Old Shadow post, weighted outs and more

[knockonwood]

I recently revisited this Shadow post from the July 06 digest.

http://forumserver.twoplustwo.com/sh...Number=6446486

Shadow goes into a lot of detail explaining the estimation of his outs after the flop. Particularly, about applying weighted averages to determine the number of outs he has.

"Everyone else calls, verifying my initial reads.
Now it’s time to evaluate my outs. We’ll start with a base of four and then work down from there. There’s likely a flush draw out there, but that doesn’t affect us. We’re against a straight, which doesn’t take up any of our outs. That leaves two passives and then a re-evaluation of MP, since his hand range does include non-flush draws that sometimes have some of our outs. The two passives likely have Ks, 9s, or As. There are either 6 straight outs or 7 straight outs, depending on Button’s hand. We can weight these distributions according to how likely it is for Button to have either hand. Just take a weighted average to determine how many straight draw outs are against you. This point doesn’t have any real bearing on our hand, but I think that it is marginally important as a note about how to calculate outs against you.

Anyway, I’m afraid that I’ve veered off on a tangent. The only hand combinations that we care about are hands in which some of our outs are taken. These hands are any Q or T hands. Since there are only four tens and queens out there, the probability of any one player holding one of our outs is severely diminished. Depending on the player, we’d have to narrow further these Q and T hands to hands in with another draw, such as T9 and KQ. This limiting provision applies only to MP in this hand, as the passives would likely play a Q or a T much the way they would play a draw, albeit less frequently (I think that this is an important, yet obvious point. The passives are more likely to throw away a pair than a draw. Additionally, there are more draw combinations out than pair combinations.) There is more intense math that you can do here to give a weighted out average based upon the hand distributions. I don’t think that it is important to do with this hand, especially since we’re drawing to the nuts. I just wanted to walk through what my mind is thinking in the five seconds or so right before I call, just making myself aware of what hand possibilities are out there, how these combinations affect my draw, my implied odds, and the possibility of not having a full four outs.

On the fly, I gave myself 3.5 outs. There are serious arguments out there to give the full four or to discount an entire out. Keep in mind, however, that if we do discount an entire out that our decision is the same on the flop, getting 15:1. The turn decision results in the same decision, although with a closer edge. "

Can someone explain to me, how the weighted outs would be calculated in this example?

[knockonwood]

bump - cmon you maths types - get your EV calculators out [img]/images/graemlins/smirk.gif[/img]

[kerowo]

From the quick scan of Shadows post (I only really read the one's on clothes) it appears that this is an extension of hand reading. So you put the folks in the hand on a hand then use the strength of your conviction to determine how good your outs are.

Since this is primarily based on your reads of the guys in the hand there isn't much to tell you as far as EV calculations go. The thing to think about is the process, the important part isn't how to calculate the equations, but on how to generate the variables.

[knockonwood]

[ QUOTE ]
From the quick scan of Shadows post (I only really read the one's on clothes) it appears that this is an extension of hand reading. So you put the folks in the hand on a hand then use the strength of your conviction to determine how good your outs are.

Since this is primarily based on your reads of the guys in the hand there isn't much to tell you as far as EV calculations go. The thing to think about is the process, the important part isn't how to calculate the equations, but on how to generate the variables.

[/ QUOTE ]

Yep, i understand what the variables are, the hand ranges and the likelihood of the villain having those hands. But, i'm not a maths person and im wanting to see a calc that includes the hand ranges and the % chance that the villain holds these hands, worked into the calc to get a more accurate estimation of the number of outs that hero has. I can remember Dr Matt and 00Snitch using the same variables to work out similar calcs for EV in posts a while back.

[Fantam]

[ QUOTE ]
Can someone explain to me, how the weighted outs would be calculated in this example?

[/ QUOTE ]

That was a very interesting and good post.

To calculate Shadow's estimate for his outs, you first need to put his opponent(s) on hand ranges where possible.

Shadow only felt that he could put the MP TAG on a tight hand range. Let us estimate that range to have been say AT,A9s,KQ,KJ,KT,QJ,QTs,JTs,T9s & 99.

Those hands had the following possible number of combinations, given that Shadow held Q [img]/images/graemlins/club.gif[/img]T [img]/images/graemlins/club.gif[/img] and the flop was T [img]/images/graemlins/heart.gif[/img]J [img]/images/graemlins/heart.gif[/img]Q [img]/images/graemlins/spade.gif[/img] :

(AT-8),(A9s-4),(KQ-8),(KJ-12),(KT-8),(QJ-6),(QTs-1),(JTs-2),(T9s-2),(99-6).

So, there were 57 different possible hand combinations that MP could have had with our estimate for his hand range.

Of these, 34 hand combinations contained either a Q or a T, and there was 1 hand combination that had both a Q and T.

So, the likelihood that MP had a Q or T was as follows:

(34/57 * 1) + (1/57 * 2) = 0.63 (of an out)

So, Shadow's estimate for his outs (adjusting for the likelihood or weighted average of MP holding 1 or more of his outs) would have been:

4 - 0.63 = 3.37 outs. Or, slightly less than 3.5 outs, as Shadow quickly estimated when playing.

Of course, if you adjust the hand range, which I assigned for MP, then your results will vary. Also if you can not estimate hand ranges for your other opponents, then you can not make a similar calculation for the likelihood of them holding one (or more) of your outs.

I hope that this helps. [img]/images/graemlins/smile.gif[/img]

[shadow.]

take the number of combos for each hand, take your outs against each set of combos, do a weighted average.

like:

against 6 combos- 3 outs
against 3 combos- 9 outs

(6/9)(3) + (3/9)(9) = 5 weighted outs.

holla

[shadow.]

ah, much better.

thanks fantam!

holla

[Fantam]

[ QUOTE ]
take the number of combos for each hand, take your outs against each set of combos, do a weighted average.

like:

against 6 combos- 3 outs
against 3 combos- 9 outs

(6/9)(3) + (3/9)(9) = 5 weighted outs.

holla

[/ QUOTE ]

Thanks Shadow. This method is actually more accurate than mine, because it considers whether catching one of your "outs" will be good or not.

I simply calculated the likelihood that MP had one of your "outs" and subtracted that from your otherwise assumed 4 clean outs.

However, if MP had QJ you would only have had 2 outs (not 3 as my calc. assumed), and you would have been splitting with QTs (so call that 1 out).

So revising my earlier calculation, we now have weighted out averages against the following possible MP hands:

AT =(8/57 * 3),A9s =(4/57 * 4), KQ = (8/57 * 3), KJ = (12/57 * 4), KT = (8/57 * 3), QJ = (6/57 * 2), QTs = (1/57 * 1), JTs = (2/57 * 3), T9s = (2/57 *3), 99 = (6/57 * 4).

Total = {(8*3)+(4*4)+(8*3)+(12*4)+(8*3)+(6*2)+1+(2*3)+(2*3 )+(6*4)}/57

So our weighted outs average becomes 185/57 = 3.25 outs.

[knockonwood]

Thanks guys. Great work.