AA is the nuts though

you're only interested in having the best possible hand,

I understand what you're saying but I don't think that's the way the question was being asked

So if I have A 6 on a 2 3 4 9 8 board that's not the nuts by their definition?

Yeah that's a pretty important distinction.

Really depends on how the question is meant to be interpreted.

The best possible hand on that board is 56

You can argue that you have the nuts, but they're interested in the best possible hand

They shouldn't use "nuts" then.

If I have A 6 on that board, I have the nuts. Full stop.

the puzzle setter is clearly not a poker player and it's ambiguous from what they've written what they mean

I think what they are going for is: which hand gives a lock for the most five card boards?

So lock = nuts, not best possible hand of all hands.
I'd be surprised if it's not one of: AA, AKs, JTs

I can't imagine it is JTs

Mostly because it is killed by AKs majority of the time

AKs vs AA seems tricky. Like if we say AKs do we get all 4 suits, or just 1 suit. That is a massive difference

Some sort of combinatorics problem where what is more likely a rainbow board with no straight and an ace, versus a flush board with nothing paired.

Typing that out makes me lean toward AKs

Any suited hand definitely can only mean one suit, not all 4 combos.

Is there an easy way to calculate this?

No tens, no fives.
One of KQJ, one of 234.
Two of 6789, but not all combinations are valid. (e.g. AJ982 doesn't work)

Due to the two-way nature of the ace, is it possible that KK is better? Tens and fives are legal (e.g. KT832).

So let's see if we approximate

AA trying for another ace

2*49*48*47*46*5 different ways this can happen=50850240

ATs trying for a flush

11*10*39*38*37*10 different ways this can happen=60317400

and this doesn't even include all the ways AA is beaten by a straight. ATs includes a few more hands where a straight flush is nuts vs AKs. I think I am going with ATs.

but the extra ways you can make a straight flush with ATs are not the "best possible hand"

Unless it's the royal your straight flush only uses the T and would be better with JT

if we're using their definition of the nuts

Yeah definition is flawed. By that definition I don't see where Asuited (face card)suited has any advantage over any other. But if you have Ac Tc and the board comes 9c 8c 7c 6c 2d you know you can't be beaten, even if Jc Tc would be the nuts

5c

https://fivethirtyeight.com/features...ss-the-street/

Looks like the answer was ATs. I'm surprised.

They even link to a 2+2 thread from 2010. Relevant quote here:

The craziest thing about those results is I wouldn't have guessed AKo over AA

poker is a game where you're only interested in having a better hand than your opponent. The nuts is nice but not necessary.

Let's look at the boards where AA is the nuts:

- 4 flush without pairs or straight flush draw
- AA on board
- TJQK without pairs or flush possibility
- Single A on board without straight, flush or full house possibilities.

Any time you get Aces Full someone else could have quads. None of the scenarios described above are likely at all

AKo still gets all the 4 flush possibilities and quads in some extreme cases but what I think pushes it ahead of AA is the fact that it can make broadway with only 3 cards to help it.

I also think that the puzzle misunderstands what the nuts is. When they said "best possible hand" they should have said "unbeatable hand" because you can guarantee that the 2+2 guy understands what the nuts is. Only way ATs is better than AKs is with the proper definition of what the nuts is because of the extra straight flushes you can make.

Also I would like to take credit for being the first person to suggest ATs (even if minutes later I decided that it was a bad answer).

Here's a nice puzzle that I stumbled upon today and was pleased with myself that I was able to solve it fairly quickly (So perhaps it's too easy for here!)

Puzzle

There is a pannel with a 3 digit code and you want to crack the code efficiently. You can try as many times as you want and if you've no digits in the right place then you get told "FAIL" and if you have 1 or more digits in the right place then you get told "CLOSE" (Even if all digits are correct it will say "CLOSE")

e.g. If the code is 140 then 197 would get a result of CLOSE, 140 would get a result of CLOSE and 014 would get a result of FAIL.

What is the optimal strategy to minimize the number of tries you need before knowing the code? Using this strategy what is the highest number of tries it might take you before you know what the code is?

How about this :

Minimum tries


Maximum tries

Quick answer. If it's wrong I'll give it more thought.

I'm pretty sure you're both +1 on the low end and maybe on the high end

I think I have a strategy for

I don't understand why you guys are answering min and max? I think we're just looking for one number that minimizes the maximum number of possible turns.