Probability

[kurto]

I posted this in the probability forum and no one's answered. I know people here know how to do this so...

I apologize if you've seen this a million times, but I last took a math course over 15 years ago and I've forgotten a lot...

How do you compute odds with 2 cards to come?

I know if I have 10 outs on the flop... I'm 10/47 to make it on the river, and if I miss, I'm 10/46 to make it on the river.

But what's the equation to figure out what percentage of the time you're going to hit when you combine both cards to come?

How does one add multiple statistical probabilities to get one number? (for that matter... I'd be curious how to do it with more then 2 cards. Like If I flip a coin 10 times... what are the odds you never once land heads?)

Thanks.

[gulebjorn]

Okay, i failed the only statistics course i ever took, and am currently enjoying my scotch, but i'll give it a go anyways...


Lets start with the easy part:
[ QUOTE ]
How do you compute odds with 2 cards to come?

[/ QUOTE ]
Take the odds you have with one card to come and multiply them by two. This isn't exactly correct, because if you are drawing to one out, you have a 1/47 chance of making it on the turn and a 1/46 chance of making it on the river, but it's close enough.

If i have to calculate odds fast, i take my number of outs, multiply it by two, add 1 if the number is 10-20%, add 2 if the number is 20-30%, add 3 if its +30%. This is for one card to come. Multiply this number by two for two cards to come.

[ QUOTE ]
I know if I have 10 outs on the flop... I'm 10/47 to make it on the river, and if I miss, I'm 10/46 to make it on the river.

[/ QUOTE ]

10 * 2 + 2 = 22% of hitting it on the turn, 44% of hitting it on turn or river.

I really, really hope this is more or less correct, cuz i've been calculating my odds like this for a long time /images/graemlins/smile.gif I'm pretty confident it's good enough for a guesstimate.

[ QUOTE ]
How does one add multiple statistical probabilities to get one number? (for that matter... I'd be curious how to do it with more then 2 cards. Like If I flip a coin 10 times... what are the odds you never once land heads?)

[/ QUOTE ]

Just multiply the probabilities. Let's say you draw from a deck of cards and bet on the suit. So there's four possibilities, with 25% chance each. Chance a heart will come up on the next card = 1/4. Chance the next two cards will both be hearts = 1/4 * 1/4 = 1/16.

Chance of drawing any suit but hearts: 75% or 3/4, so chance of drawing four cards without a single heart: 3/4 * 3/4 * 3/4 * 3/4.

In the example of the coinflip, the chance of not once landing heads in ten flips is the same as the chance of landing 10 times tails in a row (obviously), because both are 50%. The answer is 1/1024 (right?)

You can also do this for seperate probabilities. If you get called on a bet 70% of the time = 7/10, and when you get called, you win 50% of the time = 5/10, the chance of being called AND winning = 7/10 * 5/10 = 35 / 100 = 35%.

Hope this was any help.

[kurto]

"Take the odds you have with one card to come and multiply them by two." But that can't be... let's say I have a 50/50 chance to win someone on one try. If I try twice, I'm not 100% guaranteed to win. 2 * 1/2=1

what am I doing wrong then (or how am I confused)....

Let's say I'm betting that a coin will land heads up. And I get two tries. I have 50% (1/2) each time. 50%x50%=25%. (or 1/2 * 1/2 = 1/4) But that's actually how often we'll miss, correct?

Your shorthand quickie method is probable close enough for most calls. But the fact that I forgot how to do it longhand irks me. How does one figure out the total odds of 10/47 and 10/46?

Though I still have questions... It was helpful. I will use the quickie method. Thanks!

[SeattleJake]

If it's odds to hit on the turn OR the river, then you add the two probabilities:

7/47 + 10/46 for making a full house from your set

If it's odds to hit the turn AND the river, then you multiply the two probabilities:

9/47 * 8/46 for making a backdoor flush

A simple method, is to multiply your outs by 2 to make it with one card to come, and by 4 with two cards to come. If you want to get more precise, then x2 -1 for every 5 outs (river), or x4 -1 for every out over 8 (turn). But x2/x4 should be close enough for most purposes.

[Godfather80]

Multiply your chances of missing on both the turn and the river and subract that number from 1.

For example: flush draw on flop give 9 outs, so odds of hitting the turn are 9/47 and of hitting the river are 9/46.
To calculate your odds of hitting your flush on the turn or river if you are looking at the flop:

38/47*37/46=.65, then 1-.65=.35

So you are 35% to catch your flush by the river.

