think the chance of the actual amounts not coming out that way would be very low).

I don't know if Original Poster knows how to do it (I do not), but I assume it wouldn't be difficult (for someone who knows how) to write a program to simulate the two scenarios and see in which one Player B ends up with more chips.


* Original Poster provided the following calling range for Player B: "every pair and 2 cards over 10". Player B would still win more chips in nil holdem than in regular holdem in this scenario with this calling range, but it is not the best possible strategy. Player B would be folding a bunch of hands that are profitable calls.

I am willing to accept that there could be ranges that if applied to both games make more money in one game than another, but I'm not going to accept, without seeing some kind of calculation or simulation, that if a player gets to select his own independent calling range in each game, that one game gives that player higher EV.

OK that was phrased terribly. If we get to choose our own best range for each game, can we prove that in nil poker the "good" player wins more?

(Note that even if we can prove this, it's really only part of a proof that nil poker is "more skillful" or "benefits the skilled player more", because it's a subset of how the game is played. I think it's a question of interest, though)

Originally, I think one of the assumptions was that the same calling range was used for both regular holdem and nil holdem. I'm sure that any chosen calling range of Player B, provided that each hand in such calling range has greater than 50% equity vs. a random hand in regular holdem, will have a higher EV in nil holdem than in regular holdem.


Now, I think that you are considering Player B using different calling ranges for regular holdem and nil holdem. The rational calling range for either game would include every hand with greater than 50% equity vs. a random hand under the respective game's rules and no other hands. If Player B determined his calling range for regular holdem and nil holdem separately by utilizing such method, his calling range for regular holdem and his calling range for nil holdem would be different. I am quite certain that if Player B determined his calling range at both nil holdem and regular holdem as described in the foregoing, Player B would have a greater EV at nil holdem than he would at regular holdem in this example. I would be willing to bet on it anyway. However, I am not capable of creating a program to simulate it and I am not going to learn how to do it.



I don't know that it is any proof that nil poker is "more skillful". It does illustrate that the differential between the equity of different hands is greater in nil holdem than in regular holdem. However, I don't know why that would then necessarily lead to the conclusion that nil holdem is "more skillful" or "more skill based". I don't even know what "more skillfull" or more "skill based" mean. That is why my initial posts in these threads was asking what these terms were intended to mean. See Post #8 in this thread. If "more skillfull" or more "skill based" was intended to mean having a larger differential between the equity of different hands (which would decrease the amount of "bad beats" that would occur - the favorite would win more often since on average favorites would be bigger favorites), then I would agree that nil poker is "more skillfull" or more "skill based" than its regular poker counterpart.

Holdem and omaha have more differences in the rules than nil poker has to its regular poker counterpart, but the equity differential between playable hands seems to be greater in holdem than it is in omaha. Is holdem "more skillfull" or more "skill based" than omaha?


EDIT:

I do think it may benefit the "skilled player" more, at least in the current playing environment. Most fish play badly in such a way that they call too much and chase too much. In nil poker their bad calls are even worse because they have less equity vs. the "skilled player's" "good hand" than in its regular poker counterpart.

Hello, Lego05.

Larger gap in pre-flop equities in nil poker vs poker is not the only thing that is important in my example(this is not only thing that example talks about). It is wrong if someone assums that only larger gap in pre-flop equities in nil poker vs poker(52-card deck) matters when we are trying to determine more skill-based poker variant. There are other things that are important, too. But as I mentioned, it is not important to search for cause of why some game is more skill-based than the other.

Answer to a question from different post: I do not know how to write programs.

Thank you for your post.

Hello, RustyBrooks.

I thought that this is obvious. I will try to calculate this.

Thank you for your post.

Hello people, I am trying to calculate this example(post no 91). I have a bit of a problem to calculate two things:

1. What are the odds for getting(as hole cards) any pair or two 10+ cards at poker(52-card deck) and at Nil poker(65-card deck), like mentioned in example(played as holdem)? Or at least how to calculate this?

2. What is average % equity vs random cards(also from example), if calling all-in with above mentioned cards at poker(52-card) and Nil Poker? If you do not know the exact number you can also write what you think that number would be(you can also write from which number to which you think)? I will probably calculate for more numbers(Nil poker number from this question can be calculated from poker(52-card deck) number from this question).

