The first lesson in the NLHE Foundations Course has been posted at this link. Please post the answers to your exercises in this thread.
Also, post your thoughts on this lesson, and ask me anything you want about it. =)
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The first lesson in the NLHE Foundations Course has been posted at this link. Please post the answers to your exercises in this thread.
Also, post your thoughts on this lesson, and ask me anything you want about it. =)
You hold 8s7s on a board of Ks8h5s.
Find the number of starting hand combinations for the following hands:
AK, AQ, AA, KK, A8, 76.
AK - 4 * 3 = 12
AQ - 4 * 4 - 16
AA - 6
KK - 3
A8 - 4 * 2 = 8
76 - 3 * 4 = 12
AK(12), AQ(16), AA(6), KK(3), A8(8), 76(12)
AK(12), AQ(16), AA(6), KK(3), A8(8), 76(12).
:)
My Answers
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK(12)
AQ(16)
AA(6)
KK(3)
A8(8)
76(12)
I had to do this a few times before getting the right answers it seems pretty complicated at first to remember all the calcuations, although I am going to do it a few more times could I have another example or two to work through today and I can PM you my answers I find it off putting seeing the answers in this thread
For this one, I was hoping that the people taking this course would collaborate a bit and talk about how they got their answers. If you want more examples to do, then feel free to make up your own and just post them here, and I'll check over your answers for you (and others can chime in to talk about it to).Quote:
Originally Posted by MrFerguson91
Also, let's make sure not to over-complicate what we're doing here. For non-pair hands, decide on the number of each card left in the deck and multiply those numbers. For paired hands, follow the 6310 shortcut where if there are none of the card shown it's 6, if one of the cards is shown it's 3, if two of the cards are shown it's 1, and if three of the cards are shown it's 0.
I'll give a few more examples here and include the answers in spoiler tags so you can look at them when you're ready:
We hold JT on QT9. Find the combinations for the following hands:
AK -Spoiler:There are four aces and four kings left in the deck, so we do 4 * 4 to get 16.
AQ -Spoiler:Four aces and three queens left in the deck (there's a queen on the flop). That gives us 4 * 3 = 12.
AT -Spoiler:Four aces, two tens (there's one on the flop and one in our hand), so it's 4 * 2 = 8.
QT -Spoiler:Three queens (there's one on the flop) and two tens (since there's one on the flop/one in our hand). 3 * 2 = 6.
AA -Spoiler:There are no aces out of the deck, so using the 6310 shortcut, there are 6 ways to be dealt AA.
JJ -Spoiler:There is one jack out of the deck, so using the 6310 shortcut, there are 3 ways to be dealt JJ.
TT -Spoiler:There are two tens out of the deck, so using the 6310 shortcut, there is 1 way to be dealt TT.
Cheers man I get it now seems really simple once it clicks into your brain, I'll practice more tonight while playing my MTTS and SNGs to get it to sink in even further, this is to basically narrow down ranges and to give a bit more of an insight into what villain might possibly be holding is that correct?
Well the idea is that it's more likely for someone to have some hands than others just based on which cards have been dealt. If we have AA on QT4 then it's easier for our opponent to have KK(6) than QQ(3), for example. Soon we're going to use this idea to do some fun stuff like decide if certain bluffs are profitable.
I'm doing 10-15 min drills per day of this exercises. It's not difficult, but i want to be really fast doing this.
Didn't explain my results because it's a very automatic process, but i'll do it next time.
Looking forward to the next lesson. Really hope to be soon.
Thanks and sorry for my english :p
Cool no probs :) Looking forward to the next part
Thanks Spoon. I have always struggled counting combos so this is really useful.
Are there any similar methods for quickly counting the different ways that an opponent could have 2 cards of a particular suit in the case of flush draws? The best I've come up with is:
Using spades for the example ((13-#of spades on board)*((13-#of spades on board)-1))/2
Not quite as easy to swallow!
Hey! We're going to get into that soon. It's easier to actually just count out the different hands that our opponent could have based on his play than trying to think about every single possible combination of cards that would give the guy a flush or flush draw.Quote:
Originally Posted by 1QDOG7
ou hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK - 4 aces x 3 Kings, 12
AQ - 4 aces x 4 Queens , 16
AA, 4 aces ... 6 ways using the 6310 rule
KK - 3 ways using the 6310 rule
A8 - 4 aces x 2 8s, 8
76 - 3 7s x 4 6s, 12
I'm completely distracted by the fact that when spoony quoted the '#' it became a != sign.
Show me how to use this magic!
(This course is amazing and I'm following along.)
