I agree with everything in this quote of yours that I bolded. But what I have to say is:

So??


The mean of the chances of any one particular card in the y axis being a 3 for all possible y axes arrays is 1/3. That is all that we care about as long as we cannot see any of the cards. If we can see any of the cards, then we have more information and can adjust based on the additional information.






If we can see any of the cards, then we have more information and can adjust based on the additional information. This works exactly the same when trying to state the chances of the top card of one deck of cards being a particular value:

If we have 1 randomly shuffled deck and have not looked at any of the cards, the chance of the top card being the Ad is 1/52. If we flip the top card up and it is the 3c, then we can adjust based on this new information and conclude that the chance of that card being the Ad is 0. or if when we flip the card up it is the Ad, then we can conclude that the chance of it being the Ad is 1.

Or if we look at the bottom of the deck and see that card is the 3c, then we can adjust and conclude that the chance of the top card being the Ad is 1/51.

But so long as we do not know the value of any of the 52 cards, our best answer is that the chance of the top card being the Ad is 1/52.

You keep using ostensible as a noun. It isn't a noun. It is an adjective. The way you are using it doesn't make any sense. It has to be modifying some other word.

Probably the best course of action. Gravity is more important and controversial than using multiple decks at one poker table, or for online poker having one random number generator that generates values representing cards for each "table" in use at any one time in whatever order that such "tables" need a "card" to be "dealt."

I actually think that PKDK's issue is something everyone here has probably heard a lot, but probably never heard described in any sort of mathematical model context.

Say we have 4 tables. Say we have pre-shuffled Deck 1 through Deck 100. Each table plays a hand. Table 1 uses Deck 1, Table 2 Deck 2, Table 3 Deck 3 and Table 4 Deck 4.

Table 2 finishes first and Deck 5 gets sent to Table 2 for Hand 2. Table 1 finishes and Deck 6 gets sent to Table 1 for Hand 2.

This is "wrong". Table 1 used Deck 1, then Deck 6. It is not going in order. There should be a row of decks just for Table 1. Table 1 should get whatever the cards in Deck 2 are, but it is getting the cards in Deck 6, which are the wrong cards. At the very least it should get the cards in Deck 5 and everything should be in order.

Yes, all the cards in Deck 2 and the cards in Deck 6 are both in a random order, but it is probably a different random order, so it matters which deck Table 1 gets. And Table 1 should get Deck 2. It shouldn't skip around to Deck 6. Table 1 is getting the wrong cards. The cards in Deck 2 (or at least Deck 5 if we are going to go in order 1-4 for each table) belong to Table 1; not the cards in Deck 6.


You might see people talk about this a lot in blackjack when the last player to act hits. The other players might verbally attack such player because he took the dealer's card. The card that player took belonged to the dealer and now that player has messed everything all up and everyone is getting the wrong card.


This at least is how I understand what PKDK is really seeing as the problem. Certain cards in certain decks belong to certain people and when you change the deck you are taking away cards that belong to people and giving them cards that they shouldn't be getting. Even if both decks are randomly shuffled, they are not randomly shuffled into the same order, so changing decks gives people the wrong cards.

Wow - have I finally managed to get you to come out with the real issue here? Yes, the cards would be different, but they would still be random. Just like if the dealer paused 10 seconds, or moved his hand slightly, the next shuffle could be different.

A totally unrelated question, if whosnext hasn't scared you away. If you are playing blackjack, do you worry about what the player in the "anchor" seat (last player from the dealer) does? IE if you have a player there who plays incorrect strategy, and ends up taking the dealer's bust cards, do you get upset with the person and/or choose to play at a different table, as he's causing you to lose?

Edit to add: LOL, I swear I didn't read Lego's post before posting this. Great minds think alike?

I'm going to do this one more time for fun. PKDK, please answer questions 1 through 4, then read the rest and then answer questions 5 and 6.


There are four boxes: Box 1, Box 2, Box 3 and Box 4. There are two prizes. Two of the boxes each have one prize inside them. The other two boxes are empty. The prizes were placed such that each box had an equal chance of receiving a prize (no box was permitted to receive both prizes).

Question 1: What is the chance that Box 1 has a prize inside?

Question 2: What is the chance that Box 2 has a prize inside?

Question 3: What is the chance that Box 3 has a prize inside?

Question 4: What is the chance that Box 4 has a prize inside?








We take Box 2 and Box 3 and put them into a rocket ship and launch them into the sun.








Question 5: What is the chance that Box 1 has a prize inside?

Question 6: What is the chance that Box 4 has a prize inside?

Tends to happen.

holdemace486 posted extensively on this "problem" some years back. I don't remember whether he posted about it in this thread.

pkdk

Once you've given Lego05 his requested 6 answers, please give me 6 for this slightly amended test:

Please answer questions 1 through 4, then read the rest and then answer questions 5 and 6.

There are four boxes: Box 1, Box 2, Box 3 and Box 4. There is one prize. One of the boxes has the prize inside it. The other three boxes are empty. The prize was placed such that each box had an equal chance of receiving the prize.

Question 1: What is the chance that Box 1 has the prize inside?

Question 2: What is the chance that Box 2 has the prize inside?

Question 3: What is the chance that Box 3 has the prize inside?

Question 4: What is the chance that Box 4 has the prize inside?


We take Box 2 and Box 3 and put them into a rocket ship and launch them into the sun.

Question 5: What is the chance that Box 1 has the prize inside?

Question 6: What is the chance that Box 4 has the prize inside?

Elegant.

