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#1
08-01-2002, 10:56 PM
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Odds/Outs Question

I read on this forum many times how one can take possible outs and multiply those outs by 2 or 4 to get the rough % that you will make your hand on the turn and river respectively.

Example - you have 6 outs (overcards) so you roughly have a 12% chance of hitting your hand on the turn and a 24% on the river.

Now, if I do the card division:

I hold 2 cards - flop shows 3 so there's 47 cards left out there.

I need to hit 6 to make my hand.

47/6 = 7.83

What does that number mean? Wouldn't that roughly translate into the % that I would see one of my 6 cards on the next round?

I'm sure I'm just missing or overthinking here - but this is driving me nuts.
#2
08-02-2002, 12:06 AM
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Re: Odds/Outs Question

I'm trying to get a grasp on this stuff too, so this new forum is really timely. I'm sure I'll muck some of this stuff up, but here goes.

I think you've got this one backwards; it should be 6 of 47, or 6/47 = .128 or 12.8%.

And I think to get the chance of making it in two cards, you work with the chance of not making it, and subtract it from 1, like so:

1 - 41/47 * 40/46 = .241 or 24.1%

Which is pretty darn close to your estimate.

#3
08-02-2002, 12:42 AM
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Re: Odds/Outs Question

ah................

never tried it backwards.

Yes, I had read here where someone said that you can multiply your outs by 2 for the turn and 4 if you're going to the river and it works out great.

To take that one step further...

You can turn that information into how many bets you need in the pot to justify a call. It's pretty easy and once you learn a few you just memorize them.

Anyway, here's how it goes.

Let's say you have those same overcards. (not the greatest example as this is pretty much a no-brainer situation) AK to a rag board. You have a ton of callers - say 4 people in who all called your pre-flop raise so there's 9 sb's in the pot (4.5 BB's).

On the flop someone bets and you get a couple callers and it's your turn to act. You figure you may lose the player behind you so you can count on 7.5 sb's in the pot if you call.

You have 6 outs which = 12% to hit your hand on the flop. In order to make the 'right' call you want to know how many bets need to be in the pot to correctly justify this.

So, how many times does 12 go into 100?

100/12 = 8 that's how many bets you need in the pot to justify your call - now subtract your current bet and it makes 7.

So, you need 7 bets in the pot to correctly call here. You're getting 6.5 without your call that round so you can call with comfort. :-)

Not the most scientific method and of course there are often many other things to take into account besides just crunching some numbers, but it works pretty well.
#4
08-02-2002, 01:04 PM
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simplified

The odds of an event occurring are the same as the inverse of the probability of it occurring.

Ex.: An event has a 25% (or 1/4) chance of success. Then, the odds of it occurring are 4-1. So, in application, if you're facing a calling situation where success is 25% likely and are getting 5-1 odds on a given call, then it's a wise investment.

#5
08-02-2002, 02:36 PM
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Re: simplified

Actually 25% = 3-1.
#6
08-02-2002, 03:01 PM
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Yup.

If you know the probability and want to convert to odds, the relationship is as follows:

"Odds-1" = (1/Probability) - 1

If your hand has a 25% chance to win, its fair odds are 3-1, because 25% = .25 and 1/.25 = 4 and 4-1 = 3. If a hand has a 10% chance to win, its fair odds are 9-1.

#7
08-04-2002, 01:00 PM
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another super easy way....

Say you have 5 outs. Use this formula (where 'x' is your # of outs):

(50-x)/x

(50-5)/5 = 9

So you are a 9-1 dog here.

Easy enough!

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