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[Galzria]

This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.
I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,
1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

(This is basically identical to the problem in the Venture thread, heh.) Anyway, the relevant parts here are 2b/2c, because we already know your first hand is CCCCE, which tells us your remaining cards are CCCEE (in some order).

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%. The information is predetermined with the original shuffle - This is much more clear when you think of revealing the BOTTOM card, rather than the NEXT card. Shuffle a deck (26/26, or 7/3), and remove the bottom card, without looking at it. What are the odds that it's red/black (or Estate/Copper)? Now reveal X cards, one at a time from the top of the deck. The odds on the removed card don't change. Yes, you could reveal all the other cards, and know with certainty what that last card IS, but it's odds at any given moment are predefined.

Because of that, when evaluating what %chance you have to hit $5, the only relevant information is that bottom card, which since you havn't changed it's odds since the original shuffle, when it had 30%, is still 30%.

[Voltgloss]

To clarify my last post (I first made edits, then saved, then saw 8 or so posts after it - my bad):


The issue is that 18.5(d) asks you to somehow change your "winning" probability from 50% without yet having any information about the shuffle results (i.e., without having yet dealt any cards from the deck).  You are therefore confronted with a universe of possible outcomes in which exactly half will "win" and half will "lose."  Of course your chances of "winning" are therefore 50%, and nothing can be done to change that.

BUT, once you flip some cards from the deck, you eliminate a number of outcomes from the possible universe of results.  If you eliminate more "losing" outcomes than "winning" outcomes, your probability of winning - knowing that information - goes up.  And vice versa if you eliminate more "winning" than "losing" outcomes.

Back to the Dominion example, where "winning" = Estate on the bottom, drawing your first hand of 5 cards eliminates all "losing" outcomes in which (1) the 3 Estates are in the top 5 cards; and (2) 2 of the 3 Estates are in the top 5 cards.  Whereas the only "winning" outcomes eliminated are those in which none of the Estates are in the top 5 cards.  More losing outcomes were eliminated than winning outcomes, and that is why the chance of winning goes up.

But notably, this is NOT what happened. You didn't take CCCEE shuffled, and then ADD CCCCE to the top. You took CCCCCCCEEE, shuffled, and then revealed CCCCE. This produces different odds, even though we both know that the bottom 5 cards are the same set of CCCEE in some order.

No, these two situations produce the same odds, AFTER having revealed CCCCE.  In the second scenario, while the ORIGINAL probability of Estate-on-the-bottom is 30% (vs. 40% in the first scenario), the CONDITIONAL probability of Estate-on-the-bottom, after revealing the top 5 cards, is 40%.

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%

No, it's not.  Per 18.5(c), the odds that the next card is black in your example (with 22 black and 14 red remaining in the deck) are 22/36, or 61.1%.

Say you flip over 26 red cards and 0 black cards.  Are the odds of your next card being black still 50%?

[blueblimp]

This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.
I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,
1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

(This is basically identical to the problem in the Venture thread, heh.) Anyway, the relevant parts here are 2b/2c, because we already know your first hand is CCCCE, which tells us your remaining cards are CCCEE (in some order).

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%

No. Imagine you flipped over 26 red cards and no black cards (so there are only black cards left). Is the chance that the next card is black 50%, in that case?

(To be fair, conditional probability is counter-intuitive.)

[Toskk]

Or, think of it this way: Take 3 coppers and two estates, and shuffle them. Then pick up 4 coppers and an estate from the supply. (You do this because if your first hand is not CCCCE you can't buy the Nomad Camp). What is the probability that the bottom card of the deck is an estate?

But notably, this is NOT what happened. You didn't take CCCEE shuffled, and then ADD CCCCE to the top. You took CCCCCCCEEE, shuffled, and then revealed CCCCE. This produces different odds, even though we both know that the bottom 5 cards are the same set of CCCEE in some order.

Oh god, my head.. it hurts.. this thread is insane, and people seem to be attempting to calculate two totally different things.

