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Author Topic: Math request: Nomad Camp  (Read 35674 times)

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Axxle

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Re: Math request: Nomad Camp
« Reply #75 on: April 17, 2012, 04:43:42 pm »
0

The pop quiz is actually a different problem...it's about game theory or logic more than probability, and the reasoning behind it is quite different.  The most common set up for it is as the Paradox of the Unexpected Hanging.

Yes, that would be it almost exactly. And like I said, they aren't the same problem. But the idea behind the prisoners hanging, or popquiz, was to illustrate a point, disregarding the fallacies of the logic. The key being "surprise", rather than semi-controlled randomness. It was still designed to get you to "the most likely day is tomorrow".

But following that logic, the Lightning Tower is a cleaner problem, because it erases the unknown factor. It still brings you to the most likely day, because anything after would be less probable. And that's, at the basic level, what the Prisoner's Hanging / Popquiz was trying to show. It just leaves too many invariables, and a loose definition "surprise". It draws out to conclusions that can't be made with certainty.
There's still the very real difference of Lightning Tower being about a chance of something happening each day for an infinite amount of time, with the probability of it happening not changing even with known information about previous days, and Popquiz being about something definitely happening in a finite space.  Like people have been saying this entire thread.  It's the difference between flipping coins and drawing cards.
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toaster

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Re: Math request: Nomad Camp
« Reply #76 on: April 17, 2012, 04:45:12 pm »
0

Except that the "solution" of the pop quiz/unexpected hanging problem *isn't* "have it tomorrow", nor is it trying to make a point anything like the lightning problem.  The entire point of the unexpected hanging problem is the loose definition of surprise and how that leads to a superficially "airtight" logical inference that turns out to be wrong.  The paradox isn't a problem with the unexpected hanging, it's the entire point of the scenario.
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Voltgloss

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Re: Math request: Nomad Camp
« Reply #77 on: April 17, 2012, 04:45:45 pm »
+2

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

Sorry, I'm too busy waiting for the basket of cabbages to figure out whether it's wearing a red hat or a white hat, based on the fox and the goat being unable to deduce same.
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eHalcyon

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Re: Math request: Nomad Camp
« Reply #78 on: April 17, 2012, 04:50:28 pm »
+3

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

Sorry, I'm too busy waiting for the basket of cabbages to figure out whether it's wearing a red hat or a white hat, based on the fox and the goat being unable to deduce same.

The bear that eats them all is white, because they are at the North Pole.
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Re: Math request: Nomad Camp
« Reply #79 on: April 17, 2012, 04:55:18 pm »
0

Except that the "solution" of the pop quiz/unexpected hanging problem *isn't* "have it tomorrow", nor is it trying to make a point anything like the lightning problem.  The entire point of the unexpected hanging problem is the loose definition of surprise and how that leads to a superficially "airtight" logical inference that turns out to be wrong.  The paradox isn't a problem with the unexpected hanging, it's the entire point of the scenario.

Exactly. The Lightning Town and the Pop Quiz are very different. The Lightning Town, I think, is supposed to illustrate why random events appear to occur non-randomly. The lightning storms tend to look planned, rather than random, because they happen in clusters. But they look like clusters to us for good reason, because the day of the next lightning storm is more likely to be tomorrow than 1,000 years from now.

It's similar to the coin flip. People expect roughly the same number of heads as tails, but a random series of coin flips could very well come out TTTTTTHHTT, and so on.
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michaeljb

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Re: Math request: Nomad Camp
« Reply #80 on: April 17, 2012, 04:58:20 pm »
0

Fair enough that I havn't been able to provide a proof for my reasoning yet, but I've done my best to explain it in Lamens terms without it.

I've figured it out! You just need to find your magical pendant, then you'll be able to give us this elusive proof.

edit: FWIW, if the starting hand is dealt, but no one looks at it to discover that it consists of 4 Coppers and 1 Estate, I do believe the odds of the bottom card being an Estate are indeed 30%. As long as you don't look at that starting hand.
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toaster

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Re: Math request: Nomad Camp
« Reply #81 on: April 17, 2012, 05:12:08 pm »
0

Exactly. The Lightning Town and the Pop Quiz are very different. The Lightning Town, I think, is supposed to illustrate why random events appear to occur non-randomly. The lightning storms tend to look planned, rather than random, because they happen in clusters. But they look like clusters to us for good reason, because the day of the next lightning storm is more likely to be tomorrow than 1,000 years from now.

It's similar to the coin flip. People expect roughly the same number of heads as tails, but a random series of coin flips could very well come out TTTTTTHHTT, and so on.

