How accurate is Clonie Gowen's estimate?

[nightlyraver]

I read this little tip in Cardplayer the other month and I was wondering exactly how accurate it is. So here it goes:

I'm pretty certain that this tidbit was authored by Clonie Gowen. She advises that you can estimate your probability of making your hand using a very simply technique. Simply multiply your outs, assuming they are all good, by 2 and then multiply the product by the number of cards to come. For example, if you have a flush draw on the flop and nothing else, multiply your 9 outs to the flush by 2 (18% to make it by the turn) and then multiply that by 2 (36% to make it by the river). This sounds pretty head on for the flush draw since every says you have about a 35% chance to make your flush by the river. Does this technique work all the time assuming your outs are good? Exactly how accurate is this in other cases? Can you use this technique as-is if you have discounted outs?

[LetYouDown]

It's a decent, beginner's rule of thumb. It does give you a reasonable approximation...assuming all of your outs are live and there are no redraws. For instance, if you have an open ended straight on a flop with 2 of a suit, and someone holds two overcards of that suit, you're presented with a problem.

Long story short, if you're just looking for an estimation of making whatever hand you're drawing to...it's helpful. If you're looking for it to be the best without another occurence (another flush card coming, or a board pair, etc.)...it's likely to be off by a significant margin.

[nightlyraver]

OK, so let's take this one step further. Going back to my example with the flush draw on the flop, could we use this technique to approximate the probability of the flush being made by the river AND winning with the flush (assuming we lose when the board shows a 4-flush)? I guess we would also need to lower that percentage a little since we lose as well when top pair makes a runner-runner full house, but that can't happen enough to really affect this approximation, can it?

[LetYouDown]

[ QUOTE ]
but that can't happen enough to really affect this approximation, can it?

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The problem is that it doesn't take into account what other players likely hold. If you're up against a set and a higher flush draw, you could very well be drawing dead. So I'd say from 36% to 0% is pretty significant.

[nightlyraver]

I think you misunderstand. What I meant was the following: How, if possible, can we adapt this estimation to reflect 1 card of your suit hitting but not 2. For example, say the flop is Ks7h2h and Villian has AhKd while you have something like Th9h. We win when 1 heart falls, but we we see runner-runner hearts. What I meant when I said, "that can't happen too often" was the same scenario except that the board shows Ks7h2hKhAc where we also loose. Like I said, I can't see this hapenning often enough for it to affect any sort of approximation, correct?

[LetYouDown]

Assuming you hold a four flush...I think this is right.

Odds that ONLY one of the remaining cards will be of your suit:
[C(9,1) * C(38,1)]/C(47,2) = 342/1081 = 31.6%

Odds that BOTH of the remaining cards will be of your suit:
C(9,2)/C(47,2) = 3.33%

Which would put her estimate off by a few percent...but that's obviously not what the rule of thumb is for.

[jccookjr]

The only time it won't be real close is when you have lots of outs.

[nightlyraver]

OK, so tell me more. Say you flop a monster draw - an open-ended str8-flush draw w/ 2 over cards. That's 21 outs I believe to top pair or better. Assuming they are all good outs, do we not have about an 84% chance of making top pair or better (not including redraws)?

[LetYouDown]

You hold K/images/graemlins/heart.gif-Q/images/graemlins/heart.gif and the flop is: J/images/graemlins/heart.gif10/images/graemlins/heart.gif2/images/graemlins/club.gif

I come up with

1 - C(26,2)/C(47,2) = 1 - 325/1081 = ~70% that you'll make at least a pair of Q's.

[nightlyraver]

OK, so I'm a law student and we are renowned for being bad at math (else we would be med. students). Can you briefly explain how to calculate these probabilities? Thanks!

[LetYouDown]

Typically for questions like the hand above, it's much easier to calculate the probability of missing and then subtract from 1. There are 21 cards that will improve you to a pair of Q's or better. There are 47 cards left in the deck.

So basically, to miss entirely, the turn/river need to be 2 of the remaining 26 cards that don't improve your hand. There's C(26,2) ways for this to happen. There's C(47,2) possible ways for the turn/river to come.

So, in English it's:

1 minus the probability that I'll get 2 cards from the 26 cards that don't help me divided by the total number of ways the remaining cards can come.

[Bulldog]

You need a heart for a flush, but your opponent needs two hearts for a higher flush, correct?

You have 9 outs twice, or a 36% chance (estimate based on this method) to get there. Odds of runner runner flush are about 4%*. So 36% - 4% equals 32%.

*Actually 4.16% when you know five cards (the three on the flop and the two in your hand) and three of the suit are accounted for in those five. In this specific scenario, where we know seven cards and five of that suit are accounted for, it's 2.83%.

[Absolution]

Pretty sure this is right, but I plotted this very quickly:

http://www.freestandingentertainment.com/trueest.jpg

i.e. With 2 cards to come you start really overvaluing your hand after about 8 outs. With 1 card to come you're always undervaluing your hand.

[ismisus]

Very accurate. The figures will be correct if you count your outs right. If you have a flush draw, you have 9 outs, 9*2*2 = 36%. If the board is paired, I give myself 5-8 outs depending on the amount of people/betting patterns.

If you have a straight draw, you have 8 outs, 8*2*2 = 32% after the flop. After the turn you have 8*2*1 = 16% chance of hitting your straight.

If you have a pair, you have 5 outs of improving to a 3 of a kind or two-pair. 5*2*2 = 20% after the flop.