The original poster asked, providing your opening hand is CCCCE and you buy a Nomad Camp, what are your odds of getting $5 on turn 2. In this case, we absolutely *do not* care what the odds are of drawing the opening CCCCE. Randomdragoon and others are very correct that the odds of the $5 T2 hand is 40%.. the problem is precisely identical to picking one card from CCCEE (this picked card is the one you don't get for your T2 hand).

But people like Galzria are apparently attempting something totally different.. they're instead talking about the odds of drawing first CCCCE and *then* drawing Nomad Camp + CCCE. The odds of these two occurrences together are *not* 40%.

[michaeljb]

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%.

So part c is asking you to prove true something that is really false? 22/40

edit: cut the quote down to the relevant size, and realized my 22/40 probability isn't right. 22/40 would be if you flipped over 12 total, 4 being black: (26-4)/(52-12). I misread it; with flipping 12 red and 4 black, it's actually (26-4)/(52-12-4) = 22/36, much more than 50%

[timchen]

But people like Galzria are apparently attempting something totally different.. they're instead talking about the odds of drawing first CCCCE and *then* drawing Nomad Camp + CCCE. The odds of these two occurrences together are *not* 40%.

He is not attempting something totally different. He is just totally wrong.

[Toskk]

But people like Galzria are apparently attempting something totally different.. they're instead talking about the odds of drawing first CCCCE and *then* drawing Nomad Camp + CCCE. The odds of these two occurrences together are *not* 40%.

He is not attempting something totally different. He is just totally wrong.

.. I was trying to give him the benefit of the doubt..

[O]

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%.

Don't want to be rude, but you need to retake your statistics class. A finite sized deck is not a head or tails coin. The odds are nowhere near 50%.

[Galzria]

This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.
I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,
1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

(This is basically identical to the problem in the Venture thread, heh.) Anyway, the relevant parts here are 2b/2c, because we already know your first hand is CCCCE, which tells us your remaining cards are CCCEE (in some order).

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%. The information is predetermined with the original shuffle - This is much more clear when you think of revealing the BOTTOM card, rather than the NEXT card. Shuffle a deck (26/26, or 7/3), and remove the bottom card, without looking at it. What are the odds that it's red/black (or Estate/Copper)? Now reveal X cards, one at a time from the top of the deck. The odds on the removed card don't change. Yes, you could reveal all the other cards, and know with certainty what that last card IS, but it's odds at any given moment are predefined.

Because of that, when evaluating what %chance you have to hit $5, the only relevant information is that bottom card, which since you havn't changed it's odds since the original shuffle, when it had 30%, is still 30%.

So part c is asking you to prove true something that is really false? 22/40

Not exactly. Yes, you have a better chance of winning - That is, there are more cards remaining in your favor than against - But the odds of the next card, or any given, regardless of revealed information, remain 50%.

[jonts26]

This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.
I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,
1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

(This is basically identical to the problem in the Venture thread, heh.) Anyway, the relevant parts here are 2b/2c, because we already know your first hand is CCCCE, which tells us your remaining cards are CCCEE (in some order).

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%. The information is predetermined with the original shuffle - This is much more clear when you think of revealing the BOTTOM card, rather than the NEXT card. Shuffle a deck (26/26, or 7/3), and remove the bottom card, without looking at it. What are the odds that it's red/black (or Estate/Copper)? Now reveal X cards, one at a time from the top of the deck. The odds on the removed card don't change. Yes, you could reveal all the other cards, and know with certainty what that last card IS, but it's odds at any given moment are predefined.

Because of that, when evaluating what %chance you have to hit $5, the only relevant information is that bottom card, which since you havn't changed it's odds since the original shuffle, when it had 30%, is still 30%.

So part c is asking you to prove true something that is really false? 22/40

Not exactly. Yes, you have a better chance of winning - That is, there are more cards remaining in your favor than against - But the odds of the next card, or any given, regardless of revealed information, remain 50%.

That's just ... no. No, I'm sorry.

[Galzria]

Alright, look - I can't find the source material right now, despite looking through everything I can find, so I'll abstain for the time being.

I was hoping that somebody else would have encountered this particular paradox, and could help me if they had the proof on hand (I imagine mine got recycled at some point, but I'm usually pretty organized and don't get rid of anything).

If I can find the information, I'll bring it out, but for now, follow the intuitive answer of %40 - However, it's still really %30. 