Your answers will really help me a lot. Thank you.

1. That's easy, there are 13*6 combinations of pairs and (4+3+2+1)*16 combinations of non pairs in the broadway square. Total of all combinations is 52*51/2. Total of all combinations with blanks added is 65*64/2.

2. There are a whole lot of diluted combinations added which should have less equity so the number can only go up unless that new rank is adding enough equity.

1). Did I get those numbers right?

- Odds for getting(as hole cards) any pair or two 10+ cards at poker(52-card deck):17,95%
- Odds for getting(as hole cards) any pair or two 10+ cards at nil poker(65-card deck):11,44%

If I calculated those odds wrong, please correct me.

If question 1 is answered, I would like to thank you for helping me.

That leaves me wondering just about question number 2. I have a subquestion to this question. Than I can finally calculate example(post no. 91).

2b)Subquestion to #2.)

If we pick perfect card selection(with which we will call other person´s allin) against allin-every hand at holdem (52-card deck), I want to know what kind of average % equity we would have against random cards(when calling allin)? If you do not know the exact percent number you can also write what you think that number would be(you can also write from which number to which you think)?

Thank you for all the help. Really appriciate it.

When I'm bored, I can tweak my simulator to answer #2.

Hello, heehaww.

You do not have to. Enjoy your free time instead But thanks man, I really appriciate it. I just find out how to do the first part of question 2(how to calculate exact number for poker(52-card deck)).

Thank you for your help.

Hello people, I am trying calculate answer to this question number 2.


Assuming we know first part of question no. 2(the exact number for poker(52-card deck); average % equity vs random cards(from example), if calling all-in with above mentioned cards at poker(52-card)). I am trying to calculate average difference for allin-player(player A from example) to win a hand(between poker(52-card deck) vs nil poker with all the rules from example).


1.)No of combinations at poker:1326=52*51/2

2.)No of combinations at nil poker:2080=65*64/2

3.)No of exactly same combinations at poker and nil poker=1326

3b): No of exactly same combinations at poker and nil poker vs all combinations:1326/2080=0,6375

4.).No of combinations with at least one nil card(at nil poker):2080-1326=754

5.)No of NN combinations:13*12/2=78

5b) No of NN combinations vs all combinations:0,0375

6.)No of combinations with exactly one nil card:754-78=676

6b.)No of combinations with exactly one nil card vs all combinations:0,325

Addition to 3)

(13 nil cards in the deck)/61(65-4(2 players have 2 cards))=0,213
1-0,213=0,787=rounded on 0,79 (because of nil quads combination)
Change of average% of winning for player A at nil poker vs poker :79% of existing at regular poker

Addition to 5.)
NN to complete to nil quads:25%
Change of average% of winning for player A at nil poker:75%

Addition to 6.)

Average from no. 1.) is 0,79.0,79/2(because allin player has one card instead of two;half less chance to win)=0,395… Because of nil quads combination we add 6%.. 0,395+0,06=0,455
Change of average% of winning for player A at nil poker:45,5%

Conclusion:

All the combinations:

1=point #3b)+ point #5b)+ point #6b)

1=0,6375+0,0375+0,325

0,6375*0,79+0,0375*0,75+0,325*0,455=
0,504+0,028+0,148=0,68

Average %Equity for poker(52-card deck;from example;with any pair or two cards over 10)=x
Y=1-x
Y*0,68= Change of average% of winning hands for player A at nil poker vs poker(under the rules from example)

Not related to example(from post 91), but just for better explanation:
If player A won at poker(52-card deck) 20%(not real number) of hands, he wins at nil poker 20%*0,68=13,6%

Is this correct? Have I done something wrong with my calculations? Maybe I used some term wrong, but I am wondering about a method.

Did you aprox. expected this number=0,68?


Thank you for help.

I do not know if this is correct, so I calculated it in a bit different way.I also changed some equity percents (rounded to favour poker(52-card deck)).

