Pocket Tens POLL: Raise, Call, or Fold?

[Malachii]

According to PT, I'm losing money with pocket tens.. I either don't have a large enough sample size or am misplaying them... I suspect the latter.

Party Poker 0.5/1 Hold'em (10 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)
UTG+1 is aggressive, but has reasonably solid starting hand requirements.

Preflop: Hero is UTG+2 with T/images/graemlins/heart.gif, T/images/graemlins/diamond.gif.
UTG calls, UTG+1 calls, <font color="#CC3333">Hero raises</font>, <font color="#666666">3 folds</font>, CO calls, Button calls, SB calls, BB calls, UTG calls, UTG+1 calls.

Flop: (14 SB) 3/images/graemlins/club.gif, 6/images/graemlins/heart.gif, K/images/graemlins/diamond.gif <font color="#0000FF">(7 players)</font>
SB checks, BB checks, UTG checks, UTG+1 checks, <font color="#CC3333">Hero bets</font>, CO folds, Button folds, SB calls, BB calls, UTG folds, UTG+1 calls.

Turn: (9 BB) A/images/graemlins/club.gif <font color="#0000FF">(4 players)</font>
SB checks, BB checks, <font color="#CC3333">UTG+1 bets</font>, Hero....

[reubenf]

[ QUOTE ]
I either don't have a large enough sample size or am misplaying them... I suspect the latter.

[/ QUOTE ]

I'm willing to bet it's both.

[stlip]

This is one of the top 10 + EV hands, slightly stronger even than AKo.

In my PT database I've got 110 PP Ts, which have won 56 percent of the time and netted an average of 1.2 BB per hand.

But with so many people in a raised pot any overcard is likely to hit someone and A is by far the most dangerous since so many people called the original raise that at least a few of them had to have an A as the reason they stayed.

[Malachii]

Yeah, my database only has 22 instances of pocket tens in it. I've only got about 5,000 hands in it. Thus far, I'm losing approximately .57 big bets a hand with it.

[cmwck]

I suspect you don't have a large enough sample. Unless you just happen to have 200K hands of pocket tens laying around somewhere.

[Kumubou]

In a field of four players, where you know 6 of the 52 cards, the odds of one player holding one of the three remaining aces is:

(43/46) * (42/45) * (41/44) * (40/43) * (39/42) * (38/41) =
.935 * .933 * .932 * .930 * .929 * .927 =
.581

Roughly, even if your opponents are playing any two cards, you are sunk roughly 58% of the time. However, this does not take into account the fact that goofballs that these small limits love to play Ace-rag, s00ted or not (even good players will typically play Ace-rag suited... in an unraised pot, anyway). However, the K/images/graemlins/diamond.gif is also out, making it about 67% likely (roughly) that you are drawing to two outs.

67% of the time you are drawing to two outs, so you win this pot ~90% of the time when you hit a T on the river (the T/images/graemlins/club.gif gives someone an unlikely backdoor flush, and sometimes someone will be holding QJ for the straight).

.67*.04*.09 to win 15 net bets (17 minus 2, and that's if everyone just calls). Value of .002 BBs.
.67*.96 to lose two bet. Value of -1.286 BBs.
.67*.04*.01 to lose two bets minimum (and you would be raising this river, right?). Value of &lt;-.001 BBs.

33% of the time you win this pot.. uh... 70% of the time? I'm not quite sure what the odds are, as many goofy things could happen (backdoor flush; a 2,4,5, 7, or T for a random straight; a Q or J for another overcard, the board pairing to give someone trips or a boat; etc. None of these are likely but they all have an adverse affect.)

.33*.7 to win 15 net bets (that's with everyone calling). +3.47 BBs
.33*.3 chance to lose two bets. -.198

Comes out to a value of 1.988 BBs -- but that's under near-optimal conditions (and some very faulty math, probably). Given the affinity that people have for ace-rag, and the fact that UTG+1 woke up with the ace (not as much of a concern as it seems though -- I've seen lots of players bluff at aces and just check-call with weak aces), I would figure that you are behind much more often than not. However, you may be ahead here just enough to push on... but you are in serious trouble if someone likes their hand enough to raise.

