1. Lol, thats confusing.

2. Actually - It could...
... with a different set of assumptions.

Now I take a card from the stack and open it - suppose - it is a King
However, this is not what the host does: the host does not select at random (except when the choice is arbitrary because all the available options are equivalent).

In this case, the first choice does not matter, which is what Mario F. meant, and which I misinterpreted and thus misstated.

EDIT:
Because we are talking here about 2 different probability spaces - one is unconditional space P(.), second is conditional space p(.|C) where C is the known event that the door opened by host does not contain price... so probabilities in these two spaces can be different...
Yes, but for the purposes of the game we are assuming a set of rules, so unconditional space does not apply.

3. Still 1/13th. You drew the second card from the reduced pool of 63 cards with possibly only 3 kings. The math is ugly, but I think it would still come out right.

4. Still 1/13th. You drew the second card from the reduced pool of 63 cards with possibly only 3 kings. The math is ugly, but I think it would still come out right.
Mind providing the maths? I have not yet taken my statistics module in university, and my 'A' level maths is terribly rusty concerning statistics and probability.

It may be easier to simplify the problem to 4 cards: 2 red, 2 black. I pick one of the cards, vart picks another at random, and it turns out to be a red card. Then we compute the probability of my card being red.

5. Originally Posted by CornedBee
Still 1/13th. You drew the second card from the reduced pool of 63 cards with possibly only 3 kings. The math is ugly, but I think it would still come out right.
No.
I do not ask what is the probability for me to draw a King. I said - I have DRAWEN a King.

Now A - is event that You have drawen a King
B - is event I have drawn a King

P(A) = 4/52
p(B) = 4/52
P(B|A) = 3/51

P(AB) = P(A)P(B|A) = P(B)P(A|B) (= 4/52 * 3/51)

So P(A|B) == P(B|A) == 3/51

6. so the general rule is, if the game host opens a bad door, we switch; otherwise we stay?

i heard the best way to tackle the original problem is to imagine there are 1000 doors. The host opens 998 doors for the players, and ask the player whether to change door. the player will instantly grab the new door because the odds of the player being right in the first place is far less than the odds after he switch. kind of circular, but i can see why now.

kind of curious what the smart dude would think of this?

--TING

7. >> so the general rule is, if the game host opens a bad door, we switch; otherwise we stay?
In the original scenario, the host always opens a bad door (and you always switch). In other scenarios, I was thinking it depends on the ratios, but maybe not. This all assumes the host purposefully opens a particular door.

>> i heard the best way to tackle the original problem is to imagine there are 1000 doors.
I almost posted that thought (or maybe I did). It is a common way to think about the problem in another way.

Now, what is the probability, that the card that you have choosen before, but haven't open yet is a King?
Note the "before" in your sentence. My draw happened before your draw, thus it is P(A), not P(A|B).
P(A|B), the probability that I draw a king under the assumption that you have drawn a king, is only relevant if I draw after you did.
It's a case of the maths being too abstract for the real-world case you describe.

Although I have to admit I'm not quite comfortable with my line of thinking. I might be proven wrong yet.

Still trying to extend the scenario ... for example, you follow my draw by drawing four cards, and they're all kings. Now obviously the probability that I have a king is zero. Which disproves my line of thought and proves yours. But that's under the assumption that I haven't drawn a king in the first place, because if I have, you cannot draw four kings. So it might be said that claiming to have drawn four kings would be invalid.

9. Originally Posted by CornedBee

Note the "before" in your sentence. My draw happened before your draw, thus it is P(A), not P(A|B).
P(A|B), the probability that I draw a king under the assumption that you have drawn a king, is only relevant if I draw after you did.
It's a case of the maths being too abstract for the real-world case you describe.
What in the conditional probability makes you believe that the order of event is important?

The formula I have shown you has no indication of order of events... And it works both ways equally

10. What in the conditional probability makes you believe that the order of event is important?

The formula I have shown you has no indication of order of events... And it works both ways equally
I never was happy with this even when I learned it. Which would explain why I'm having such a hard time with it now.

11. The question is, after you know that the other person has drawn a king, what are the odds that your card is a king also. At that point, there are 51 unknown cards, 3 of which are kings, so the answer is 3/51.

12. So it might be said that claiming to have drawn four kings would be invalid.
Even when you saw 4 king lying before me you will still doubt if this is possible? Just because you could draw a king before me?
Obviously - if I had shown you 4 kings - you HAVEN't drawn one...

13. Originally Posted by CornedBee
Still 1/13th. You drew the second card from the reduced pool of 63 cards with possibly only 3 kings. The math is ugly, but I think it would still come out right.
Think of it this way. Draw a first card, don't look at it. Then draw a second card and look at it. If it turns out to be the ace of spades, I'm sure you agree that the probability of the first card being the ace of spades is 0.

EDIT:I guess the same reasoning was posted before me...

14. Here's another (harder) switching problem:

Suppose instead there are two doors both having a sum of money behind them. One of the doors contain twice the amount as the other one. Suppose you choose one door, open it and the sum of the money behind it is X. Now you're given an option to switch to the other door. Should you do it?

If the door you switch to is higher, you gain X. If you switch to a lower door, you lose 0.5X. Therefore you gain 0.5X on average by always switching.

Where's the problem?

15. @Sang-drax: A variation of Pascal's Wager, except of course he had a big flaw in his calcs.

Bit late but.........

Originally Posted by mike_g
I was talking to this guy that loves gambling the other day, and he had this theory about how to win lots of money.

Using a roulette example (ignoring 0 for simplicity sake) he said that each time one colour follows a colour of the same type the probability of the ball landing on the opposing colour doubles. EG: 3 blacks in a row means that on the next spin theres a 1:8 probability that by betting red you would win.

<snip>

Anyone know whats right here?
Tell his guy he should bet WITH the run, not against it.

This is because the run (of the same colour) can continue (theoretically) forever. The run can end only once (and he has to pick that one spin).

'Back your luck' is the saying.

I used to deal AR at a casino.

The spin is varied by;
training the dealer putting different amount of force when flicking the ball,
spinning the wheel opposite directions,
and spinning with different hands.

The AR wheel is divided into three sections, Voisins du Zero, Tiers and Orphelins.
Many punters are looking to see if the spins (or the alternating spins) hit the same section, and then have a pattern bet on that section (ie a call bet on the 'Orphans' one spin, then the Neighbours' the next).

Dealers who hit the same area of the wheel (each or alternating spins) are called 'section spinners' and are quickly removed by the casino.

Card counting on BJ is your best method of pushing the odds into your favour, but auto shufflers have stopped that here (dealer's card here is visable).