How much have you got?


I am saying that:

--- If we see a Jack less than 550 times or more than 850 times, then you win.
--- If we see a Jack 550 times, 850 times or any number of times between 550 and 850, then I win.

You get 700 numbers. I get 300 numbers.


Do you accept?

?/3

When will you start?

if we turn the array to


xxx


1 jack, then yes

I have tried to start the discussion several time with a question or a scenario, but people have failed to ''play along'' to reach an answer. People are not being completely objective .

For the last time, we have two sets of two , box 1 and box 2


12
12

In either box two is a prize.


Now we are going to remove the numbers , shuffle the boxes and remove two of the boxes.

xx

Your chance of winning a prize is ?/2 or to put it more technically https://wikimedia.org/api/rest_v1/me...e97dfafea9fd26

It is hard to be completely sure what you are saying here. But this appears to be a significant change. If we would make this significant change, then I would want to make other significant changes also before accepting.

Why don't we just go with the bet that I already laid out in detail?


DO YOU ACCEPT THE BET I PROPOSED (WHICH IS SET FORTH BELOW) THAT IS BASED ON A SCENARIO THAT YOU CREATED YOURSELF EARLIER IN THIS THREAD???

My scenario earlier was showing that an empty column could be formed after the shuffle. It was not creating a bet, but I bet you in this array

aqk
aqk
aqk


That the left column of the array will sometimes have no queens. ?

If you are not confident enough in your "theories" to take the bet based upon a scenario that you posted in this thread yourself, then just say so and say you do not want to take the bet. You don't need to propose completely different bets.


I agree with you that in the above if the rows are each shuffled randomly it will sometimes result in the first column having no Queens. Therefore, I decline that bet.

I will not take the bet because the odds are ?/3 .

By agreeing that at times the first column may have no queens agrees with me that the probability of y is not equal to x, x is always 1/3 for a queen in this scenario.

I'm starting to wonder if what you're getting at here is that once you remove those cards/boxes/what have you, if you knew what had been removed, the odds have changed.

In the example above, I assume you mean that each box 2 has a prize. So if it turns out that both box 2's were removed, you have 0 chance of winning a prize. If it turns out that both box 1's were removed, you have 100% chance of winning a prize. And perhaps that's where you're getting your ?/2.

But that's not how it works. We don't know what's been removed, and our chance of winning a prize is 1/2. I know it's not an easy thing to get your head around sometimes, but it's true.

PKDK:

I have changed the terms of the bet proposed. The bet is restated below in its entirety with the changed terms in bold. Let me know whether you accept the bet with these new changed terms.


If you win the bet, I will give you $100,000. If I win the bet, you will give me $50,000 (we can decrease the bet amounts somewhat if you don't have $50,000 as you said). We have to escrow the money before arranging to effectuate the bet.

The bet will be based upon a scenario you laid out earlier in this thread.


We will take 3 Jacks, 3 Queens and 3 Kings and arrange them into a 3 by 3 grid as follows:

JQK
JQK
JQK

Then we will shuffle each row so that the Jack, Queen and King is arranged randomly in each row. We will then have the following grid:

xxx
xxx
xxx

where we do know that each row contains exactly 1 Jack, 1 Queen and 1 King, but the columns could contain more than 1 or less than 1 of each card.

We will then flip over all of the cards in the first column and note whether or not we see at least one Jack. (For the record, earlier in the thread you said that the chance of seeing a Jack is ?/3. I disagreed with that and said the chance is roughly 70.4%. You said 70.4% is wrong and the correct answer is ?/3.)

We will do the above 1,000 times.

The winner is determined as follows:

--- If we see at least one Jack less than 640 times or more than 760 times, then you win.
--- If we see at least one Jack 640 times, 760 times or any number of times between 640 and 760, then I win.

You get 880 numbers. I get 120 numbers.

I am giving you 2 to 1 on the money ($100,000 vs. $50,000) and I am giving you 880 numbers to my 120 numbers.

Over 1,000 trials of something that has odds of occurring of ?/3, I am somehow predicting that the number of occurrences will fall within a range of only 120 numbers. Even though, according to you, we cannot possibly know how often this event will happen (the odds of it are ?/3), I am willing to bet money that over 1,000 trials it will happen 640 to 760 times.

What do you say? Do you accept?

If you accept, then if you would agree to do so with a simulation rather than actually dealing cards, we could be done relatively quickly.

Ok, you have almost understood , you just need to realise why it is not 1/2


If you state it is 1/2 , that is stating you know the hidden values in the boxes, you are stating for 100% there is definitely a prize in one of the boxes.

This is where the uncertainty calculation comes into ''force''.

I have done another piece of abstract math to show you

52*52
All x = 52
All y = 52
all x=1/52
all y=?/52

The answer to y is y = var (1/52x)

y having a probability function range of 0 - 52

First of all, to be clear, it's not the problem I've "almost understood" - I understand the problem. It's you that I've "almost understood".

There's no need to go off to your abstract math, we have a very simple example, from your post.

12
12

In either box two is a prize (by which I assume you mean each box 2 has a prize).

Now we are going to remove the numbers , shuffle the boxes and remove two of the boxes.

xx

You ask what the odds are of getting a prize if you select 1 box. The answer is 1/2. Hopefully you'll be able to understand why that is at some point, as I'm not sure how to explain it any better than has been done many times already.

Every box has a 50% chance of being a prize, no matter whether you remove boxes or not. The odds don't change when you remove a box at random - how could they?

