Poker Puzzle

Bozzer

Playing a heads-up NLHE freezeout with your very own bot.
You have Qc 5h

Board looks like this on the turn

Flop: As Td Jc

Turn : Kd (200)

You both have $X behind.
Bot open pushes.

You know from writing it’s code that it definitely has a Q, but that’s all you know.

You fold.

What’s the least amount (roughly) you had behind assuming you made the correct laydown.

No programs/calculators. You have 1 minute....bit dramatic?

ok... Can I ask why the laydown?

ArmaniJeanss

Hate doing quick maths in my head.
My answer is you had 200 behind to make the correct laydown.
Anything less and its a call.
Will doubtless be embarassed when I get out the calculator.

Bozzer

Papa Smut wrote:

ok... Can I ask why the laydown?

think about the freeroll

Bozzer

ArmaniJeanss wrote:

Hate doing quick maths in my head.
My answer is you had 200 behind to make the correct laydown.
Anything less and its a call.
Will doubtless be embarassed when I get out the calculator.

Ability to do quick head math is very important as you move up in stakes imo.
Not sure where you got 200 but the answer is much >.

Crumbs

Quick heads math gives me 388 ?

Lafortezza

Don't we need to know stack sizes and pot sizes?

Bozzer

lafortezza wrote:

Don't we need to know stack sizes and pot sizes?

pot size is 200. probably could have made that clearer.
the question is "how big do the stack sizes have to be for the fold to be correct?", which you can figure out.

Dave

Do you not have the nuts?

Amaru

You do, but its not the "pure nuts", Qx of diamonds would be the pure nuts at this point.

Bozzer

Dave wrote:

Do you not have the nuts?

we have the nut straight, but can you think of a hand you'd rather have on this board?

DapperGent

Bozzer wrote:

we have the nut straight, but can you think of a hand you'd rather have on this board?

I'd like AdQdAc2s.

Everything is more interesting when it's Omaha.

Rnger

Your fold is assuming he has two diamonds? If this is true, it makes calculations much easier.

9/45 = 1/5 the bot will win with his flush. You only stand to win $100 at best.
So if he raises... $401+ all-in you will be correct in folding?

When you dont know he has two diamonds, $X can be much higher to warrant a call.

Ste05

Bozzer wrote:

we have the nut straight, but can you think of a hand you'd rather have on this board?

The only hand I'd be afraid of would be Qxd, (although I'd love to have AQd on this board ) other than that there's nothing else I'd even dream of folding to. Trips etc. drawing to a house I'd call because of the statistical advantage you have to win the tournament. As you'll be unlikely to get a better chance to win the game against an opponent of the same skill level. (he knows all your moves and you his - after all you programmed it!)

So without doing any mathematical equations, or models, I'd look at the point that it's not worth giving him a chip lead. Obviously this isn't the way this question was meant to be approached because blind levels, etc. haven't been specified, but anyway. I'd say about x would have to be about 1000. Any less and the chip lead you'd be giving away would be very difficult to catch without catching cards (which I never like to rely on).

If he has a flush draw to go along with his straight then he's about 4:1 to catch it, so again about 1000 would make this fold correct????

Am I in the ball park, I'm actually looking forward to hearing the real answer properly reasoned out.

Rnger

Bozzer wrote:

You know from writing it’s code that it definitely has a Q, but that’s all you know.

My thoughts before are wrong, its all about figuring out the chances that it will have these two diamonds. For me not to call this... $X would have to be greater than 1000

jd

you don't know he has a flush.
1/3 he has qd and 10/45 he has another d
and 9/44 river is a d approx

66/1 he makes flush approx

so about 13k in chips?

Bozzer

jd wrote:

you don't know he has a flush.
1/3 he has qd and 10/45 he has another d
and 9/44 river is a d approx

66/1 he makes flush approx

so about 13k in chips?

these are the type of calc's i was looking for.
but can you see where you went wrong with your 13k figure?

jd

the most you win is 100 (not 200 silly me), but 1 in 66 times you lose the lot
so x=65*100= 6500

Ste05

Very good jd, nice calcs, although if you have anything like 6.5k, is it worth any risk to split this 200???

That's why I went alot lower??? Let him have this 200, I know no bot I programmed would make this massive overbet :rolleyes: :rolleyes:

Bozzer

jd wrote:

the most you win is 100 (not 200 silly me), but 1 in 66 times you lose the lot
so x=65*100= 6500

nice.
i used 1/13.5 of time he has diamonds and 1/5 of the time he hits
and rounded (1/13.5)*(1/5) to 1/68
so you needed 6700 behind.
your answer is closer to the correct one but tougher to do in your head.
I'm afraid this was just a boring old maths question alright, but you may have to apply it sometime.

A situation like this one came up in a heads-up game i was playing.
My opponent bet 3000 into a pot of 360 and I had to figure how often he had to be freerolling to make a fold correct
I got 30% doing the math in my head and folded!

I was 99% sure he had a queen and figured that he'd nearly always push with Qdxd, to get me to make a bad call with just a Q.
I thought that he'd push some % with just the Qd, just incase I was the type to fear the freeroll(sort of contradicts the last point).
And I thought that he'd almost never push just a Q because he's only get called by another Q and he'll be a split at best.
If you add all that up, then a fold was probably right.

ArmaniJeanss

Nice question as always Bozzer, keep 'em coming.
In my defence of my appalling bad answer, I read the board as showing 3 diamonds already. (which admittedly still wouldnt make my answer close to correct )

Bozzer

ArmaniJeanss wrote:

Nice question as always Bozzer, keep 'em coming.
In my defence of my appalling bad answer, I read the board as showing 3 diamonds already. (which admittedly still wouldnt make my answer close to correct )

Thanks, i'll try to keep 'em coming.
I figured you thought 3 diamonds were out.
Either that or you were doing some fvcked up 5th level thinking .

mewso

Nice question but perhaps bearing in mind newbies etc. you might ask how large does your stack have to be instead of how much do you have to have behind. I for one having played the game for 3 years basically guessed you were asking about stack sizes.

jd

I am quite good at maths (in my head) but am crap at poker...

Bozzer

musician wrote:

Nice question but perhaps bearing in mind newbies etc. you might ask how large does your stack have to be instead of how much do you have to have behind.

didn't even realise I wrote that until you pointed it out.
sorry for any confusion.

jd

Bozzer wrote:

nice.
i used 1/13.5 of time he has diamonds and 1/5 of the time he hits
and rounded (1/13.5)*(1/5) to 1/68
so you needed 6700 behind.
your answer is closer to the correct one but tougher to do in your head.
I'm afraid this was just a boring old maths question alright, but you may have to apply it sometime.

.

I guess 1/3*2/9*2/9=4/253=1/60 approx
or rrealistically 1/3*1/4*1/5=1/60 (aqveraging out)
*100= 6k