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Author Topic: Math request: Nomad Camp  (Read 43787 times)

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GendoIkari

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Math request: Nomad Camp
« on: April 17, 2012, 12:00:01 pm »
0

If you have $4 in your opening hand, and buy a Nomad Camp, what are the odds that you will have $5 on turn 2?
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Fabian

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Re: Math request: Nomad Camp
« Reply #1 on: April 17, 2012, 12:02:44 pm »
+7

40%
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Higara

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Re: Math request: Nomad Camp
« Reply #2 on: April 17, 2012, 12:08:06 pm »
0

40%

What he said. To expand, you're looking for the probability that the bottom card of your deck is an estate. Before you buy the Nomad Camp, you have five cards in your deck, of which two are estates. Since buying the Nomad Camp doesn't change the order of cards beyond the top, you have a 2 (estates)/5 (cards) probability of drawing three coins.
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Galzria

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Re: Math request: Nomad Camp
« Reply #3 on: April 17, 2012, 12:08:13 pm »
+2

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

Voltgloss

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Re: Math request: Nomad Camp
« Reply #4 on: April 17, 2012, 12:10:11 pm »
+4

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.
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DStu

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Re: Math request: Nomad Camp
« Reply #5 on: April 17, 2012, 12:11:59 pm »
+1

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.
I would guess they would rise to 100%...
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Voltgloss

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Re: Math request: Nomad Camp
« Reply #6 on: April 17, 2012, 12:12:59 pm »
0

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.
I would guess they would rise to 100%...
How do the odds of the last card in your deck being Estate rise to 100% if you had all three Estates in your starting hand?
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DStu

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Re: Math request: Nomad Camp
« Reply #7 on: April 17, 2012, 12:16:09 pm »
0

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.
I would guess they would rise to 100%...
How do the odds of the last card in your deck being Estate rise to 100% if you had all three Estates in your starting hand?
Oh, I was thinking about the probability that you have $5 in hand 2.  But of course context of "it" changed in between.
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Galzria

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Re: Math request: Nomad Camp
« Reply #8 on: April 17, 2012, 12:18:26 pm »
+1

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.

Give me a minute to pull up the necessary information. You're intuition is correct - And if I were to SHOW you those remaining 5 cards, and THEN reshuffle them, it would be 40%. But because the information entered into the system was 7 Coppers and 3 Estates at the first shuffle, the odds of the BOTTOM card have NOT changed, and were determined at that time to be 30%.

And edit while I find what I need:

The original problem was stated thusly:

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.
« Last Edit: April 17, 2012, 12:27:08 pm by Galzria »
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

DStu

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Re: Math request: Nomad Camp
« Reply #9 on: April 17, 2012, 12:22:18 pm »
+1

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.

Give me a minute to pull up the necessary information. You're intuition is correct - And if I were to SHOW you those remaining 5 cards, and THEN reshuffle them, it would be 40%. But because the information entered into the system was 7 Coppers and 3 Estates at the first shuffle, the odds of the BOTTOM card have NOT changed, and were determined at that time to be 30%.
Only because something did not change does not mean the probability did not change, see T1 CCEEE. The buttom card does also not change when you draw this hand, and still you won't try to convince me that there's still a probability of 30% that the 10th card is a 4th Estate, or?

Edit: Just because not ALL of the information got leaked in the first turn, that doesn't mean that NONE has leaked...
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Thisisnotasmile

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Re: Math request: Nomad Camp
« Reply #10 on: April 17, 2012, 12:22:42 pm »
0

30%.

The bottom card of your deck was decided upon first shuffle, and no new knowledge changes it's original odds of being an Estate.

No, 40% is correct.  The new knowledge that changes the odds is the fact that the first 5 cards of your deck are 4 Coppers and 1 Estate.  It's the same principle by which the odds plummet to 0% if your first hand is CCEEE.

Give me a minute to pull up the necessary information. You're intuition is correct - And if I were to SHOW you those remaining 5 cards, and THEN reshuffle them, it would be 40%. But because the information entered into the system was 7 Coppers and 3 Estates at the first shuffle, the odds of the BOTTOM card have NOT changed, and were determined at that time to be 30%.

If you can SHOW me those five remaining cards and they are not Copper, Copper, Copper, Estate, Estate, (in any order, it doesn't matter because we're going to be reshuffling them) then I'll believe you. Until then, 40% is correct.
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Fabian

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Re: Math request: Nomad Camp
« Reply #11 on: April 17, 2012, 12:23:08 pm »
+5

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Voltgloss

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Re: Math request: Nomad Camp
« Reply #12 on: April 17, 2012, 12:27:48 pm »
+2

Oh, I was thinking about the probability that you have $5 in hand 2.  But of course context of "it" changed in between.
Sorry about the confusion; it started out hungry but then wanted to take a walk instead.

Galzria, the question is not "in a shuffle of 7 Coppers and 3 Estates, what is the chance Estate is the bottom card."  Rather, the question is "in a shuffle of 7 Coppers and 3 Estates, knowing that the first five cards are 4 Coppers and 1 Estate, what is the chance Estate is the bottom card."

Are you familiar with the Monty Hall problem?  That's effectively what this is.
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DStu

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Re: Math request: Nomad Camp
« Reply #13 on: April 17, 2012, 12:33:13 pm »
+1

Quote
The original problem was stated thusly:

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.
You have to be very carefull with probabilities, small changes in the conditions can change a lot.
But as stated here, I tell you the answer is yes, because I will just wait 51 cards and count and thus know what the last card will be.