The conclusion that the analysis gave me struck me as non-obvious -- my first instinct with two overcards and a LP-P waking up like that is to dump it. :shrug: Maybe there's something grossly off with my math which, when found, will show this to be very much -EV. Or not.

Looking at the poll results and the thread, everyone is saggesting to dump it (which I agree with). So now I am curious as to where my math/analysis is off. Is putting everyone on any two /way/ too lose? Should I redo this with only average or better hands, or just any two suited + two big cards + ace-rag?

-K

[Dave G.]

Your calculations here are slightly off. The correct way to calculate this is...

p(at least one ace) = 1 - p(no aces)

p(at least one ace) = 1 - (43/46) * (42/45) * (41/44) * (40/43) * (39/42) * (38/41) = 0.349144

The multiplication you did above comes out to be 0.650856, or 1 - 0.349144. Not sure where the .581 came from. So the true answer is 35%.

The fact that people love to play ace-rag doesn't factor in here. They have to be dealt it first before they can play it.

I'm not sure what's going on with the rest of your calculations here. How do you get the 67% figure?

The probability that someone has a K here is 0.605797. this comes from using 6 opponents. 6 opponents saw the flop and the flop will have affected their decision to stay in the hand. So whether they have a K on the turn or not is no longer a purely random event. On the flop however, it is.

Put these two together and there is, at minimum, a 71% chance that you're beaten. I for some stupid reason picked "call" in the poll, but I just woke up. Good reminder of why not to play poker when you're tired. /images/graemlins/wink.gif

[Kumubou]

[ QUOTE ]
The multiplication you did above comes out to be 0.650856, or 1 - 0.349144. Not sure where the .581 came from. So the true answer is 35%.

[/ QUOTE ]

.581 is the odds of five players not having an ace. Oops, twice. For miscounting the number of players and forgetting to subtract that from one.

[ QUOTE ]
The fact that people love to play ace-rag doesn't factor in here. They have to be dealt it first before they can play it.

[/ QUOTE ]

It should -- it means that the fact that they are in the hand at all is not a random event. With this model, you're right it does not matter -- but in reality a random Party Poker .5/1 player is much more likely to come in with Axo than, oh, 53o.

[ QUOTE ]
I'm not sure what's going on with the rest of your calculations here. How do you get the 67% figure?

[/ QUOTE ]
It was a lazy approximation from the previous number -- the odds of no ace and/or no king being dealt just being the initial number (for aces) being multiplied by itself and then subtracted from one. In reality, the odds are a bit worse (40/46*39/45...*35/41).

[ QUOTE ]
I for some stupid reason picked "call" in the poll, but I just woke up. Good reminder of why not to play poker when you're tired. /images/graemlins/wink.gif

[/ QUOTE ]

And I've been up for about 30 hours. &gt;_&lt; That probably explains a lot.

-K

[cmwck]

I don't see how you guys are calculating this stuff.

The probability that none of 5 players has an ace on the turn is:
43/46*42/45*41/44*40/43*39/42 = .70224, correct?

[Kumubou]

I think I did this just because some pooh-bah in another thread complained about the lack of EV calculations, and decided to give it a shot.

Then I ran it through PokerStove, and I need to lay off the crack:


<font class="small">Code:</font><hr /><pre> equity (%) win (%) / tie (%)

Hand 1: 13.6669 % [ 00.14 00.00 ] { ThTd }
Hand 2: 28.8015 % [ 00.28 00.01 ] { Any pair, any two suited, any two broadway, any Ace }
Hand 3: 28.7802 % [ 00.28 00.01 ] { Any pair, any two suited, any two broadway, any Ace }
Hand 4: 28.7514 % [ 00.28 00.01 ] { Any pair, any two suited, any two broadway, any Ace }</pre><hr />

2 bets to win ~12-15 bets? Fairly safe to say you can dump it. Now if these people are really playing any two, you actually have a pot equity edge... but I doubt that is the case, even at these low limits.

-K

[Dave G.]

[ QUOTE ]
It should -- it means that the fact that they are in the hand at all is not a random event. With this model, you're right it does not matter -- but in reality a random Party Poker .5/1 player is much more likely to come in with Axo than, oh, 53o.

