Odds question...
 

Hi guys, I am trying to explain to my friend pot-odds, and I can't give him a good explanation of why when you calculate pot odds, you don't include the chances that someone else at the table either has one of your outs or it was already folded. Can someone give me a quick and easy explanation for why you do not include cards that were dealt to other people to adjust your outs, and please dont say because there is an equally likely chance that they do have the card as if they don't. Thanks.

-Obi-

Re: Odds question...
 

OK...let's do a flush draw. You have T [img]/images/graemlins/diamond.gif[/img]9 [img]/images/graemlins/diamond.gif[/img], board is A [img]/images/graemlins/diamond.gif[/img]Q [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/diamond.gif[/img]2 [img]/images/graemlins/heart.gif[/img].

There are 9 diamonds left in the deck and 46 unknown cards. So you have a 9/46 chance of hitting.

Your friend is confused about counting outs because your opponent might have one of your diamonds?

OK...he MIGHT have one of your diamonds. He MIGHT have two of your diamonds. He MIGHT have none of your diamonds. You have no idea what he has, so you just have to count the cards your opponent has with what is left in the deck.

Re: Odds question...
 

He understands pot odds Kyro. Your answer didn't really explain why you don't take into account the probability that other people at the table could have been dealt one of your outs.... He really wants to know why there is not adjustment for the possibility that since there are so many cards that are out of the deck in the muck pile or in someones hand, why you just ignore that...

Re: Odds question...
 

Yes I did.

[ QUOTE ]
OK...he MIGHT have one of your diamonds. He MIGHT have two of your diamonds. He MIGHT have none of your diamonds. You have no idea what he has, so you just have to count the cards your opponent has with what is left in the deck.


[/ QUOTE ]

Let me try it another way.

Let's say you have 9 outs with 46 cards to come. It's a 10-handed game. So on average, the other 9 people at your table will have 3.5 diamonds in their hand. So there are on average 5.5 diamonds left, but now you only have 28 cards left to come. 5.5/28 = 9/46. Is that easier?

Re: Odds question...
 

it doesn't really make a difference whether the cards are in someone else's hand or not. what is the difference between some of the outs being in someone else's hand and some of the outs being at the bottom of the deck? either way, you are not going to hit those outs, and you should think of them in the same way. you'd only take into account what other people are holding if you actually KNOW what they have for a fact.

in other words, saying "maybe somebody else is holding some of my diamonds" is no different than saying "maybe some of my diamonds are at the bottom of the deck." so you just count them both.

Re: Odds question...
 

28x2=56, not 46
haha

Re: Odds question...
 

you are good at math. so obviously you'll be able to agree with me when i say that 28+18=46.

Re: Odds question...
 

He explained it as best as it could be explained I think. Ask your friend this -- how should the probability be adjusted when *nothing is known* about the cards that are in other player's hands or in the muck? Sure, they are dead and are not in the deck, but you know nothing of what those cards are. If you don't know what they are, you have no idea how the remaining content of the deck has changed. Therefore, you can't make any adjustments to the probability because you have no new information.

Re: Odds question...
 

I would also agree with you that 1+1=2, what is your point?

Re: Odds question...
 

Well this would explain why you can't help your friend out.

Go read my second post again. Especially the last part of it. If you can't figure out how that applies to the situation, then I give up and I hope that someone else will have better luck with you.