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...
 

The quick and easy answer is that Pot odds, by definition, calculates known vs unknown cards. Opponents cards are unknown. Any read on an opponents holdings can be used to calculate implied odds, but not standard pot odds.

Re: Odds question...
 

Not good enough, need a better answer...

Re: Odds question...
 

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Not good enough, need a better answer...

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What exactly would you like to hear? You do not know what cards are in your opponents' hands, which ones are at the bottom of the deck, or which ones are going to come up. As far as you're concerned, every card you haven't seen is still in the deck.

If your friend thinks he can adjust pot odds because he can get super accurate reads on other people and thus take those cards "out" of the deck, he's wrong (for the most part - obviously there are rare situations where the hand range is so small, you might be able to downgrade your outs because there's a good chance your opponent has them).

Re: Odds question...
 

Look, if your friend insists on adjusting the probability of a flush draw based on the idea that his opponent might have some of his outs, that his opponent has a given card, then just encourage him to do that calculation and compare it to the calculation you "claim" the answer is.

Example:

You: 2s 3s Him: unknown Board has 2 spades after 4 cards

To calculate the probability you make your flush on the next card, you simply take 9 outs out of 46 unknown

or
P(Os) = Probability opponent has 0 spades
P(1s) = Probability opponent has 1 spade
P(2s) = Probability opponent has 2 spades

P(f0) = Probability of making flush given opponent has no spades = (9/44)

P(f1) = Probability of making flush given opponent has 1 spade = (8/44)

P(f2) = Prob of making flush given opponent has 2 spades = (7/44)

So the probability you will make your flush is :

P(0s)*P(f0) + P(1s)*P(f1) + P(2s)*P(f2) = 9/46

note: This is not the probability you will win the hand
Clearly in this example if you have 23, and your opponent has 2 higher spades, you are out luck

If your friend can't figure out the probability his opponents hold 0, 1 or 2 spades given that he holds 2 spades, stop trying to explain this concept to him.

If he does this calculation, he'll be able to understand why your answer is correct...

Hope this helps

Re: Odds question...
 

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there is an equally likely chance that they do have the card as if they don't.

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Re: Odds question...
 

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[ QUOTE ]
there is an equally likely chance that they do have the card as if they don't.

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technically, this isn't true / what is meant, right

There is a much better chance that they don't have the card, than they do. But there is an equal chance that any given card will appear in any given slot in the deck after shuffling...just mincing words...

I've got a second solution, too:

Suppose you are just sitting at a table, you shuffle the deck completely and you only deal one hand. Let it be suited. Next, turn over the next four cards. At this point, if there are 2 cards of your suit showing, you can calculate the odds of catching a fifth card of your suit on the river.

Now, repeat this shuffling, but instead of giving yourself the first 2 cards, give yourself the first and third cards, and place cards 5-8 inclusive on the table. What are the odds the ninth card will complete your flush?

The odds are the same. If your opponent happens to have cards number 2 and 4, it doesn't affect your chances of making the flush, unless, of course, you know the specific cards your opponent is holding.

~BK