[/ QUOTE ]

This is true to an extent, but it's very difficult to quantify this in a meaningful way. I think going with straight probabilities is the only way to get a useful result. Beyond that it's guessing, and if we start guessing like that we might as well not do the math.

[ QUOTE ]
It was a lazy approximation from the previous number -- the odds of no ace and/or no king being dealt just being the initial number (for aces) being multiplied by itself and then subtracted from one. In reality, the odds are a bit worse (40/46*39/45...*35/41).

[/ QUOTE ]

Ah righto. That's where I got the 71% number from.

[ QUOTE ]
And I've been up for about 30 hours. &gt;_&lt; That probably explains a lot.

[/ QUOTE ]

Jeez. I hope you're not playing at the moment. (psst... If you are, which table? ha ha, kidding...)

[Dave G.]

[ QUOTE ]
I don't see how you guys are calculating this stuff.

The probability that none of 5 players has an ace on the turn is:
43/46*42/45*41/44*40/43*39/42 = .70224, correct?

[/ QUOTE ]

You have the right idea, but you're forgetting that each player gets dealt two cards. Everyone gets two chances to be dealt an ace.

So the probability that noone has an ace out of 5 people on this board is (43/46 * 42/45) * (41/44 * 40/43) * (39/42 * 38/41) * (37/40 * 36/39) * (35/38 * 34/37) = .470356

[cmwck]

[ QUOTE ]
[ QUOTE ]
I don't see how you guys are calculating this stuff.

The probability that none of 5 players has an ace on the turn is:
43/46*42/45*41/44*40/43*39/42 = .70224, correct?

[/ QUOTE ]

You have the right idea, but you're forgetting that each player gets dealt two cards. Everyone gets two chances to be dealt an ace.

So the probability that noone has an ace out of 5 people on this board is (43/46 * 42/45) * (41/44 * 40/43) * (39/42 * 38/41) * (37/40 * 36/39) * (35/38 * 34/37) = .470356

[/ QUOTE ]


Ahhh... I see
So now we can find the probability that at least one person out of three has an ace or a king on the turn:

1-p(nobody has an ace AND nobody has a king)=
1-( (40/46*39/45) * (38/44*37/43) * (36/42*35/41)
= .59

Therefore our hand is only good 41 percent of the time on the turn against 3 random hands.
Did I do this right?

[Dave G.]

Your calculations are spot on.

However, you have to take into account the fact that 6 opponents saw this flop. Whether these opponents continue onto the turn is going to depend on the flop.

That is, the probability that somebody has a king on the turn is no longer a random event. Someone with a king will have seen the king on the board, realised they have top pair, and are now almost certainly going to continue to the turn. Someone who has a pair of 7s may see the king and decide to throw it away.

For these situations I normally just calculate the probabilities based on the number of people who saw the flop. You can do a sort of weighted probability for the turn, I guess. I might spend a bit of time figuring out the details of that.

[KingOtter]

[ QUOTE ]
Your calculations are spot on.

However, you have to take into account the fact that 6 opponents saw this flop. Whether these opponents continue onto the turn is going to depend on the flop.

That is, the probability that somebody has a king on the turn is no longer a random event. Someone with a king will have seen the king on the board, realised they have top pair, and are now almost certainly going to continue to the turn. Someone who has a pair of 7s may see the king and decide to throw it away.

For these situations I normally just calculate the probabilities based on the number of people who saw the flop. You can do a sort of weighted probability for the turn, I guess. I might spend a bit of time figuring out the details of that.

[/ QUOTE ]

Actually you need to take into the fact that 10 people were dealt cards pre-flop. And you need to add in the chances of a K being out there, too, because you lose to it, too.

KO

[Dave G.]

[ QUOTE ]
Actually you need to take into the fact that 10 people were dealt cards pre-flop.

[/ QUOTE ]

No, we don't have to take this into account. All we are determining here is the probability that a certain card can be found at a certain position in the deck before any of the cards are dealt.