This is the direction I thought pkdk was headed...but neither I nor pkdk could express as cleanly.

Good chance he is permanently off to the land of ostensible, so odds are he will not acknowledge what was said above (or if he does it likely will be in the "clearly you guys cannot understand this" style he overuses).

However, assuming this is the thing that has him all worked up, he also talked about how this could in theory be used for advantage (in roulette for example), and the only thing holding him back from riches is a lack of interest in money, or something. I suppose I am curious if any of you guys can figure out how someone like that can have his set pf beliefs when the cards or roulette spin(s) are all random.

As I said somewhere in the noise before, this sort of reminds me of the broken logic some people try to use when they mix -EV bets to think they created a +EV spot, along with the martingale debates. They create logic loops in their head that make perfect sense to them, even if they are not accurate.

This one seems a bit more grounded in simple reality, even with that guy's inability to ever clearly explain himself, where I would ask - who cares if you got someone else's card as the first card dealt from a randomly shuffled different deck. Where is the edge coming within the mind of someone like that who believes this changes the odds of the games themselves. Where are they tricking themselves?

Endless and heated discussion has taken place in the past (on 2+2 and elsewhere) about "fixing the online deck" at the beginning of each deal vs. "continual shuffling of the online deck" when dealing later streets. IIRC Pokerstars did it one way and Full Tilt did it the other way.

But let's get to the real crux of the matter:

Did pkdk ever explain the impact that not burning cards has? Surely that has to be an even bigger deal than giving different decks to "wrong" tables.

Please no, that would keep him busy for at least another 3 years while saying that x isn't y.

Enough out of you - you're in the penalty box for scaring poor pkdk away!

answered in red

answered in red

6%~

Well, that pretty much ends this conversation for me. pkdk is either a troll, or clearly has a very poor understanding of probability while thinking the same of everyone else.

Don't think I've ever seen a more obvious case of Dunning-Kruger.

Why do people think guys like this are trolls? He has pitched the same obsession off and on here for years in forums that have basically no trolling.

Why he thinks burning a card impacts the expected outcome of a game is curious only in how does someone twist their thinking into actually believing it, almost as a mix of superstition and religion.

6% impact (whatever that means) when not burning a card. Can that be exploited? Sure, if it was real. Has he exploited it? Of course not. Can he explain with a simple hand how to exploit it? Of course not. However, nothing will change his belief on it.

This is not trolling on his part, nor do I think it is fully a Dunning–Kruger effect. Seems more like a bit of a genuine mental condition, one he will clearly have for his lifetime.

He can be both. I think he's just a ******ed troll.

Sometimes posters like that do a bit of trolling when they run out of things to say, but the core belief remains. Saying this guy is a troll is like saying a guy standing on a crate at a street corner preaching his manifesto to the clouds is trolling the people walking by.

pkdk if you honestly believe this then you need to go back to statistics 101.

Your first 4 answers are correct - 2/4 - which can be expressed as 1/2. There is a 50% chance that a box contains a prize.

Your last 2 answers however fail to grasp something very fundamental to probability - which is that the fact that some unknown choices have been removed has zero impact on the probability.

In the case where 4 boxes exist, 2 of them randomly selected, have a prize, and 2 of them, randomly selected independently of the first selection, are obliterated by launching them into the sun, the destruction of 2 of the boxes has no effect on the odds of the other 2 having a prize.

(This isn't a quantum mechanics problem - this is a "standard reality" problem)

4 boxes exist, each box has a 50% chance of having a prize. 2 boxes are destroyed. The other 2 boxes still have a 50% chance of having a prize.

Just like in poker - you get dealt 2 cards - you "know" 2 values. You do not know the other 50. Thus, calculations of the odds of what other players hold are based on the 50 remaining values. Lets say you have KK

When the dealer burns a card, but hasn't dealt the flop yet, the probability of someone else having an ace is 4/50 or 12.5%. The probability of someone else having K is 1/50 or 2%. The probability of someone else having K is 0% (0/50) because you KNOW where that value is.

When the flop comes out, 3 new values are known. the unknowns drop to 47. Those unknowns are still all the players cards, the burn card, and the stub. The known cards are on the flop and in your hand.

That's all that matters when we talk about poker probability. None of the above has any bearing on the next hand we play, or any other hand we play, nor is it influenced by any hands that have previously been dealt or by any hands yet to come.

You contradicted yourself.

You said Box 1 has a 2/4 chance of having a prize (most people would have written this as 1/2 by the way; you’re really supposed to simplify the fraction as much as you can; I think most high school math teachers would have deducted a few points for that). Then you said Box 1 has a ?/2 chance of having a prize.

It’s the same question, knowing the same information about the contents of all of the boxes, so the answers should be the same. You gave two different answers. They can’t both be right.

Physically moving 2 of the boxes doesn’t change anything about the contents of any of the boxes.





How about this:

There are 4 boxes. Box 1, Box 2, Box 3 and Box 4.

Box 1 and Box 2 are on Earth. Box 3 and Box 4 are on a space station on the moon.

Two of the boxes each have one prize inside them. The other two boxes are empty. The prizes were placed such that each box had an equal chance of receiving a prize (no box was permitted to receive both prizes).

Question 1: What are the chances there is a prize in Box 1?

Question 2: What are the chances there is a prize in Box 2?

Question 3: What arethe chances there is a prize in Box 3?

Question 4: What are the chances there is a prize in Box 4?

So, by extrapolation of your methodology, 2 + 2 = ?

Little wonder you failed every math exam you ever took.