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

[jackaaron]

It's just, thinking of it conceptually as marbles (earlier post analogy by another), with 30 blue marbles and having to get two, and 15 red, and Phil only having to get one of the 15, I feel like I like my chances of picking two marbles when 30 of 45 marbles are blue is pretty good.
Maybe this is okay to say?...
Phil has a one in three chance of winning, twice, and the villian has a one in three chance of LOSING, twice.

[jackaaron]

Okay, on this a little more...in the example, essentially you're saying the villian needs to get running cards.

So, if you have to get two running cards, but there are THIRTY available, under what circumstances are you NOT going to bet?

[Siegmund]

[ QUOTE ]


In this example, Phil has 15 outs. 8 outs on the draw, plus the remaining 7 outs of the suit.However, the villian has 30 outs.

[/ QUOTE ]

An out is a card that *improves* your hand. It does not follow that any card that is not an out for you is an out for someone else.

Right now, the pair of aces is ahead. 15 cards can come that will make the straight-flush draw hand better than the pair of aces.

Of the other 30 cards, most of them do absolutely nothing for either player. The few cards that improve villain's hand - king of clubs for example - still don't improve his hand so much as to make him unbeatable. In your example villain has, in effect, no outs at all.

Change it slightly and give villain [img]/images/graemlins/club.gif[/img]A [img]/images/graemlins/diamond.gif[/img]A . Now there are 14 cards that make hero's hand better than villains (the same 15 as before, minus the ace of hearts), but there are also 7 turn cards that destroy hero's chances by improving villain to quads or a full house, and leave hero only two outs on the river to make the straight flush.

If you wanted you could call this "14 outs vs. 7 outs" and approximate hero's chances by ( 14x38 + 24x14 ) / (45x44) = 43.8% - pokerstove says they are 42.1% - but usually you will only hear "outs" used when one player has a good but beatable hand and the other player is behind but has a chance to catch up.

[jedi]

[ QUOTE ]

Maybe this is okay to say?...
Phil has a one in three chance of winning, twice, and the villian has a one in three chance of LOSING, twice.

[/ QUOTE ]

Phil has a 1 in 3 chance of winning, twice.

Villian has a 2 in 3 chance of getting another 2 in 3 chance of winning. He needs to dodge Phil's 1 in 3 chance twice.

[Marm]

You are counting the outs wrong too.

Yes there are 15 total outs, but there are NINE for the Flush, and SIX for the straight.

EDIT: I misread the originalpost at first. Yes, the count is right, but it is confusing to some. Always count outs to the stronger hand first, jsut easier that way.

[jackaaron]

But, if the villian doesn't need to improve his hand, then those 30 cards actually do help him because they don't help the other guy.

[jedi]

[ QUOTE ]
But, if the villian doesn't need to improve his hand, then those 30 cards actually do help him because they don't help the other guy.

[/ QUOTE ]

Yes, but he has to hit them twice. If you have a 2/3 chance of hitting something, but have to hit it twice, what's the odds that you will do so?

[Mike Haven]

here's another way to think of it

because Phil has 15 outs out of 45 cards he has a 33% chance of hitting one on the turn

if he misses, he has another 33% chance of hitting one on the 66% remaining hands on the river, or another 22%

therefore, when he's all in, he has 33% + 22% = 55% total chance of hitting one of his 15 outs on the turn or the river

(in round figures)

[Eclypse]

[ QUOTE ]
Phil says this (backed by online calculators at, for example, twodimes.net et al):
If Phil has:
J [img]/images/graemlins/heart.gif[/img]T [img]/images/graemlins/heart.gif[/img]
and the villian has:
A [img]/images/graemlins/spade.gif[/img]K [img]/images/graemlins/diamond.gif[/img]
and the board is:
A [img]/images/graemlins/club.gif[/img]9 [img]/images/graemlins/heart.gif[/img]8 [img]/images/graemlins/heart.gif[/img]

Phil says he has a 56.3 percent chance of winning. Otherwise, a little more than half the time he should win, overall.

In this example, Phil has 15 outs. 8 outs on the draw, plus the remaining 7 outs of the suit.

However, the villian has 30 outs. Why 30? Because we realize there are seven cards in play, 52-7=45. Phil has 15 outs, therefore, every other cards aside from his outs helps the villian since the villian currently is winning with a pair of aces. If cards from our 30 outs come, we win.

Please explain to me what I'm missing (or why I'm a moron), and tell me why the villian doesn't have 30 outs since all of those cards that don't help Phil helps the villian. I appreciate your help on this!

Also, if you could keep it with what is currently on the board, that would help. I realize that if a T comes on the turn, the villian loses a few outs because a J on the river would give him two pair.

[/ QUOTE ]

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 the 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

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.

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 are 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 Tens—there are nine hearts, but since we’ve already counted the Seven-of-Hearts and the Ten-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, Ten or Heart.

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

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

Ah Ad which gives us a flush but our opponent four aces
Ah Ks which gives us a flush but our opponent a full house
Ah Kh which gives us a flush but our opponent a full house
Kh Ks which gives us a flush our opponent four kings

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

555 – 4 = 551

The total combinations that help us is 551 + 6 (the running pairs of jacks or tens) = 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%

[mindflayer]

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.