[gulebjorn]

[ QUOTE ]
But that can't be... let's say I have a 50/50 chance to win someone on one try. If I try twice, I'm not 100% guaranteed to win. 2 * 1/2=1


[/ QUOTE ]

Probabilities don't work like that. It means that if you try an infinite number of times, you will hit exactly 50% of the time.

[ QUOTE ]
Let's say I'm betting that a coin will land heads up. And I get two tries. I have 50% (1/2) each time. 50%x50%=25%. (or 1/2 * 1/2 = 1/4) But that's actually how often we'll miss, correct?

[/ QUOTE ]

You have a 50% chance of flipping heads. So 1/2 * 1/2 = 1/4 is the chance of flipping heads two times in a row. This is in fact the same as not hitting heads once in two tries.

Total odds of 10/47 and 10/46 is just 10/47 + 10/46. For simplicity's sake, i say multiply by two. The error can be neglected for practical purposes.

A simple rule is this: if you have two probabilities and want to know what the odds are of them both happening, so you say: what's the chance of A AND B, you multiply the chance of A happening with the chance of B happening. If you want to know the odds of either A or B happening, you add the probabilities.

[SeattleJake]

[ QUOTE ]
[ QUOTE ]
But that can't be... let's say I have a 50/50 chance to win someone on one try. If I try twice, I'm not 100% guaranteed to win. 2 * 1/2=1


[/ QUOTE ]

Probabilities don't work like that. It means that if you try an infinite number of times, you will hit exactly 50% of the time.

[/ QUOTE ]

Right. For the coin-flip, 50% to hit heads on the first try, the 50% to hit on the second try (when you miss the first time):
.5 + .5*.5 = .75

If you flip it again:
.5 + .5*.5 + .5*.5*.5 = .875
(The same as 1-.5*.5*.5, or not hitting tails 3 times in a row.)

The only time you'd add .5 & .5 is when they are mutually exclusive, like what are the chances of hitting heads or tails.

[Godfather80]

[ QUOTE ]
[ QUOTE ]
But that can't be... let's say I have a 50/50 chance to win someone on one try. If I try twice, I'm not 100% guaranteed to win. 2 * 1/2=1


[/ QUOTE ]


Probabilities don't work like that. It means that if you try an infinite number of times, you will hit exactly 50% of the time.

[ QUOTE ]
Let's say I'm betting that a coin will land heads up. And I get two tries. I have 50% (1/2) each time. 50%x50%=25%. (or 1/2 * 1/2 = 1/4) But that's actually how often we'll miss, correct?

[/ QUOTE ]

You have a 50% chance of flipping heads. So 1/2 * 1/2 = 1/4 is the chance of flipping heads two times in a row. This is in fact the same as not hitting heads once in two tries. Let's say you try four times. On average, 1 in 4 times you will hit heads, 1 in 4 times you won't hit it, and 2 in 4 times you will hit it just once. As it should be.

Total odds of 10/47 and 10/46 is just 10/47 + 10/46. For simplicity's sake, i say multiply by two. The error can be neglected for practical purposes.

A simple rule is this: if you have two probabilities and want to know what the odds are of them both happening, so you say: what's the chance of A AND B, you multiply the chance of A happening with the chance of B happening. If you want to know the odds of either A or B happening, you add the probabilities.

[/ QUOTE ]

You couldn't be more wrong. Please stop trying to teach people how to calculate probabilities.

The odds of hitting at least one head on two coin flip is calculated by doing what I said before: take the odds of missing, multiply them, then subtract that from zero.

New example: odds of hitting at least one head on two coin flips.

1/2 * 1/2=1/4 1-1/4=3/4

You have a 3/4 chance to flip at least one head on two coin flips.

Again, please don't try to teach people probabilities anymore.

[gulebjorn]

I think SeattleJake is a lot better at this than i am, so i'm gonna shut up now /images/graemlins/cool.gif

[dtbog]

[ QUOTE ]
How do you compute odds with 2 cards to come?

[/ QUOTE ]

(EDIT: Godfather already said this. He's right too.)

The technically correct way is to calculate the probability of MISSING your draw, and subtract it from 1. This is easier to do than setting up multiple conditions.

Say you have 10 outs on the flop... and ignore redraws and what-have-you.

Assuming you don't know any more cards than the three on the flop and the two in your hand:

37/47 * 36/46 = ~0.61
(chance of missing on the flop) * (chance of missing on the turn, given that you missed on the flop)

So.. subtracting this from 1 gives you slightly over 39%.

Of course, if you're figuring this out at the table and don't need it to be EXACT, you should use one of the estimation methods.

I never calculate this anymore though.. how many different situations are there? OESD on flop = 8 outs = just over 32%. etc etc.

-DB