If, as I can not read in post, but anyway, the cards are not a equal number of red and blacks, but equally likely red or black, that's a completely different situation, and it's also a different situation than we have here. Because the cards of a Dominion deck are not Copper with 70% and Estate with 30%, but there are exactly 3 Estates and 7 Coppers. In the first case, if you know the first 9 cards, you don't know more about the 10th card, it is still 30% Estate and 70% Copper. In the second case, if you know the first 9 cards, you know the 10th, its Copper if you've just found 6 Coppers and Estate if you have found 2 Estates.

But in the first case that would also mean that the startung deck could consist of 10 Estates with probability 0.3^10.
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michaeljb

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Re: Math request: Nomad Camp
« Reply #14 on: April 17, 2012, 12:39:36 pm »
+2

Isn't it just as simple as this?

Original odds of bottom card being Estate 3/10.

Odds of bottom card being Estate, given exactly one Estate is in the top 5 cards, would be 2/5.

Odds of bottom card being Estate, given exactly two Estates are in the top 5 cards, would be 1/5.

Odds of bottom card being Estate, given exactly three Estates are in the top 5 cards, would be 0.
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Tables

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Re: Math request: Nomad Camp
« Reply #15 on: April 17, 2012, 12:39:45 pm »
0

The original problem was stated thusly:

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?

The ultimate question here is a trick, and you're being caught out by it. This asks if there's some strategy to improve the odds above 50% (actually, I think you might be able to improve the odds above 50%, although I'd need to revisit the drunkards walk to be certain, and that isn't relevant here). But that isn't the same as what's being asked here. Let me take an extreme example in your question. I can let you reveal any number of cards, correct? I'll let you reveal 51. At this point I can be 100% certain of what the bottom card is. BUT having revealed those 51, it's still a 50% chance the card will be black. Revealing 51 cards doesn't increase the odds the bottom card will be black, but after seeing 51, I'm not going to turn around and say the last card has a 50/50 chance, am I?

The same principle occurs here. Suppose I draw cceee as my first hand. What's the chance my bottom card is an estate now? It's not 30% any more, is it? Similarly, suppose I draw cccce. Well, the last 5 cards of my deck are cccee in some order, so there's a 2 in 5, 40% chance, it's an estate on the bottom.
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Galzria

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Re: Math request: Nomad Camp
« Reply #16 on: April 17, 2012, 12:52:19 pm »
+1

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.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

Kuildeous

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Re: Math request: Nomad Camp
« Reply #17 on: April 17, 2012, 12:58:11 pm »
0

When I have questions, I throw together a quick Excel sheet and see what distribution I get with 20k iterations. I'll go more if necessary.

It is indeed 40%. I wanted to verify that before cranking out the numbers, but it boils down to conditional probability. The question asked "If first hand has exactly 4 coppers then…" This is more than just calculating the probability of the last card. It's the probability of the last card, GIVEN that the first five cards are arranged a certain way. That's why it's not as simple as 30%.
 
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timchen

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Re: Math request: Nomad Camp
« Reply #18 on: April 17, 2012, 12:58:41 pm »
+1

This shouldn't need to be explained. Consider the probability of drawing a $4 hand in the second turn (without buying stuff like NC). Whatever you draw in the first turn, the probability becomes either 0% or 100%. It is just silly to say the probability does not change.
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Voltgloss

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Re: Math request: Nomad Camp
« Reply #19 on: April 17, 2012, 01:01:48 pm »
+6

Galzria, your error is that the question you are answering - 18.5(d) - isn't the question that we're all answering (or the question that was posed by the original poster).  We're all answering 18.5(c), with r = 3 (Coppers) and b = 2 (Estates).
« Last Edit: April 17, 2012, 01:13:32 pm by Voltgloss »
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randomdragoon

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Re: Math request: Nomad Camp
« Reply #20 on: April 17, 2012, 01:02:17 pm »
0

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 correct, but it is not the same situation here. In order to buy Nomad Camp you must have CCCCE in your first hand. This is akin to, in the above problem, saying you don't even get to play the game unless the first card turns up red; in this case, you will easily have higher than a 50% probability of winning.


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?
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Galzria

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Re: Math request: Nomad Camp
« Reply #21 on: April 17, 2012, 01:05:12 pm »
+1


Quote
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.
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Quote from: Voltgloss
Derphammering is when quickhammers go derp.

Faust has also been incredibly stubborn this game. In other news, it's hot in the summer, and water falls from the sky when it rains.


Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
SCUM Wins: M2, M19, M23, M100, DM3, RM1, RM2, RM48, RM50
SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

michaeljb

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Re: Math request: Nomad Camp
« Reply #22 on: April 17, 2012, 01:07:04 pm »
0

My intuition tells me if you can't guarantee better than 1/2 odds of winning it's because only black lets you win, not because you don't get new info. For example, if my strategy is to only take the next card when more reds have been revealed then blacks, I can't win if all the blacks land on top. But if I can win by predicting red or black, then my strategy just needs to include seeing at least one of each color and that stacked deck case becomes trivial.

The challenge in that problem is being restricted to black, not that the new info you get from revealing cards doesn't help you.
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O

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Re: Math request: Nomad Camp
« Reply #23 on: April 17, 2012, 01:09:53 pm »
0


Quote
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.

Except it doesn't produce different odds. This isn't the revealing door puzzle, where a conditional choice is made.
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blueblimp

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Re: Math request: Nomad Camp
« Reply #24 on: April 17, 2012, 01:10:37 pm »
+1

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).
« Last Edit: April 17, 2012, 01:15:20 pm by blueblimp »
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