Cards dealt to other players when calculating odds

Okay, thanks. I get the point although it still seems like something is missing. I suppose this is one of those case where you have to trust the maths and not your intuition.

You feel like something is missing because you think that there must be a way to calculate the the effect of the opponents cards to the total equation. Actually there is. But it makes no difference.

To clear your mind lets work on a simple situation;

You have two cards and none of them are Ace of Spades, opponent has two cards and we don't know what they are. We deal one card. What is the probability of that card to be As.

Common calculation:
There are 50 unseen cards left and only one of them is As so the probability of As is 1/50. Easy...

If we want to take opponents cards into account:
What is the probability of the opponent not to be dealt As?
- (49/50) * (48/49) = (48/50) = 96%


96% of the time, opponent doesn't have As (now we have 48 cards left in the deck and one of them is As), What is the probability of As to be dealt?
- 1/48

4% of the time, opponent does have As (now we have 48 cards left in the deck and none of them is As), What is the probability of As to be dealt?
- 0/48 = 0


And in total, the probability of As to be dealt is:
- [(1/48) * 96%] + [0 * 4%] = [(1/48) * (48/50)] + 0 = 1/50


So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it?

Originally Posted by Belt

You feel like something is missing because you think that there must be a way to calculate the the effect of the opponents cards to the total equation. Actually there is. But it makes no difference.

To clear your mind lets work on a simple situation;

You have two cards and none of them are Ace of Spades, opponent has two cards and we don't know what they are. We deal one card. What is the probability of that card to be As.

Common calculation:
There are 50 unseen cards left and only one of them is As so the probability of As is 1/50. Easy...

If we want to take opponents cards into account:
What is the probability of the opponent not to be dealt As?
- (49/50) * (48/49) = (48/50) = 96%


96% of the time, opponent doesn't have As (now we have 48 cards left in the deck and one of them is As), What is the probability of As to be dealt?
- 1/48

4% of the time, opponent does have As (now we have 48 cards left in the deck and none of them is As), What is the probability of As to be dealt?
- 0/48 = 0


And in total, the probability of As to be dealt is:
- [(1/48) * 96%] + [0 * 4%] = [(1/48) * (48/50)] + 0 = 1/50


So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it?

LOL... schrödinger's cat.. GG sir

Originally Posted by Belt

So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it?

Never open the Box.

It's well known that that Cat is double dealing and not double dealing both at the same time and skewing the statistical data and not skewing the statistical data at the same time.


Remember the cat can be alive and dead at the same time as long as you never open the box.

Just Never Open the Box.

For those of you who never, ever even thought of studying quantum mechanics, and have noooo f'n idea about the 'inside joke' going on itt, I present the inevitable link to wikipedia:

Schrödinger's cat - Wikipedia, the free encyclopedia

The gist of it -- and how it applies to the above -- is as follows: in the 'schrodinger's cat' thought experiment, the cat could either be alive or dead; however, due to some obscure equation in quantum mechanics, the cat must be considered both alive and dead at the same time (i.e. it exists in a 'superposition' of both states). However, we could, at any time, "open the box" the cat is in and 'observe' whether it is indeed still alive.

The idea is analogous to our situation with whether or not the 'next card dealt' is the ace of spades. In probability theory, the card has a probability assigned to it, based on known information, that it is indeed the ace of spades. However until we actually observe it is that exact card, we can merely assign a probability.

(At least that's my rudimentary understanding of it. I looked it up in wikipedia, for god's sake. Go me.)

And I thought there'd be no way to learn something from this thread.Props to you sir for looking that up and explaining it to the rest of us mere mortals.

???Who makes the Quantum and do their mechanics make as much as say a Ferrari mechanic?

Originally Posted by Belt

So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it?

Awesome, awesome quote. Love it!