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View Full Version : help me with pot odds...does my bet count in the total?


whiskeytown
04-05-2005, 05:41 PM
don't know why after all these yrs. I'm still screwing this up.

Blinds are 100/200 -

whiskeytown: raises to 525
Anxiety: raises 1140 to 1665 and is all-in

Now, I read this as I have to risk 1140 to win 3555 (the blinds, plus all the money that will be in the pot if I put in 1140) - I read that as a little over 3-1 pot odds, so I called.

But someone else stated I'm risking 1140 to win 2400, (the total without my 1140)- so you are getting 2.1 - 1 odds.

(math numbers don't totally add up, but you get the jest)-

I thought I counted the money I was putting in the pot as part of the final pot odds total. Have I been screwing this up for yrs or what?

RB

gasgod
04-05-2005, 06:39 PM
Don't count the money you have to put into the pot to call. You can't win that money because it is already yours. Count only the money already in the pot. (Include all money you have put in previously.)


GG

mostsmooth
04-05-2005, 08:30 PM
[ QUOTE ]
Don't count the money you have to put into the pot to call. You can't win that money because it is already yours.
GG

[/ QUOTE ]
but once he puts it in the pot, its not his!!! this is quite the paradox!!! /images/graemlins/cool.gif
ok im kidding, dont count your bet as part the pot

BruceZ
04-05-2005, 09:19 PM
[ QUOTE ]
But someone else stated I'm risking 1140 to win 2400, (the total without my 1140)- so you are getting 2.1 - 1 odds.

[/ QUOTE ]

Correct. Pot odds are always (money already in the pot) : (the current bet to you).

[ QUOTE ]
Have I been screwing this up for yrs or what?

[/ QUOTE ]

Possibly, but it really depends on how you use this number. If you have 9 outs with 47 unseen cards, do you call that odds of losers:winners = 38:9 = 4.22:1, or do you look at it as a probability of total/winners = 47/9 = 5.22:1? If you use odds of 4.22:1, then you would compare that to your pot odds which does not include your bet as part of the pot, as described above. But if you use the probability of 5.22:1, then you could compare that to the "pot odds" (pot probability?) that you have been computing by including your own bet, and this would be an alternative correct method, though an unusual one.

RocketManJames
04-05-2005, 09:28 PM
Sometimes when numbers look daunting, I will think of things in simpler terms... what some would say obvious terms.

You have $1, and I have $1. No blinds in this heads up game. And, say that at this instant, you will win 45% of the time and I will win 55% of the time. I push you all in for $1.

If you were to include your money in the 'pot odds' then you are getting 2 to 1, which means you should call. But, you probably see that this is clearly incorrect.

By not including your money in the pot odds calculation, you are getting 1 to 1 (or even money), which means you should definitely not call. But, at this point, you probably know this.

-RMJ