Calculations of example from post #91

Rules:

- 50bb each

- Every hand re-buy to 50bb

- Person A is allin every hand

- Person B calls (Person A´s allin), if he has any pair or two cards over 10

- Played as holdem(texas holdem and nil holdem)

- Rake is not included in calculation

- 10.000 hands played


Method:

1.)No of combinations at poker:1326=52*51/2

2.)No of combinations at nil poker:2080=65*64/2

3.)No of exactly same combinations at poker and nil poker=1326

3b): No of exactly same combinations at poker and nil poker vs all combinations:1326/2080=0,6375

4.).No of combinations with at least one nil card(at nil poker):2080-1326=754

5.)No of NN combinations:13*12/2=78

5b) No of NN combinations vs all combinations:0,0375

6.)No of combinations with exactly one nil card:754-78=676

6b.)No of combinations with exactly one nil card vs all combinations:0,325

Addition to 3)

(13 nil cards in the deck)/61(65-4(2 players have 2 cards))=0,213
1-0,213=0,787=rounded on 0,8 (because of nil quads combination)
Change of average% of winning for player A at nil poker vs poker :80% of existing at regular poker

Addition to 5.)

NN to complete to nil quads:25%
Person B wins 75 percent of hands.

Addition to 6.)

Average from no. A3.) is 0,8.0,8/2(because allin player has one card instead of two;half less chance to win)=0,4… Because of nil quads combination we add 5%..

Av. Equity:

Av. Equity for poker(52-cards) if calling with any pair or two over 10 cards vs random was calculated from total of all the mentioned hand equities vs random(against 1 player) and than devided by 23(number of calling hands). Mentioned average equity is 65,28%(for poker (52 card deck)).

NOTE:All the equities that I rounded are in favour of poker(52-card deck). If you see any wrong calculation(or if I missed anything), please feel free to tell. I will be more than happy to correct it. Thank you.

If you're giving them both the same range, then it's not a remotely interesting calculation. By definition the player playing nil poker needs to have a different range, because it's a different game.

Like, it should not be a surprise to you that playing the same range in stud high and stud hi/lo will not be a good idea?

Additions to calculations:

1.)With 25bb each



Diff EV(buy-in):4,218

2.)With 50 bb each



Diff EV(buy-in):18,202

Difference in buy-in(25 vs 50) is multiplier 2. If the poker would be same skill-based as NPoker there would be exactly x2 Diff EV(buy-in)for player B at Nil Poker(difference between A and B). If there is more than x2, the game is considered more skill-based(because both players are making ev+ calls; criteria from example). If this number is lower than 2, this means that Npoker is less skill based than poker.


18,202/4,218= 4,315;4,315 is more than 2

Nil Poker is more skill-based than poker (as holdem).

Hello, RustyBrooks.

Player B is making better selection(closer to GTO) at poker, than at Nil Poker.


Player B has more money in mentioned situation at Nil Poker vs poker. And that is why Nil Poker is more skill-based than poker.

Anyway, I agree that would be interesting to see results calculated in the way you mentioned, too(same good hand selection).

Thank you for your post.

I don't know if you're being deliberately obtuse or what, but looking at a single particular strategy, forcing the player to play the same strategy in 2 different games, and then trying to draw conclusions from that is pointless at the very least. Different games have different EVs on the same strategy. So what? This says nothing interesting about the game.

OP doesn't want to play a different game, he wants to play Holdem but doesn't want to get sucked out on so much. So to him, Nil Poker is just Holdem, only better (to him).

Right but if I'm playing someone heads up I am allowed to pick a different strategy than the one I'd play if I was just playing holdem.

Cmmon RustyBrooks.Be nice. I am here to discuss things.

If I do understand what you are saying. Player B should play optimal against all-in?
Player B should call every hand that has more than 50 percent equity vs random(in both cases:nil poker and poker)?

For example: For poker(52-card deck) I pick hands that have more than 50% vs one player… Total devided by number of hands..to get average equity?From this site:
http://www.natesholdem.com/pre-flop-odds.php

Anybody has any idea what hands should I take at nil poker that have more than 50% equity(same as in poker and which else?)? I would assume all the hands that have more than 45% at poker, they have more than 50 percent equity at nil poker?

I have to tell you that in order to calculations from example(post no#117 and #120) result in the way they did, there has to be prerequisite that what you are asking me to calculate will be even better for player B that plays nil poker(more bbs for him; because player B has worst hand selection at nil poker, that same player has at poker(52-card deck) at example post #117). And that is the only way that calculations from post #117 and #120 would result in the way they did.

Thank you for your post.