Basic Calc Question

[Forbillz]

Okay, everyone always does probability calcs based on all their outs being live. My question is this:

Let's assume I'm at a game of 10, and I have 2 spades and see the flop. 2 spades come up. Calcs are usually done assuming 9 outs.

I'd argue that, with 9 other players out there, there are 19 dead cards (including the one burn card prior to flop). Wouldn't logic say that there should be 4.75 spades already gone? (25% of 19 cards). And therefore, isn't it more correct to assume 4-5 outs are more likely out there than 9?

Obviously, in any given situation, as many as 9 more spades could be there, but over time, wouldn't it be more likely this way?

Thanks.

[jason1990]

Where's the FAQ? Is this on there? Seriously, is there a FAQ and if not has anyone considered compiling one from previous posts?

As j goes from 0 to 19, let E_j be the event that there are 19 dead cards containing exactly j spades. Let F be the event that you make your flush. Compute

\sum_{j=0}^19 { P(F|E_j)P(E_j) }.

This would be the rigorous way to do what you're suggesting. The "miracle" is that you get the same answer by assuming there are 0 dead cards. So you don't really have to go through all this hassle -- you can just work with the fact that there are 9 spades among the 45 unseen cards, not ? spades among the 26 cards in the remainder of the deck.

A FAQ style answer should probably have more detail than this. [img]/images/graemlins/wink.gif[/img]

[BruceZ]

This thread has my answer to a similar question.

[jason1990]

[ QUOTE ]
you can just work with the fact that there are 9 spades among the 45 unseen cards, not ? spades among the 26 cards in the remainder of the deck.

[/ QUOTE ]
Sorry that should be:

[ QUOTE ]
you can just work with the fact that there are 9 spades among the 47 unseen cards, not ? spades among the 28 cards in the remainder of the deck.

[/ QUOTE ]
(Who said mathematicians can't do basic arithemtic?) [img]/images/graemlins/blush.gif[/img]