Am I flawed, or is Phil Gordon and the calculator...

[Eclypse]

[ QUOTE ]
Excellent explanation.
You should change your name to skalansky jr.

BTW i think some of the calculators out there
such as at cardplayer.com dont actually do the calculations the way you did. They just run a simulation of 10k hands or so and give you an approximation. I have used the calculator before with identical problems and the solution had a fractional change in the answer. such as 54.3% one time and 54.1% the next.

[/ QUOTE ]

We might want to hold off on the Sklansky jr thing--I missed 6 more combinations that give us a flush and our opponent a full house:

Ah 9s
Ah 9c
Ah 9d

Ah 8s
Ah 8c
Ah 8d

...reducing our total of winning combinations from 557 to 551 which means the corrected winning percentage for us is 551/990 = .55657 or approximately 55.66%

D'oh!

[revots33]

I think you're getting confused by the fact that the pair of Aces is currently ahead. A drawing hand can still be a favorite if there are enough outs (in this case 15) and enough chances to get them (in this case 2).

Imagine if the game of holdem had 10 board cards instead of only 5. Now, Phil would have SEVEN more chances to get one of his 15 outs. The villian still has 30 "outs" to win, and Phil only 15. Even by the last card, the villian would still have more outs than Phil - but Phil would be a huge favorite to hit one of his outs at some point along the way.

Don't get confused by counting "outs" for the hand that's ahead. Just focus on the outs for the drawing hand. In this case the drawing hand has approximately a 56% chance of hitting one of his outs by the river. This leaves the villian approximately 44% chance of dodging those outs on both the turn and river.

[Eclypse]

I decided to run this through PokerStove and it came up with EXACTLY my original calculation which was 56.263% (see my first post above: “Clear Explanation”)

This was before I realized I had left out 6 more hands that make us a flush, but also make our opponent a full house. I posted these in the post immediately above this one.

So, now I don’t know what to think. I am pretty sure these 6 hands must be removed from our winning combinations, but that would put our winning percentage at 55.657% which doesn’t agree with PokerStove.

Could PokerStove be making the same mistake I was making in my original calculation, or am I missing something?

[Eclypse]

[ QUOTE ]
I decided to run this through PokerStove and it came up with EXACTLY my original calculation which was 56.263% (see my first post above: “Clear Explanation”)

This was before I realized I had left out 6 more hands that make us a flush, but also make our opponent a full house. I posted these in the post immediately above this one.

So, now I don’t know what to think. I am pretty sure these 6 hands must be removed from our winning combinations, but that would put our winning percentage at 55.657% which doesn’t agree with PokerStove.

Could PokerStove be making the same mistake I was making in my original calculation, or am I missing something?

[/ QUOTE ]

Never mind, I see ALL KINDS of mistakes I made. I will repost the corrections when I get it done.

Sorry everyone!

[Eclypse]

Sorry about all my previous mistakes (I should never do this kind of calculating before I wake up in the morning). Anyhow, I re-wrote it with the correct additions and subtractions--AND it agrees with PokerStove (great program!).


You have two cards to come, so the first thing we need to do is calculate how many 2-card combinations there are, and then determine how many of these 2-card combinations give us a winning hand.

There are 52 cards in a deck. We’ve seen seven, so that leaves 45 cards that can be dealt off the deck to give us the first of two cards (the turn card). Once that card has been dealt, we have 44 cards that can be dealt to give us the second of two cards (the river card).

So, we have 45 * 44 = 1,980 ways to get the final two cards. However, if you get say, 2h 4s, it is the same two cards and same value as 4s 2h, it’s just switched in the order it came off the deck, so we must divide 1,980 by two to give us 990 total 2-card combinations.

Of these 990 combinations, how many will deliver a win for us?

Well, we can get any of the following:

Two running Jacks to make three Jacks
Two running Tens to make three Tens
Any Seven for a straight
Any Queen for a straight
Any Heart for a flush
A TJ gives us two pair

Anything else would not help us and give the win to our opponent.