I don't see what you're talking about and/or it didn't do it for me. PM me a screenshot?Quote:
Originally Posted by MadMojoMonkey
Hi, sorry I arrived late at class, but I will do the excersice anyways:
We hold 7s8s and the board is Ks8h5s:
AK(12): there's one K so there are 4 A in the deck and 3 K in the deck so 4*3=12
AQ(16): there are four A and four Q in the deck so 4*4=16
AA(6): there are 4 A in the deck so there are 6 combinations (6310 rule)
KK(3): there are 3 K in the deck so there are 3 combinations (6310 rule)
A8(8): there are four A in the deck and two 8 so 4*2=8
67(12): there are four 6 in the deck and three 7 so 4*3=12
I had no problem doing the excersices, and I find it really easy doing the math in my head quickly. How can we put this into practice at the tables? Should we do this after we range our opponents and do a qualitative equity, or there's a math process to calculate our equity?
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK(12)
AQ(16)
AA(6)
KK(3)
A8(8)
76(12)
I know I'm late to the game and answers are already posted, but I figured I'd take part anyways lol.
Hey I just want to point out that you shouldn't be trying to count combinations out for a range like this at the table. I tried to reference that in the actual lesson, but I just want to reiterate that here.Quote:
Originally Posted by matiusaa
However, we are going to use this as a tool during our study, and I'm going to show you how to study in a way that will give you a much better feel for poker. That's actually one of the major skills I want to give you for this course because so many people aimlessly study random topics without having a system in place for getting better and making more money.
Hey there's not really a time limit on this or anything, so don't feel like you're super behind or something. :cool:Quote:
Originally Posted by Kashkar
Feel free to just complete them at your own pace. I'm estimating that there will be 8-10 of these lessons for this initial course, and I'm going to do more stuff after that on different topics.
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK(12) - Four Aces, & Three Kings left in the deck - 3*4=12
AQ(16) - Four Aces, & Four Queens left in the deck - 4*4=16
AA(6) - 6310, No Aces blocked - 6
KK(3) - 6310, One King blocker - 3
A8(8) - Four Aces, & Two Eights left in the deck - 4*2=8
76(12) - Three Sevens, & Four Sixes left in the deck - 3*4=12
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK(12), AQ(16), AA(6), KK(3), A8(8), 76(12)
Nothing new to contribute here just moving along :P
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
1.) 4*3 = 12 AK(12)
2.) 4*4 = 16 AQ(16)
3.) AA(6) because of the 6310 shortcut
4.) KK(3) because of the 6310 shortcut
5.) 4*2 = 8 A8(8)
6.) 3*4 = 12 76(12)
You hold 8s7s on a board of Ks8h5s.
Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76
AK =12 (4x3)
AQ =16 (4x4)
AA =6 (6310 rule, nothing on board)
KK =3 (6310 rule, 1 on board)
A8 =8 (4x2)
76 = 12 (3x4)
Hey guys, these answers look good. Keep up the good work.
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK(12) 4-A left x 3-K left in the deck
AQ(16) 4-A left x 4-Q left in the deck
AA(6) using 6310 method no A showing leaving 6 combos
KK(3) using the 6310 method 1 K is showing leaving 3 combos
A8(8) 4-A left x 2-8 left in the deck
76(12) 3-7 left x 4-6 left in the deck
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands: AK, AQ, AA, KK, A8, 76.
AK - 4 As and 3 Ks (4*3) = 12 combos.
AQ - 4 As and 4 Qs (4*4) = 16 combos.
AA - Using 6310 = 6 combos of AA.
KK - Using 6310 = 3 combos of KK.
A8 - 4 As and 2 8s (4*2) = 8 combos.
76 - 3 7s and 4 6s (3*4) = 12 combos.
Quote:
Originally Posted by Rambam77
Good job guys. I know that this is something that some people have seen before, but it's pretty critical to be able to do this, and that's why I've included it as the first lesson.Quote:
Originally Posted by Cobra_1878
AK:12
AQ:16
AA:6
KK:3
A8:8
76:12
Hey if you think you've got this, then that's awesome. If you don't, then let me know in this thread.Quote:
Originally Posted by othd13
yeah, i think i got this, hope you check my exercises on course 2 soon. Very thank you soonoitnow!
AK: 12
AQ: 16
AA: 6
KK: 3
A8: 8
76: 12
looking over everyone elses answers - looks like i got them right :) weeeeee - its like being at school again
also thanks for all the effort youve put into this mate, i really appreciate it
No problem. I hope you enjoy it and make it through all of the parts of the course that are up.Quote:
Originally Posted by siriusisness
You hold 8s7s on a board of Ks8h5s. Find the number of starting hand combinations for the following hands:
AK(12), AQ(16), AA(6), KK(3), A8(8), 76(12).
AK(12), AQ(16), AA(6), KK(3), A8(8), 76(12)
AK(12)
AQ(16)
AA(6)
KK(3)
A8(8)
76(12)
off to #2
Good stuff guys.
I know that this lesson seems kind of straight-forward and maybe even trivial because of how relatively simple it is, but it's something that you really need to be able to do some pretty incredible analysis that we'll get into.
8s7s on a board of Ks8h5s.
AK = 12
AQ = 16
AA = 6
KK = 3
A8 = 8
76 = 12