[michaeljb]

Not exactly. Yes, you have a better chance of winning - That is, there are more cards remaining in your favor than against - But the odds of the next card, or any given, regardless of revealed information, remain 50%.

What is this I don't even

[Voltgloss]

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%. The information is predetermined with the original shuffle - This is much more clear when you think of revealing the BOTTOM card, rather than the NEXT card. Shuffle a deck (26/26, or 7/3), and remove the bottom card, without looking at it. What are the odds that it's red/black (or Estate/Copper)? Now reveal X cards, one at a time from the top of the deck. The odds on the removed card don't change. Yes, you could reveal all the other cards, and know with certainty what that last card IS, but it's odds at any given moment are predefined.

Because of that, when evaluating what %chance you have to hit $5, the only relevant information is that bottom card, which since you havn't changed it's odds since the original shuffle, when it had 30%, is still 30%.

I think the issue is that you are defining "odds" differently from the rest of us.  You appear to be defining "odds" as "the chance this card is X when the deck is first shuffled."  We are all defining "odds" as "the chance this card is X, knowing what we know about the rest of the deck."  In other words, you are talking about initial probability; we are talking about conditional probability.  See section 18.3 of the textbook you cite.

Importantly, the example you raise is NOT from the "conditional probability" section of the textbook you cited.  A better example is problem 18.14:

Problem 18.14.
A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards.
(a) If you have one ace, what is the probability that you have a second ace?
(b) If you have the ace of spades, what is the probability that you have a second ace?
Remarkably, the two answers are different. This problem will test your counting ability!

Last two lines are straight from the textbook.

[Robz888]

This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.
I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,
1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

(This is basically identical to the problem in the Venture thread, heh.) Anyway, the relevant parts here are 2b/2c, because we already know your first hand is CCCCE, which tells us your remaining cards are CCCEE (in some order).

And yet, if I've flipped over 12 red cards, and 4 black cards, the odds that the next card is black is STILL 50%. The information is predetermined with the original shuffle - This is much more clear when you think of revealing the BOTTOM card, rather than the NEXT card. Shuffle a deck (26/26, or 7/3), and remove the bottom card, without looking at it. What are the odds that it's red/black (or Estate/Copper)? Now reveal X cards, one at a time from the top of the deck. The odds on the removed card don't change. Yes, you could reveal all the other cards, and know with certainty what that last card IS, but it's odds at any given moment are predefined.

Because of that, when evaluating what %chance you have to hit $5, the only relevant information is that bottom card, which since you havn't changed it's odds since the original shuffle, when it had 30%, is still 30%.

So part c is asking you to prove true something that is really false? 22/40

Not exactly. Yes, you have a better chance of winning - That is, there are more cards remaining in your favor than against - But the odds of the next card, or any given, regardless of revealed information, remain 50%.

They absolutely do not remain 50%. If there is more of one card than the other, the odds of drawing that card are higher.

[Fabian]

Galzria,

Just want to make sure I'm getting your argument, please answer the following!

1. I shuffle my deck and draw my starting hand. My starting hand is 3 Estate and 2 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?


2. I shuffle my deck and draw my starting hand. My starting hand is 2 Estate and 3 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

3. I shuffle my deck and draw my starting hand. My starting hand is 1 Estate and 4 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

[timchen]

I think your problem might be that you are confused by the idea of conditional probability.

Let us talk about a very simple deck with 1 copper and 1 estate. Sure, each card in the deck will have 50% being either.

I would assume that you agree that after I flip the first card to be an estate, the remaining card has to be a copper.

 Yet in this case you claim the second card has 50% to be an estate. The only plausible explanation to me is that you have some sort of idea of "prior probability" in mind: that is, you are thinking about this 50% chance in the beginning, or in an ensemble, that is if you do this many times, there will be 50% chance for the second card to be an estate.

But that is not the question we are interested. The question is what is the chance of the second card being an estate, GIVEN THAT the first card was flipped and shown to be an estate. And the answer certainly is not 50%.

[jonts26]

Alright, look - I can't find the source material right now, despite looking through everything I can find, so I'll abstain for the time being.

