Repeat values over time is a lack of randomness.

Exactly like if I dealt you the first card, put it back, shuffled the deck, dealt you the first card, until you had seen 10 cards.

No, there is still a 1/52 chance of it being any given card.

No, they don't. They're now set, because you're done shuffling, but what's on top of one deck is in no way affected by what's on top of another.

Yes they do,


If you are getting the first card off the top of the deck, do you agree that the rest of the deck is irrelevant?

That card is any one of 52 individual cards.

I am sorry your maths is poor. If I only have ten cards how can you have something out of 52?

it would be out of ten.

Hmm. Maybe I finally see what you're getting at. You're still wrong, but maybe now I can see the way to explain it.

Is what you're getting at that you can have the exact same card multiple times, whereas if you take 10 cards from one deck, there can be no repeats?

The comparison doesn't work, and I'll explain why, but I want to make sure this is where your head is first.

Yes exactly,

Well, the comparison doesn't work, because when you're dealt the cards live, the deck is shuffled every hand, and is effectively a new deck. So if you shuffle 10 decks and take the starting card from each, or shuffle it and take the starting card, put it back, shuffle, and repeat until you've seen 10 cards, the odds of seeing repeats is exactly the same.

Nice try but no. I understand the way you are considering it and it is completely different to the discussion.
Go back to your earlier post , I thought you had it for a minute.



Consider this


123
123
123


consider you are getting the first value on the left


so it looks like this when i offer you a choice of row.

1
1
1


The chance of a 1 is 3/3 yes?

Just get to your point.

I was showing you the point, it is obvious 3/3


2
2
2

The chance of a 1 is 0/3 in the above

so in this next one i will hide the values , shuffle them , and give you a choice of row,

xxx
xxx
xxx


you get the first value , what is your chance of a 3?

x
x
x

If I'm understanding you correctly, each opportunity is 1 in 3.

When you compound them, I believe it is 70.4% of getting at least one 3.

The question was not an individual question, the question was a choice of the whole of rows. You are answering the question as if one deck followed each other in a sequence and you had a different one each time.


I also asked you about a specific value, a specific value that might not be in your column at all. There could be 3 number 1's in your column.

In a standard game it would be 33.3%


1,2,3,

Wait, what?

In what "standard game" would 3 independent 1/3 opportunities only give you a 33.3% outcome?

Each opportunity gives you a 33.3% possibility of a particular result. Three such opportunities compounded isn't still 33.3%. Just like when you toss a coin 3 times, the odds of getting at least one head isn't 50%.

In this game we are ''playing''.



1/3 is a third of the time. which is about 33.3% .

I would not like to even attempt to do the percent difference in poker, but I assume what works on a small scale works on a large scale


So I think online poker is producing approx twice the action , twice the downswing , everything bad related with twice as much.


In a 3 variable game , you will receive one of the variables for approx a third of the time you play over a large sample scale.

In a 52 variable each value should show about 1.92307692308% of the time over a large sample.

I'm not even sure what you're asking anymore. Here's your post again:

You put 3 x's at the bottom, so I was assuming you were asking me what the odds were that I would get at least one 3, given 3 opportunities. The answer is 70.4%, I believe. If I only get 1 opportunity, it's 33.3%.

I'm off to dinner, so I'll look forward to catching up on your next 50 posts when I get back.

Well I will reveal the values I had in my mind and speak later. Good night

1
1
1


You had 0 odds all along.


The full array

132
123
123

Your 3 never aligned, good night

The total number of cards doesn't have to match the denominator in the probability fraction. You can make the denominator any number as long as you adjust the numerator accordingly. People tend to set the numerator at 1 and then adjust the denominator accordingly to make it easier to visualize.

But if you want the denominator to match the total number of cards, then in that situation it is (rounded): 0.192/10 or written as a percentage, roughly a 1.92% chance.




70.37037037% to be a little more exact.

So f:x = 33.3% ~

and f:y = 70% ~

so x is not equal to y

The total number of cards doesn't have to match the denominator in the probability fraction. You can make the denominator any number as long as you adjust the numerator accordingly. People tend to set the numerator at 1 and then adjust the denominator accordingly to make it easier to visualize.

But if you want the denominator to match the total number of cards, then in that situation it is (rounded): 0.192/10 or written as a percentage, roughly a 1.92% chance.


Consider weather forecasts:

It can either rain or not rain. That is 2 items or events. Say there is a 20% chance of rain. If that were to be written as a fraction it could be written 1/5. There are only 2 items or events, but the denominator is not a 2. It doesn't need to be. If you wanted the denominator to be a 2, then you could write it that way and it would be 0.4/2. 0.4/2 is still 20%; the same as 1/5. We could also write it 12/60 if we felt like it for some reason or 117/585 or 0.05/0.25.

I think in the world you would more often see 1/5 written than 0.4/2 (or any of the other examples above) to notate a 20% chance. I think most people find it easier to visualize and immediately have a good understanding of a 1/5 chance than a 0.4/2 chance, so they use the 1/5.

Yea.

Why would it be equal?

What you are calling f:x is randomly arranging three distinct values, looking at one and seeing if it is a particular one of those three values. What you are calling f:y is effectively doing f:x three times and seeing if you see that certain value any of the three times. Obviously we would have a greater chance of seeing the certain value if we try three times than if we try one time.

So this is just a troll account then?

Exactly.

And yes, 70.370 repeating - I rounded it. Just glad to have it confirmed that I was doing it right.

What in the world is going on?

Is this as ridiculous as it appears to be??

(Please don't ask me to read the entire discussion.)

Well no, I put 70% because I thought you had worked something out, but the real answer is ?/3 .

x=1/3

y = ?/3

Yeah it is ridiculous that some of the members cant work out the answer is ?/3

If you are still suggesting that the odds of picking one number out of three in three independent trials is other than 70.37%, you are incorrect. But if you still believe otherwise, it's time to show your work in a comprehensible way - a little less condescension would be a welcome bonus.

?/3 is not an answer, and if that's all you have, I'm inclined to think that others may be right that you are trolling.