Is online poker flawed, fundamentally? - Gambling and Probability

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Is online poker flawed, fundamentally?

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02-28-2018 , 02:25 PM

pkdk

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Online poker is not rigged, but there is a fundamental flaw in the process that is ostensible. Ostensible is something that first appears to be true, but when looked at in great depth , turns out to be false.

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02-28-2018 , 02:35 PM

Freewill2112

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Quote:

Originally Posted by pkdk

I don't think you're using that word quite correctly. It sounds like you're saying that the process outwardly appears flawed, but is actually not when looked at more in-depth.

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02-28-2018 , 02:39 PM

pkdk

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Quote:

Originally Posted by Freewill2112

I don't think you're using that word quite correctly. It sounds like you're saying that the process outwardly appears flawed, but is actually not when looked at more in-depth.

ostensible
ɒˈstɛnsɪb(ə)l/Submit
adjective
stated or appearing to be true, but not necessarily so.

At the moment we look at online probability function in being the same as a live game. All the probability functions seemingly working the same and true. However the function of online probabilities is different. Originally seemingly to be true, but when looked at in more depth , ostensible is shown.

Is that better?

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02-28-2018 , 02:47 PM

EvilGreebo

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Ostensible doesn't mean "appears to be true but ends up false" it means "appears true but ends up being in question" (ie: not necessarily so - which means it still might be true but it might not)

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02-28-2018 , 02:54 PM

pkdk

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Quote:

Originally Posted by EvilGreebo

Ostensible doesn't mean "appears to be true but ends up false" it means "appears true but ends up being in question" (ie: not necessarily so - which means it still might be true but it might not)

Ok, that is what I was saying , it may still be true , but I already know because it is my work, it is not true , but agree , to be demonstrated to be false.

f:x is not equal to f:y

demonstrated, where f=function and x =array rows and y=array columns.

123
123
123

x=1/3

y=?/3

When all x is randomly shuffled then stopped.

When x delta y , which means if you are getting the left array value from x, but switch to choice from y, the probability changes.

Last edited by pkdk; 02-28-2018 at 03:02 PM.

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02-28-2018 , 03:17 PM

Kelvis

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Just so everyone knows, pkdk made this floater of a thread, among many other aids threads:

https://forumserver.twoplustwo.com/3...rator-1705516/

He gonna fk online poker now.

Edit: I seem to already have replied in one of these:

https://forumserver.twoplustwo.com/3...38/?highlight=

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02-28-2018 , 03:20 PM

pkdk

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Quote:

Originally Posted by Kelvis

Just so everyone knows, pkdk made this floater of a thread, among many other aids threads:

https://forumserver.twoplustwo.com/3...rator-1705516/

He gonna fk online poker now.

That was just a figure of speech of course, it is already fked .

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02-28-2018 , 03:25 PM

pkdk

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Quote:

Originally Posted by Kelvis

A doubter? let's I and you play a friendly game of odds.

We will have 3 sets of 1,2,3 set out in an array

{1,2,3,}
{1,2,3,}
{1,2,3,}

You will receive the left aligned value out of one of the sets, when the random shuffle of each set stops.

{x}
{x}
{x}

What is your odds of getting a number 3?

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02-28-2018 , 03:27 PM

Kelvis

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1/3

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02-28-2018 , 03:33 PM

#10

pkdk

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Quote:

Originally Posted by Kelvis

1/3

I think you need to go back to school, try again or come back when you do the maths.

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02-28-2018 , 03:38 PM

#11

EvilGreebo

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So you're saying - shuffle 3 sets of 3 cards, A, 2, 3 - lets say hearts, spades and clubs, each suit in its own pile

Draw the top card from each pile

What are the odds at least one is a 3?

Is that the question?

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02-28-2018 , 03:42 PM

#12

Kelvis

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I was given one of these piles, 1/3 of them is a 3.

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02-28-2018 , 03:44 PM

#13

pkdk

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Quote:

Originally Posted by EvilGreebo

More or less, what is the odds all 3 are a 3 , all 3 are a 2 or all 3 are an A?

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02-28-2018 , 03:48 PM

#14

Kelvis

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I think you need to go back to school, try again or come back when you can make proper sentences that actually describe what you mean.

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02-28-2018 , 03:50 PM

#15

EvilGreebo

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I mean if I understand the question correctly than, my rudimentary understanding of probability says we need to calculate the chance of no draws being a 3 across all 3 draws:
Draw 1: 2/3 chance of not getting a 3
Draw 2: 2/3 chance of not getting a 3
Draw 3: 2/3 chance of not getting a 3

Total chance: 2/3 * 2/3 * 2/3 = 29.6296~% chance of not getting a 3.

The chance of getting a 3 is thus 100% - 29.6296~ = 70.3703~% chance of getting a 3.

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02-28-2018 , 03:51 PM

#16

pkdk

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Quote:

Originally Posted by Kelvis

I think you need to go back to school, try again or come back when you can make proper sentences that actually describe what you mean.

The answer is ?/3 , you should know that because you do not know how many 1,2,3, have fell into the left aligned array position.

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02-28-2018 , 03:52 PM

#17

EvilGreebo

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However the odds that you get a 3 on each chance is simpler: 1/3 * 1/3 * 1/3 = 3.70370~% chance of getting all 3s.

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02-28-2018 , 03:55 PM

#18

Kelvis

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Actually the answer is 50/50. Either it happens or it doesn't.

Also how do you explain oceans? They can't be water because else the fish would drown. I think we're not getting told the whole story.

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02-28-2018 , 03:56 PM

#19

pkdk

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Quote:

Originally Posted by EvilGreebo

When it should be 33.333...%

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02-28-2018 , 03:58 PM

#20

EvilGreebo

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Quote:

Originally Posted by pkdk

When it should be 33.333...%

No, it should no be 33.333. It should only be 33.333 when you take ONE chance.

If you spent all this time on the "science forums" trying to learn this - you have failed to understand probability.

The odds of something occurring on a single chance do not simply add together when you have multiple chances with clean starts.

This is why it is not 100% that you will, on 6 rolls of one die, roll a 6 one time.

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02-28-2018 , 03:58 PM

#21

pkdk

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Quote:

Originally Posted by Kelvis

Actually the answer is 50/50. Either it happens or it doesn't.

Also how do you explain oceans? They can't be water because else the fish would drown. I think we're not getting told the whole story.

LOl where did you get 50/50 from when using 3 linear arrays?

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02-28-2018 , 03:59 PM

#22

Kelvis

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Quote:

Originally Posted by EvilGreebo

You do realize you're stepping in a troll right? Nobody is this ******ed.

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02-28-2018 , 04:00 PM

#23

pkdk

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Quote:

Originally Posted by EvilGreebo

I think you read my post wrongly. If you pick 1 of 3 from x there is a 33.3333... recurring% chance of the left array alignment of the value 3.

3,2,1

1/3

But if I pick one from y there is 70.3703~% chance of getting a 3.

2/3 approx

x is not equal to y.

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02-28-2018 , 04:07 PM

#24

pkdk

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Quote:

Originally Posted by Kelvis

You do realize you're stepping in a troll right? Nobody is this ******ed.

Quite clearly your math is poor if you cant work out the answer is ?/3.

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02-28-2018 , 04:09 PM

#25

EvilGreebo

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Quote:

Originally Posted by Kelvis

You do realize you're stepping in a troll right? Nobody is this ******ed.

I'm allowing for the slim probability that this is really just someone who phenomenally failed to understand probability.

What can I say - I like hopeless cases sometimes.

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