I was hoping that somebody else would have encountered this particular paradox, and could help me if they had the proof on hand (I imagine mine got recycled at some point, but I'm usually pretty organized and don't get rid of anything).

If I can find the information, I'll bring it out, but for now, follow the intuitive answer of %40 - However, it's still really %30. 

I think no one else is helping you out here because you are quite obviously wrong. Either you are using a definition of odds which is not the standard definition or you have some sort of fundamental misunderstanding of conditional probability.

And it's been said many times already but let me try again. Probability and odds are a measure of unknown information. Once I reveal to you new information (like the first 5 cards) the odds can and do change. This is called conditional probability, the odds of something happening given a certain amount of information. The odds start at 30% but once you draw your hand with 1 estate they change to 40%.

Also, ss someone mentioned before, I highly recommend you read up on the monty hall problem for a particularly mind bending example of this. http://en.wikipedia.org/wiki/Monty_Hall_problem

[Galzria]

Galzria,

Just want to make sure I'm getting your argument, please answer the following!

1. I shuffle my deck and draw my starting hand. My starting hand is 3 Estate and 2 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

The odds that the bottom card is an Estate IS still 30% - BUT - You know that it will be 0% of the time, since you have 3 of 3 in hand. This only holds true, however, because you've revealed 100% of the remaining. Thusly:

2. I shuffle my deck and draw my starting hand. My starting hand is 2 Estate and 3 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

The odds that the bottom card is an Estate IS still 30% - BUT - In this case, intuitively, 1 in 5 remain to be an Estate, so it would SEEM to be 20%, but it isn't, because it's initial probability hasn't changed. Until you reach 0% or 100%, it remains 30% because that was the information the system was given to start. EVERY card is 30% until all are revealed. That is, even when you can know 100% of the time, if it IS or ISN'T, IT'S odds are still 30%.

3. I shuffle my deck and draw my starting hand. My starting hand is 1 Estate and 4 Copper. Is the probability of me having an Estate on the bottom of my library a) 0% b) 30% c) 40% d) other?

As above, The odds of the bottom card is an Estate IS still 30% - BUT - In this case, intuitively, 2/5 remain to be an Estate, so it would SEEM to be 40%, but it isn't.

-- Either way! -- I can't provide the proof right now, so I'm happy to abstain and let it stand with the intuitive answer until I can.

[GendoIkari]

40%

Thanks. I have poor Nomad Camp luck; it seemed like I never managed to hit $5. To prove cognitive bias though, I just went through all my games that had Nomad Camp. Turns out I bought it on turn one 12 times; hit $5 4 of those. So worse than average; but only a bit.

[Voltgloss]

Or, Galzria, how about this variant situation:

- Your deck consists of 3 Provinces, 1 Tactician, 1 Wishing Well, 8 Coppers, and 2 Silvers.
- Your opponent has 4 Provinces, 1 Curse, and no other victory point cards.  Buying the last Province will win you this 2-player game.
- You played Tactician last turn, discarding the rest of your hand.  Your deck was then empty, so you reshuffled (this reshuffle contains 3 Provinces, 1 Wishing Well, 8 Coppers, and 2 Silvers).
- You then drew 3 Provinces, 6 Coppers, and 1 Wishing Well.  You play your Wishing Well and draw another Copper.
- Your discard pile is empty.

Here are the questions:
- What are the odds your next card is Copper?
- What are the odds your next card is Silver?
- To have the best chance of winning this turn, what should you wish for?

[Galzria]

Alright, look - I can't find the source material right now, despite looking through everything I can find, so I'll abstain for the time being.

I was hoping that somebody else would have encountered this particular paradox, and could help me if they had the proof on hand (I imagine mine got recycled at some point, but I'm usually pretty organized and don't get rid of anything).

If I can find the information, I'll bring it out, but for now, follow the intuitive answer of %40 - However, it's still really %30. 

I think no one else is helping you out here because you are quite obviously wrong. Either you are using a definition of odds which is not the standard definition or you have some sort of fundamental misunderstanding of conditional probability.