Lets say we have a 4 handed game. If there is a king in deck positions 1 to 8, someone will be dealt a king. Now move to a 10 handed game. If there is a king anywhere within deck positions 1 - 20, someone will again be dealt a king. It's more likely that a king was dealt overall because more cards are being dealt.

However, the probability that an individual player gets a king is not affected by this. A king is no more likely to appear in deck position 1, 5, 23, or 51, or anywhere else in the deck. It's a random event. It doesn't matter how many people you deal cards to; you could deal the entire deck. The probability of someone getting a king before the flop never changes.

The reason we consider the number of people who saw the flop is because once the flop is laid out, the probabilities all change significantly. Additionally people make decisions whether to fold or continue based on the flop. When we get to the turn, a persons likely holdings is no longer a purely random event (well, for most players anyway...)

One might argue that someone is more likely to see a flop with a high card. However, as I stated in a previous post, I don't believe that such conjecture is useful. We don't know anywhere near enough about our opponents to begin to even estimate how this affects the probability of them seeing a flop. Making guesses and assumptions and making up probabilities and trying to plug them into mathematical equations will yield a whole load of nothing.

[ QUOTE ]
And you need to add in the chances of a K being out there, too, because you lose to it, too.

[/ QUOTE ]

This is where the 71% figure I used comes in. 6 cards beat us (3 kings, 3 aces) with 4 opponents. I stated it's a minimum because someone with a king who sees this flop will almost always see the turn.

[k000k]

[ QUOTE ]

However, the probability that an individual player gets a king is not affected by this. A king is no more likely to appear in deck position 1, 5, 23, or 51, or anywhere else in the deck. It's a random event. It doesn't matter how many people you deal cards to; you could deal the entire deck. The probability of someone getting a king before the flop never changes.

[/ QUOTE ]

True, but you also must also consider that players dealt aces and kings are more likely to see the flop than players dealt rags.

[Homer315]

I skipped a large part of the odds discussion here because this is an example where doing it is stupid. Everyone knows people are far more likely to be playing big cards. The chances that no one has EITHER an ace OR a king is about 10 or 15 percent, MAYBE. We all know the hero is beaten here, and he only has two outs. Easy fold.

You can't use odds to describe a hypothetical situation and then act on that situation based on your hypothetical odds without taking into account reality. This reminds me of guy in vegas who had the King high flush and was raising and re-raising HU on the river to like 6 or 7 bets, because, as he put it, "there's only one card in the deck that beats me (the ace), and what are the chances my opponent has it?" Of course, the answer, statistically, is 1 - (44/45 * 43/44) which equals 4.5%. but you have to put the odds into the context of what happened in the hand. Three opponents, one guy is betting into the PF raiser with an A and K on board. Why is he doing that? purely on a flush draw bluff into three opponents?

[Malachii]

Thank you everyone for your contributions.

Party Poker 0.5/1 Hold'em (10 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)

Preflop: Hero is UTG+2 with T/images/graemlins/heart.gif, T/images/graemlins/diamond.gif.
UTG calls, UTG+1 calls, <font color="#CC3333">Hero raises</font>, <font color="#666666">3 folds</font>, CO calls, Button calls, SB calls, BB calls, UTG calls, UTG+1 calls.

Flop: (14 SB) 3/images/graemlins/club.gif, 6/images/graemlins/heart.gif, K/images/graemlins/diamond.gif <font color="#0000FF">(7 players)</font>
SB checks, BB checks, UTG checks, UTG+1 checks, <font color="#CC3333">Hero bets</font>, CO folds, Button folds, SB calls, BB calls, UTG folds, UTG+1 calls.

Turn: (9 BB) A/images/graemlins/club.gif <font color="#0000FF">(4 players)</font>
SB checks, BB checks, <font color="#CC3333">UTG+1 bets</font>, Hero folds, SB folds, BB calls.

River: (11 BB) 8/images/graemlins/club.gif <font color="#0000FF">(2 players)</font>
BB checks, UTG+1 checks.

Final Pot: 11 BB

Results in white below: <font color="#FFFFFF">
BB has Qs 6s (one pair, sixes).
UTG+1 has 9h 9s (one pair, nines).
Outcome: UTG+1 wins 11 BB. </font>