Help me figure this stat out please.

salvador9475

Hi can I create 2 usernames and pretend to ask about calculations so that later in the thread i can use the second username to recommend my product?

NO

ianmc38

2.1-1

Mellor

So he has two cards that aren't paired. His chances of pairing them are the same for any cards, except if one is a 3s. Lets take then as 10 J.
So you want his chances of pairing (or better) on the flop.
There are 48 unknown cards in the deck.
He has 6 outs.
So outs/unknowns*cards
=6/48*3
=3/8
=37.5%

The odds of this are 1.7/1

sikes

38.3%

6/48 + 6/47 + 6/46

Mellor

sikes wrote:

38.3%

guessing you got that by:

6/48 + 6/47 + 6/46

There is a slight error in that method, the fact that you dont take into account that the outs might reduce each time.
Using smaller values can highlight it.
Say chances of getting an Ace from three cards (AKK) in two goes.
The method above gives 1/3+1/2=5/6. Which is clearly too high. as it is 2/3, all cards equally could be the one left behind.

sikes

Mellor wrote:

guessing you got that by:

6/48 + 6/47 + 6/46

There is a slight error in that method, the fact that you dont take into account that the outs might reduce each time.

that only happens when you hit it. this gives the possibily of hitting at least a pair.

Mellor wrote:

Using smaller values can highlight it.
Say chances of getting an Ace from three cards (AKK) in two goes.
The method above gives 1/3+1/2=5/6. Which is clearly too high. as it is 2/3, all cards equally could be the one left behind.

Here we are using P(A) = 1 - Not(P(A))

So for the original question

P(Hitting) = 1 - Not(P(Hitting))

1 - (42/48 * 41/47 * 40/46) = .34

You have to reduce the sample space each time. BTW im not expert on this at all, and im sure someone can give reasons why one or the oher should be used, but my original post was merely pointing out that you calculation didnt reduce the sample space

Mellor

sikes wrote:

that only happens when you hit it. this gives the possibily of hitting at least a pair.
Thats what makes it confusing. But trust me on it.


Here we are using P(A) = 1 - Not(P(A))

So for the original question

P(Hitting) = 1 - Not(P(Hitting))

1 - (42/48 * 41/47 * 40/46) = .34
Thats a better way to do it. P(A) = 1 - Not(P(A)

I know it looks like the error only starts when you hit, so wheres the harm, but its always there.
Say you were to keep drawing cards from the deck when you get only a few cards left. Say ten cards left, (possible not to hit outs by then, unlikely but possible) 6/11 and 6/10 would be the last two values. Those two alone are over 1 which indicates error.

42/48 * 41/47 * 40/46 = .664
6/48 + 6/47 + 6/46 = .383
Which totals 1.047. That theres the slight error.

Anyway, long story short. P(Hitting) = 1 - Not(P(Hitting) is the better way to do it. No more for error. My original post might be off too. Paper is better than in my head.

sikes

not really know what your getting at. Why dont you reduce the number of cards in the deck?

ditpoker

my head hurts

cardshark202

Wrong forum.
Is there a probability forum on here?

RoadSweeper

i always thought you hit the flop 33% of the time?

ianmc38

RoadSweeper wrote:

i always thought you hit the flop 33% of the time?

yes so 2.1-1 that you'll hit a pair.

shoutman

Why do you need to know if he will make a pair or not, you have 33. CRAI!!!!!

HiCloy

ianmc38 wrote:

yes so 2.1-1 that you'll hit a pair.

Its 2.08 to hit a pair when you know nothing about your opponents cards

Prob of not hitting a pair = (44/50)*(43/49)*(42/48) = 0.6757
Prob of hitting a pair =1-[ (44/50)*(43/49)*(42/48)] = 0.3243 = 2.08/1

2 pair, trips, full house and quads are all included, as they all contain at least 1 pair.

If you have 33, and you know the other guy has 2 different, non-paired cards, then

Prob of not hitting a pair = (42/48)*(41/47)*(40/46) = 0.6637
Prob of hitting a pair =1-[ (42/48)*(41/47)*(40/46)] = 0.3363 = 1.97/1

(I have spent too many years studying statistics )

sikes

stats suck

theodore91k7

Killme00

salvador9475 wrote:

Hi can I create 2 usernames and pretend to ask about calculations so that later in the thread i can use the second username to recommend my product?

NO

theodore91k7 wrote:

ahhhh i see what you did there and im drunk..so what made you think one of the mods wouldnt spot it

5starpool

You are drunk at 10:43am? WP Sir.