And it's been said many times already but let me try again. Probability and odds are a measure of unknown information. Once I reveal to you new information (like the first 5 cards) the odds can and do change. This is called conditional probability, the odds of something happening given a certain amount of information. The odds start at 30% but once you draw your hand with 1 estate they change to 40%.

Also, ss someone mentioned before, I highly recommend you read up on the monty hall problem for a particularly mind bending example of this. http://en.wikipedia.org/wiki/Monty_Hall_problem

In particular Jonts, this came up (for me, years ago), following the Monty Hall problem, when after solving it, we were provided with a list of other seeming Paradox's to choose from, and provide the proofs for. This was one of them. It is just as unintuitive - That is, the easy answer of Monty Hall's problem is 33%, as is the easy answer here 40% (or in the case of the 26/26 card deck, whatever the odds would SEEM based on what you KNOW is left) - But it isn't. As Monty's is actually 66%, so here is it actually 30%, and in the case of the playing deck, 50%.

[GendoIkari]

If I have a 52 card deck, equal red and equal black, shuffled to together randomly, and I start to reveal cards to you one at a time...:
You may tell me to stop at any time, and guess what the color the bottom card will be. Can you ever increase your odds better than 50% that it will be black? What if I show you the NEXT card instead?

-- Against intuition, the answer is NO. It is always 50/50, because those odds were determined with initial information input of 26/26 - and even as you remove cards, and the total remaining may change, it doesn't change the initial odds on any GIVEN card from the remaining to be 50/50.

Um, actually the answer is yes. Given that the top card is red; you now have a less than 50% chance that the bottom card is also red.

The reason this is different than the Monty Hall problem is that the top card was flipped randomly. With Monty Hall, he knows what is where, and ALWAYS opens a door that has a goat. He does not just choose a door to open randomly. This is why the odds of your door being the car don't change. But with the card example (and with the Dominion example), it's different. The top card had a chance to be black, but it turns out it wasn't. This changes things. If, on the other hand, you said "I'm now going to show you a red card" and then remove 1 red card from the deck; the odds of the bottom card haven't changed.

[DStu]

-- Either way! -- I can't provide the proof right now, so I'm happy to abstain and let it stand with the intuitive answer until I can.

... which will be, if you want to provide a correct proof, ONCE AND FOR ALL!

[Thisisnotasmile]

For the record, I am now +1ing all of the exceptional trolling in this thread for it's exceptionalness.

[GendoIkari]

Alright, look - I can't find the source material right now, despite looking through everything I can find, so I'll abstain for the time being.

I was hoping that somebody else would have encountered this particular paradox, and could help me if they had the proof on hand (I imagine mine got recycled at some point, but I'm usually pretty organized and don't get rid of anything).

If I can find the information, I'll bring it out, but for now, follow the intuitive answer of %40 - However, it's still really %30. 

I think no one else is helping you out here because you are quite obviously wrong. Either you are using a definition of odds which is not the standard definition or you have some sort of fundamental misunderstanding of conditional probability.

And it's been said many times already but let me try again. Probability and odds are a measure of unknown information. Once I reveal to you new information (like the first 5 cards) the odds can and do change. This is called conditional probability, the odds of something happening given a certain amount of information. The odds start at 30% but once you draw your hand with 1 estate they change to 40%.

Also, ss someone mentioned before, I highly recommend you read up on the monty hall problem for a particularly mind bending example of this. http://en.wikipedia.org/wiki/Monty_Hall_problem

In particular Jonts, this came up (for me, years ago), following the Monty Hall problem, when after solving it, we were provided with a list of other seeming Paradox's to choose from, and provide the proofs for. This was one of them. It is just as unintuitive - That is, the easy answer of Monty Hall's problem is 33%, as is the easy answer here 40% (or in the case of the 26/26 card deck, whatever the odds would SEEM based on what you KNOW is left) - But it isn't. As Monty's is actually 66%, so here is it actually 30%, and in the case of the playing deck, 50%.

Here's another way to think of it. Let's say instead of revealing the top card, you reveal the top 26 cards. And, based on crazy shuffle randomness, all 26 of them were red! According to your argument, there is still a 50/50 chance of the bottom card being red. So, do you think that is correct? Given that the top 26 cards are all red, what are the odds of the bottom card being red? 0 or .5?