I have answered you and old guy, I replied to you with a quick diagram of your coin set up, sorry I thought you would assume I agreed with your set up .

Agree

If I want to pick a heads and I will be able to see the results of the 10 flipped coins when I need to pick one and I get to pick whichever one I want, then I am going to pick one that is a heads.

So, the question is effectively what are the chances that we will get at least one heads from flipping 10 coins and the answer is roughly 99.9%.

For simplicity, let's consider only two coins. Call them a penny and a nickel.

Suppose that Coinstars is in the business of "dealing" pennies and nickels.

Suppose that you play at Coinstars each and every day and get dealt one penny and one nickel. Of course, all that matters is whether a coin is heads or tails.

Do you agree that the probability of getting two heads or two tails = 50%?

Do you agree (it's actually the same question) that the probability of getting one head and one tails = 50%?

I think you do, based upon your prior statement about pennies and nickels being independent.

So what is the chance of picking the 1 heads out of 10 disclosed results?

99.9%/10?

Yes I agree with that .

This is like discussing aeronautics with somebody who has a core belief that anything over 100 pounds is too heavy to possibly fly.

There will be some points of agreement along the way, but it’s not going to end up anywhere good.

But they are disclosed , the values are hidden

?
?
?
?
?
?
?
?
?
?


What is the chance of picking a heads from the disclosed results?

Imagine you had someone create a 52 sided die, with the sides numbered 1 to 52. And you then got them to give you 52 of these. You get someone to roll each one and record the result and put the results in an array but you don't get to see any of those results. All you see are

?
?
?
.
.
.
52 times.

Here are the questions for you, pkdk.

1) What is the probability that the first one is the number 8?

2) What is the probability that the 10th one is the number 8?

3) What is the probability that at least one of the rolls is the number 8?

4) Regardless of your answers to any of questions 1-3, how will your answers be different if instead of rolling dice you have a computer randomly select a number between 1 and 52?

How about this one? I shuffle a deck of cards. I then turn the top card over to reveal that it’s the Ace of Diamonds.

Now: what are the odds that the top card is the Ace of Diamonds?

Assuming there is at least 1 heads after flipping the 10 coins (of which there is roughly a 99.9% chance), then the chances of me picking a heads is 100% because I will purposely pick a heads.

But I realized what you meant after I read your edit. Before we flip the coins, then in this scenario the chance of selecting a heads is 99.9% as stated in previous posts. If you want to see it as a fraction with 10 as the denominator, then it is 9.99/10.



Did you watch the video I posted regarding simplifying fractions?

Did you do the marbles test yet?

If I can’t see the values of the coins and I just pick one randomly, then there is a 50% chance that the coin I pick is heads.

Tougher question. I had to think about it longer. I think I am going to go with 100% though.

Then you agree that a player who plays exactly two "hands" of "poker" at "Coinstars" each and every day is playing an eminently fair game.

For his first "card" (penny) is truly random (50% heads/50% tails) and his second "card" (nickel) is truly random (50% heads/50% tails) and his coin combos are truly random (50% matching/50% non-matching).

This proves that Coinstars, which deals millions of pennies and nickels in millions of different hands to thousands of different players each and every day, provides a truly random deal to all of its players in all of its hands on all of its tables.

It's simply a matter of scale. What is true for two is true for a million.

1)1/52
2)1/52
3)1/52

4) The first 3 questions are not the question.

My answers would not be different in your computer example.

Yes I agree with your scenario because you have not covered the question in your scenario, the question does not involve the randomness.

That's the question he's been trying to ask the entire time, but his communication skills are so terrible, it takes that many posts to get there. It's the same question he keeps asking different variations of over and over and over again, thinking he'll get us to "understand" that it's not 1/2, but ?/10. And even though he's so obviously wrong, and we've explained it to him dozens of different ways, and used countless examples and analogies to show him, he's utterly convinced that we're the ones not understanding his new math.

Edit to add: Post 1K, coming soon!

I did not ask you that question, i did not ask you what is the chance the one you have picked being a heads,

You have jumped the question . I asked you what is the chance of picking a head out of the 10 unknown values?

Thats not the question

Your answer to 3 would be incorrect - that would be the probability for any one of the rolls, but not for the probability there is at least one. But your answer to 4 is interesting. You believe that a computer generating cards is different from a deck of cards doing the same thing. Why isn't the computer different from rolling dice?

Thank you for your patience.

Since online poker (the subject of your obsession) deals exclusively with probabilities and you have (sensibly) agreed that all online poker probabilities are truly random, this concludes our conversation.

Thanks again.

Oh, oh, I get it now...I can't know the odds because I don't know what the unknown values are, so it must be ?/10...I get it!!!





NO. It's 1/2. There is a 1/2 chance of each value being a head, so if I randomly pick one, the odds that it's a head are 1/2.

Just because you want to pretend that making them into an array and choosing from one column of the array means we no longer know anything about those coins, doesn't make it true. We get to carry that information forward with us.

If you told us that you placed an unknown number of heads and tails into the 10 results and we have no way of knowing how you chose them, then yes, the answer would be unknown. ?/10, if you like. But that isn't the case. We know that the 10 results were obtained by flipping a fair coin 10 times, so we can use that information to calculate the odds of choosing a head to be 1/2.

Sorry I do not think I understood your original question. A computer is different to rolling a dice. The computer is programmed to output the probabilities with standard deviation.
A dice is not programmed to do anything other than random.

You almost get it. In your second part you mention, you jump the question,
you answer the question with your first part.

Keep thinking on those lines and you will start to see it.

You are not adding the top bit where i put it in red.

It’s effectively the same question in this case because each unknown coin has the same chance of being heads or tails as all the other coins.


But, anyway, the answer is that if I am picking one of the coins at random, then I have a 50% chance of selecting a heads.