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AmericanAirlines
01-31-2003, 10:32 PM
Hi Everyone,
In a 2+2 book I had a long time ago, I seem to recall David putting forth an equation to convert percentages into odds.

Can anyone recount it for me?

Sincerely,
AA

AmericanAirlines
01-31-2003, 10:37 PM
Hi Everyone,
Was this it? Seems to work to me:

<pre><font class="small">code:</font><hr>
100 - Percent
-------------- = Odds:1
Percent

E.G. for 2%:

100 - 2 98
-------- = -- = 49 to 1
2 2
</pre><hr>

Sincerely,
AA

Ray Zee
02-02-2003, 01:23 AM
how about 1 divided by the percent chance. you get a number then deduct one from that and you have your odds to one.
example 1 is divided by .25 =4 so your odds are 3 to 1. get it.that works when you are the dog.

but the best way is to take yor chances say .25 and divide it into the difference from 100. example .75 divided by .25 =3
.60 divided by .40 = 1.5 to one

AmericanAirlines
02-04-2003, 03:39 PM
Hi Ray,
I'd never seen that first technique before!

Thanks for confirming my little formula with your second example.

Here's my current "formula list" after noodling with the numbers for a few minutes. Let me know if you see any blatant errors.

<pre><font class="small">code:</font><hr>Basic Formulas

Change Percent to odds:1

(100-%)
-------- = odds:1
%

Change Outs to Percent

Outs
---------------- = %
Cards Left - outs

Change Outs to Odds

Cards Left - outs:outs

Change raw odds to odds:1

Cards Left - outs
----------------- = Odds:1
outs</pre><hr>

Sincerely,
AA

Bozeman
02-05-2003, 02:28 AM
Actually, %=OUTS/CARDSLEFT.

Craig

AmericanAirlines
02-06-2003, 08:56 PM
Good point, guess I was getting cross eyed.

Thanks for keeping me straight.

Sincerely,
AA

lowroller
01-24-2004, 04:12 AM
Ray,

Where does the ".60 divided by .40" come from?

.60 divided by .40 = 1.5 to one

Homer
01-24-2004, 01:21 PM
For every .6 times you win, you will lose .4 times. So, your odds against are .6:.4. To reduce this to X:1, divide both sides by .4. .6/.4 = 1.5. So, .6:.4 is the same as 1.5:1.

-- Homer

lowroller
01-24-2004, 02:27 PM
I ran into a situation last night that I have run into before, and I would like to know how my new found math skills might be able to help me...

Had KK, one before the button. I raise, everyone already in calls (I'm last to act)...4 to the flop. The flop comes A-7-2. Now I am in a 3-6 game with a 10K jackpot where EVERYONE plays A-whatever, so the Ace on the board makes me uneasy.

Now I know that with me not having an Ace, it is roughly 60% that the other 3 do not have an Ace either (memorized the Caro table). So there is a better than even-money chance that no one has the Ace (better than 1:1) and there is 6BB in the pot.

Excuse my ignorance, but I am not sure how to use the %/odds information as it relates to the pot in a situation like this. I know how to compare pot/drawing odds in situations where I am drawing a flush, straight, etc.

How can I use this information to help me proceed in a hand like this??

Thanks!

wdbaker
01-24-2004, 06:46 PM
the way i would use it is by first realizing that the odds of your improving on the turn with 2 outs is 4% or by river 8% which is equivalent to 24:1 or 11.5:1 respectively. You don't have the odds to draw out so your only options are to fold if someone bets out or raise their bet to see if they come over the top at which point you fold or check-call to the river as long as know one is getting wild.

I personally would think about folding if their was any chance I was the underdog so would look for everyone to check it around to me then bet out and hope they all fold /images/graemlins/smile.gif
else drop in the muck /images/graemlins/frown.gif

That may be weak tight but with the jack pot and all, thats how I'd play it.

I'm sure someone will tell you i'm wrong and thats ok too, thats how i learn /images/graemlins/grin.gif