How do you calculate odds when there are two cards left to be dealt?

[SlickBorv]

Let's say you have a 1 in 5 chance of getting your card, but you have the turn and river left.

With one card left, it would be 20%. With 2, would it be a 40% chance? What's the formula?

[Richie Rich]

Slick- why did you post this question in this forum? (HUGE HINT) [img]/images/graemlins/tongue.gif[/img]

[kagame]

a quick way to think about it is that if you have a 50% chance of hitting on the turn you obviously dont have a 100% chance by the river.

give it like half the original percentage added on to the original percentage.

so like 75% with two to come if youre 50% to hit with one card.

or something like that

[jerrybai125]

Odds of hitting it on the first card + Odds of missing it on the first card * Odds of hitting it on the 2nd card

[Sephus]

???

the odds of hitting it on the first card + the odds of missing on the first card = 1. so youre saying the odds of hitting with 2 cards left = the odds of hitting it on the second card.

the answer is 1-[(odds of missing on the first)*(odds of missing on the second)]

[italianstang]

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Odds of hitting it on the first card + Odds of missing it on the first card * Odds of hitting it on the 2nd card

[/ QUOTE ]

Yes, I am confused as well, what the hell does this mean?

[evanski]

hes right. its just more complicated. say youre 1 in 3 to hit a draw. you could figure your probability to hit it on either the turn or river in two ways:

1. 1- (2/3 x 2/3) = 1-(4/9) = 5/9

2. 1/3 + (2/3 x 1/3) = 3/9 + 2/9 = 5/9

Basically the second way is your odds of hitting it on the turn (1/3), plus the chance you miss it it on the turn and then hit it on the river (2/3 x 1/3)

-evan

[dtbog]

[ QUOTE ]
give it like half the original percentage added on to the original percentage.

[/ QUOTE ]

And if you're 90% to hit it on the turn? [img]/images/graemlins/smile.gif[/img] 135% on the river?

[Loci]

[ QUOTE ]
Slick- why did you post this question in this forum? (HUGE HINT)

[/ QUOTE ]

Originally I had agreed with you, but there don't seem to be many people that can properly answer the question. The way too impractical answer is that to find out every card that will make you win and divide it by the number of cards in the deck, and then to add to that the save-cards divided the remaining cards in the deck again and then to convert that to the exact percentage. so, having seen your two cards and the flop, knowing that you have a flush draw to a set for instance, you have 8 cards that will save you (since one will boat your opponent), 8/43(since you have mystically known that his cards are trips)+ (assuming the board hasn't paired and killed you) 7(one more boat card in your suit)/42 then convert to percentage.
But who's going to really do that?
The quick, practical way of doing it is to count your outs, multiply it by two. That's your rough percentage. Take that number multiply it by the number of cards to come. That'll be close enough for most practical purposes that you are going to need. So, for instance you're all in preflop, your opponent has AA, you have KK, you need a K to win. 2 (kings) * 2 (percentage) * 5 (cards to come)= 20%, which means you are a 80:20 underdog, or 4:1.
On a side note, these are kind of basic knowledge q's, so while you might play high stakes, you probably should post this in probability or poker theory.
Hope this helps.

[jerrybai125]

yea did u know multiplication goes before addition?? ur awnser will be under 1. Lets say you have a 20% chance to hit it with 1 card to come. 2 cards is 20% + 80% * 20% = 36%.