Since we have the Jack-of-Hearts, we can get running Jacks three different ways:

Jc Js
Jc Jd
Js Jd

Likewise, since we have the Ten-of-Hearts, we can get running Tens three different ways:

Tc Ts
Tc Td
Ts Td

So far, that’s 6 combinations that help us.

We can also catch a Jack and a Ten to give us two pair. Here are the possible combinations:

Tc Js
Tc Jc
Tc Jd

Ts Js
Ts Jc
Ts Jd

Td Js
Td Jc
Td Jd

Which gives us 9 more winning combinations.

As far as any 7, Q, or heart (for the straight or flush respectively), we don’t have to get two running cards as long as at least one of these cards is dealt out of the next two.

The easiest way to calculate how many of the 2-card combinations contain one of these cards is to calculate how many combinations there are that DO NOT contain one of these cards, then subtract this number from 990 (our total universe of 2-card combinations).

Going back to our original 45 unseen cards, we know that four of these are Sevens and four of these are Queens—there are nine hearts, but since we’ve already counted the Seven-of-Hearts and the Queen-of-Hearts, there are only seven more hearts to be counted. This gives us a total of 15 cards to be removed from the original 45 leaving us with 30.

(30 * 29)/2 = 435 combinations that DO NOT contain a Seven, Queen or Heart.

Which means there are 990 – 435 = 555 combinations that DO contain a Seven, Queen or Heart.

But there’s one more fly in the ointment: We can make our flush and still lose if it contains the Ace-of-Hearts coupled with any of the remaining aces, kings, eights or nines as well as the King-of-Hearts coupled with either of the two remaining kings:

Ah Ad gives us a flush but also gives our opponent four aces.

The following combinations give us a flush but also give our opponent a full house:

Ah Kh
Ah Ks
Ad Kh
Ah Kc
Ah 9s
Ah 9c
Ah 9d
Ah 8s
Ah 8c
Ah 8d
Kh Ks
Kh Kc

This means we have to subtract these thirteen combinations from our total:

555 – 13 = 542

The total combinations that help us is 542 + 6 (the running pairs) + 9 (the JT combinations) = 557
(The number of combinations that don’t help us is 990 – 557 = 433)

Our winning percentage therefore, is 557/990 = .56263 = 56.263% or approximately 56.3%

[masonx]

obviously jack doenst understand general probability. there is no point.

[jackaaron]

Great explanation. I definitely agree that in this case, Phil would have the advantage. Obviously, this hand won't come up too many times. But, in the past, I wouldn't have been quite so confident with it. Still, it's almost a coin-flip. I don't play ring games, but if I did, I think about the long run in terms of winning, and I would play this hand. Since I play tournaments and sit n gos, I play in a snapshot of time, and therefore, you can't consider the odds of what will happen over the course of 100,000 hands because you're only playing 1-200 hands...lol. I do quite well in sit-n-gos, but in tourney's I haven't won yet (I've only been playing 2.5 months). In the last ten tourney's I've played, I've finished in the top 20 percent or better, so I'm getting there, just not far enough obviously. Anyhow, good job everyone.

[Crooked Paul]

Jack, I don't blame you for being confused. Unless you've ever taken a course on probability, it can be pretty unintuitive.

If you want to help your game in the long run, though, I would suggest that you do a little research into the subject, and not just about poker probability. If you understand the fundamentals of probability, you'll have a much firmer grasp on what's happening in a given poker hand. You'll notice special cases and won't overcount your outs, etc.

Here's a tidbit to get you started. At its most basic level, the probability of a given event is the number of successful outcomes divided by the total number of possible outcomes, or

P = successes/total

All the rest is just mathmatical ways of making it easier to tally that up.

Be careful with your language, too. Thinking of those 30 cards that "help" the villain as "outs" isn't doing you any favors. Really those are BLANKS, or cards that help neither player.

Did a quick Google search to get you started, and I found this step-by-step probability tutorial:
Basic Probability
(http://cne.gmu.edu/modules/dau/prob/...cprob_frm.html)

It looks like it was designed back in 1998, but the concepts are there. Hope this helps.


Crooked Paul

[jackaaron]

I'm right on it...