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  #1  
Old 09-16-2004, 01:35 AM
Vic Ferrari Vic Ferrari is offline
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Default Question on odds

I'm confused by other posters' math.

i.e. you have 10c10h and the board is A 2 3 5 and you are trying to calculate the odds of either of the two remaining tens coming on the river. Let's say your opponent turns AK face up.

Isn't the chance (2 out of the remaining 44 cards) -- 2/44?
But I often see posters calculate 2/42 here.

In other words, don't the burn cards have nothing to do with the equation because a ten is equally likely to be in the burn as any other card?

Sorry for the simple question.

Thanks
Vic
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  #2  
Old 09-16-2004, 02:25 AM
uuDevil uuDevil is offline
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Default Re: Question on odds

In the instance you describe, your math looks right to me.

Wild guess, but maybe the confusion is due to giving the result in terms of odds instead of a probability:

A probability of 2/44 is the same as odds of 42:2, or 21:1 against.
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  #3  
Old 09-16-2004, 02:31 AM
AncientPC AncientPC is offline
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Default Re: Question on odds

It's a difference between fractions and ratios (aka odds).

Edit: uuDevil beat me to the punch.
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  #4  
Old 09-16-2004, 10:14 AM
Vic Ferrari Vic Ferrari is offline
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Default Follow-up

Thanks for the response uudevil,

The most recent post I noticed that confused me (and from which I took my example) was Nightwish's TT post in the Mid-Limit forum.
" Preflop : Nightwish is UTG+1 with T [img]/images/graemlins/club.gif[/img], T [img]/images/graemlins/heart.gif[/img].
UTG calls, Nightwish raises, 2 folds, MP3 (TAG) calls, CO calls, 1 fold, SB calls, BB (Loosie) calls, UTG calls.

Flop Turn : (12 SB) 5 [img]/images/graemlins/club.gif[/img], 2 [img]/images/graemlins/diamond.gif[/img], 3 [img]/images/graemlins/club.gif[/img] (6 players)
SB checks, Loosie bets, UTG folds, Nightwish raises, TAG 3-bets, CO folds, SB folds, Loosie calls, Nightwish caps, TAG calls, Loosie calls.

Turn : (12 BB) A [img]/images/graemlins/spade.gif[/img] (3 players)
Loosie checks, Nightwish bets, TAG raises, Loosie folds "



Analysts's second response in that post is this:
[ QUOTE ]


Quote:
--------------------------------------------------------------------------------

He has AK or AQ clubs.

Fold.


--------------------------------------------------------------------------------



AK is unlikely as TAG would 3-bet PF, but even if he showed you these cards, you'd have to call.

2/42 * 450 = $21.43 for the two T outs (and you'd probably win another 1-2 bets on the rivers)

3/42 * 225 = $16.07 for the non-club 4s

Gotta call.

[/ QUOTE ]

So I'm still a little confused by the 2/42 when there are 44 remaining possible cards to come on the river.

Thanks!
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  #5  
Old 09-16-2004, 11:44 AM
AncientPC AncientPC is offline
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Default Re: Follow-up

I'm not sure where he's getting 2/42 from.

2 outs, 6 cards seen makes for 2:44 or 2/46. I think he's knocking off another 2 by making the assumption that TAG has AK and not a T in his hand.
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  #6  
Old 09-16-2004, 01:48 PM
uuDevil uuDevil is offline
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Default Re: Follow-up

Well nobody's perfect. Even BruceZ (resident probability expert) has been known to make mistakes. Books, magazine articles, and web pages have errors in them too. The answers in the back of my college probability text were sometimes wrong. Posts on these forums are not immune to error.

If you see what you think is an error in a post, it's probably best to ask the poster about it so that he can either verify that it is an error or explain why it is correct. More so of course if it matters to the discussion-- no reason to nitpick (not saying you are).
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  #7  
Old 09-16-2004, 03:00 PM
Analyst Analyst is offline
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Default Re: Question on odds

In this particular instance, "anonymous" poster just had a temporary brain-freeze. 2/44 is right.
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  #8  
Old 09-16-2004, 03:11 PM
ThaSaltCracka ThaSaltCracka is offline
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Default Re: Question on odds

here is an easy, but sort of rough estimate on odds. On the flop, count how many outs you have, then times that by 4, that is roughly the % that one of those outs will come on the turn. If it hasn't shown up on the turn, then take your number of outs and times by 2, thats the % that one of your outs will come on the river. Its a rough estimate, but close enough to be meaningful.
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  #9  
Old 09-16-2004, 06:44 PM
Vic Ferrari Vic Ferrari is offline
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Default Re: Question on odds

Thanks for the response ThaSaltCracka,

[ QUOTE ]
On the flop, count how many outs you have, then times that by 4, that is roughly the % that one of those outs will come on the turn.

[/ QUOTE ]

I assume you mean: one of those outs will come on the turn OR river. At least then the equation works out for me.

Nice little shortcut.

Thanks!
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  #10  
Old 09-17-2004, 04:42 PM
ThaSaltCracka ThaSaltCracka is offline
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Default Re: Question on odds

hehe, I swear I always leave one little thing out, but luckily for me, most people on here are smart enough to know what I mean. Yeah exactly, that would be the % that you are going to hit one of your outs on the turn and river.
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