When an event is changed, things obviously change. So if we start with 4 boxes, and knowingly each box number 2 contains a prize, we know for sure even without numbers it is a 50/50 situation.


However if I start off with only two boxes with no numbers on them, how do you know there is a prize in any of the boxes?

I have just removed a possible 100% of the prizes.

I know if I ordered a pallet of 52 random items from Xeon it would have exactly 4 ace prizes on the pallet.

I know if I ordered a pallet of 52 random items from Yeon it could have exactly 0 ace prizes on the pallet upto 52 ace prizes on the pallet.

Would those pallets melt steel?

PDKD, I guess you missed post number 84935. I did not see a response to it from you. Here it is again:




PKDK:

I have changed the terms of the bet proposed. The bet is restated below in its entirety with the changed terms in bold. Let me know whether you accept the bet with these new changed terms.


If you win the bet, I will give you $100,000. If I win the bet, you will give me $50,000 (we can decrease the bet amounts somewhat if you don't have $50,000 as you said). We have to escrow the money before arranging to effectuate the bet.

The bet will be based upon a scenario you laid out earlier in this thread.


We will take 3 Jacks, 3 Queens and 3 Kings and arrange them into a 3 by 3 grid as follows:

JQK
JQK
JQK

Then we will shuffle each row so that the Jack, Queen and King is arranged randomly in each row. We will then have the following grid:

xxx
xxx
xxx

where we do know that each row contains exactly 1 Jack, 1 Queen and 1 King, but the columns could contain more than 1 or less than 1 of each card.

We will then flip over all of the cards in the first column and note whether or not we see at least one Jack. (For the record, earlier in the thread you said that the chance of seeing a Jack is ?/3. I disagreed with that and said the chance is roughly 70.4%. You said 70.4% is wrong and the correct answer is ?/3.)

We will do the above 1,000 times.

The winner is determined as follows:

--- If we see at least one Jack less than 640 times or more than 760 times, then you win.
--- If we see at least one Jack 640 times, 760 times or any number of times between 640 and 760, then I win.

You get 880 numbers. I get 120 numbers.

I am giving you 2 to 1 on the money ($100,000 vs. $50,000) and I am giving you 880 numbers to my 120 numbers.

Over 1,000 trials of something that has odds of occurring of ?/3, I am somehow predicting that the number of occurrences will fall within a range of only 120 numbers. Even though, according to you, we cannot possibly know how often this event will happen (the odds of it are ?/3), I am willing to bet money that over 1,000 trials it will happen 640 to 760 times.


What do you say? Do you accept?


If you accept, then if you would agree to do so with a simulation rather than actually dealing cards, we could be done relatively quickly.

Maybe if they travelled fast enough through a field creating a charged electrostatic field around it to cause a quantum tunnel on impact.

I do not gamble, and an estimation is not an approximation, therefore it is ?

If you are declaring a range that shows uncertainty.

Give me an exact amount and I might try find 50k lol I have pokerstars play money lol

The event hasn't changed. There are 4 boxes, and 2 have a prize. There is a 50% chance of each box having a prize. You taking away 2 boxes, chosen at random, doesn't change anything. Yes, there could be no prizes, or they could both have prizes. The odds of each individual box containing a prize is still 50%.

Perhaps this will help. Here are all the possible configurations of the 4 boxes, with box 2 containing the winners.

11 22
12 12
12 21
21 12
21 21
22 11

The first pair are what remains once you've taken 2 boxes away, leaving you with these pairs to choose from:

11
12
12
21
21
22

4 times you will have a 50% chance of choosing a winner. 1 time you will have a 100% chance, and 1 time you will have a 0% chance. The average is...

...50%!

If only this could help you make more money in $2 tournaments...

All the best.


P.S. Try to guess ? where ? is the total amount I have actually gotten riggies to wager on prop bets over nearly a decade. Probably a hundred before you have been in your position where they were presented them or actually initiated them (hint, the fact they initiated them rarely meant they would follow through if accepted). You can put your answer in ?/3 form as well if you prefer.

What is the relevance to poker? There is no situation where 52 decks of 52 cards are shuffled together so that there is a possible outcome of 52 aces or 0 aces in one deal. The 52 decks are independent of one another.

I assume what you're getting at is that the first card dealt from 52 separate decks could have no aces, and it could have 52 aces. But what does that matter? If a deck is shuffled 52 times, it could also have an ace on top 0 times or 52 times. 52 decks shuffled and dealt from once, or 1 deck shuffled and dealt from 52 times - nothing changes.

Interesting.


What? An estimation actually is an approximation. Those two words are synonyms.


But anyway, I am not picking a range of 640 to 760 as an estimation or an approximation. The exact expected amount of occurrences is roughly 704, but the probability of the number of occurrences actually being exactly 704 is only roughly 2.66%. I picked a range of 640 to 760 because it is a range around 704 that is wide enough that there is basically a 100% chance that the number of occurrences will fall inside that range and, therefore, there is basically a 100% chance that I would win the bet.




Anyway, you didn't actually decline the bet. But I assume from your post that you decline the bet because you do not bet. Is that correct?

You contradicted yourself, you say taking away two boxes does not change anything then in the next ''breathe'' explain the possible changes. Then at the end of paragraph neglect the possible changes and re-insist it is still 50%, completely ignoring any of the now uncertainty you already mentioned.
You can not be certain that there is a prize in any of two boxes that are left . Your new choice is now 1/2 boxes not 2/4 boxes. 2 boxes that you no longer know any information about. Your only information based on a different situation of having 4 boxes and knowing there is two prizes in the boxes. There is a big difference in having information than having no information.