ELI5: I've had something that's always bothered me about calculating odds. I've noticed that odds are always calculated assuming that all of the outs are still available, rather than making statistical assumptions... an example is listed below:

Hold'em: 2 suited cards in pocket, two matching suit on board. Odds are calculated that all 9 are still available rather than making the assumption that 1/4 cards that cannot become part of your hand (turn/river, burn cards, other players' cards).

In a four handed game you know there are 8 cards in other players' hands and let's say it's a bet post turn... then you also know there are 2 burn cards that could be your matching suit. Why don't you take 2.25 from the number of outs... so calculating with 6.75 outs instead of 9? It seems like it would make sense to do so as it is a more conservative and statistically accurate calculation.

Can anyone help?

Question #1 on the FAQ
https://forumserver.twoplustwo.com/1...stions-676280/

The best thought experiment I can offer you is... after the flop is dealt, the next card is the burn, the next is the turn, then next is a burn and the next is the river.

None of the cards after those can be dealt either, and neither can the burns. Why aren't you discounting those also?

If the calcs seem counterintuitive think of a simpler game where you can count up all the cases.

Let's define a game of "Simple Poker" with the following rules:
a) There are a total of 4 cards in the deck: two cards each of two suits
b) There are exactly 2 players
c) Each player gets 1 hole card
d) Then 1 card is burned
e) Then one card is put face up
If your hole card and the faceup card are the same suit then you have a "flush"

What's your chance of getting a flush?

Calculation by the usual method:
You have one card. There are 3 unknowns cards out there (opponents card, burn card,and the last card to be turned up), one of which is the same suit as the card you have (you have 1 'out'). So your chance is 1/3

If we construct all the cases possible where we assume the other suit car is in the hand of the opponent, the burn card, etc. and count them all up then we must get the same probability as above (1/3) for the calculation to be valid.

Here we go:
There are 3 cases - each of which is equally likely:

Case 1: Our needed suit card is in the hand of the opponent. Chance of us making a flush in this case: 0%
Case 2: Our needed suit card is the burn card. Chance of us making a flush in this case: 0%
Case 3: Our needed suit card is neither the burn card nor the card our opponent has. Chance of us making a flush in this case: 100%

1/3* 0% + 1/3 * 0% + 1/3 * 100% = 1/3

Voila. Same as above.

There is also a Sticky Thread in the Probability Forum on this very topic.

https://forumserver.twoplustwo.com/2...-draw-1619362/