Actually, the Lightning Town example doesn't illustrate the events tend to happen in clusters...if you take the same problem but state that there *wasn't* a lightning storm today, the most likely date of the next storm remains the same: tomorrow. 

On the other hand, your point about what humans think of as random and how they perceive clusters where there aren't any is right on the money.
« Last Edit: April 17, 2012, 05:26:24 pm by toaster »
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WanderingWinder

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Re: Math request: Nomad Camp
« Reply #82 on: April 17, 2012, 05:22:57 pm »
+7

So I think Galzria's problem is this. Here's the proof that he's talking about that he remembers. You have the red/black card game thing. When you flip a card over, it of coursse doesn't change the distribution of cards, which were determined by the shuffle. However, the new information of this top card of course changes what we know. Specifically, every distribution of cards which has that top card at any position other than the top card has been eliminated. And so now, after that's happened, we have a better idea about each of the other cards in the deck. It doesn't make it more likely that your shuffling process produces a deck with black card on the bottom. It just changes what we know about this particular instance, this particular shuffled deck end product, because certain distributions we thought to be possible before we now know are not possible. Ok.
Now, so if the top card is revealed and we know that it's red, then that makes it more likely that the bottom card is black, not because it changes the shuffle order, but because there's more different possibilities left that have black cards on bottom (because there's more black cards left in the deck) than there are that have red on bottom. And so at this point, you CAN make predictions better than 50/50. But the clever proof he's remembering shows that you can't actually take advantage of this extra information to win more than 50% of the time, and the reasoning is, because you need the bottom card to be a particular color to win. And the only time you can actually have any choice (let's say you get to pick which color it is), is before the distribution of cards is fixed. Or let's say you can pick when to stop and look at the bottom card. Fine. Obviously, if you stop when there's more cards in the deck that are black than are red, you're more likely for it to be black. But the marginal gains you get there are totally cancelled out by the losses you get from basically never getting to the situation where you've revealed more reds than blacks. Because every black you reveal hurts your probabilities, and you're as likely to get this hurt as you are to get any help. Now, if you could choose to never go for that bottom card, then you win every time - just wait until all the cards left in the deck are what you need them to be. And this jives with our intuition that no part of looking at the top X cards of the deck will change the color of the bottom card. Because it doesn't. But that doesn't mean we can't guess better now, because we have more information.

So the analogous thing here is basically saying that that bottom card is either an estate, or it isn't. And buying the nomad camp doesn't change that. Drawing your first hand doesn't change that. BUT, there are 10! ways of distribution your 10 cards (assuming you have like the faces marked 0-9, so you can tell the difference). 30% of those have an estate on the bottom. And after you see that first hand, man, that doesn't change that 30% of the time you shuffle, you get an estate on the bottom. BUT, after you see your first hand, you know that a whole bunch of those distributions can't be right, because the first 5 cards don't conform. So you have to throw all those out as impossible, and you space of possible distributions shrinks. So in this case, all those cases that had 0 or 1 or 3 estates in the bottom 5 cards aren't viable anymore, because we know from the first hand that those bottom 5 MUST contain 2 estates. And hands containing 2 estates have an estate as the 5th card more often than when we have this distribution from 0 to 3 that we know about before we shuffle, and there you get the 40% number.

Now, I would love to get a statistics sub-forum so I can go off on this idiotic Tuesday boy birthday garbage :)

michaeljb

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Re: Math request: Nomad Camp
« Reply #83 on: April 17, 2012, 05:29:16 pm »
+5

So I think Galzria's problem is this.

...


While I really enjoyed the rest of your post, and completely agree with it, I'm not sure I agree that that's what Galzria's problem is....after all there was this:

Quote
Galzria,

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

Quote
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

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Taco Lobster

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Re: Math request: Nomad Camp
« Reply #84 on: April 17, 2012, 07:39:57 pm »
0

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

Sorry, I'm too busy waiting for the basket of cabbages to figure out whether it's wearing a red hat or a white hat, based on the fox and the goat being unable to deduce same.

The bear that eats them all is white, because they are at the North Pole.

Or on the Island...
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DStu

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Re: Math request: Nomad Camp
« Reply #85 on: April 18, 2012, 02:02:32 am »
+1

The bear that eats them all is white, because they are at the North Pole.

Is this an african or an european Polar Bear?
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jonts26

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Re: Math request: Nomad Camp
« Reply #86 on: April 18, 2012, 02:06:51 am »
+1

I...I don't knowwww aahhhhhhhhh.
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Kirian

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Re: Math request: Nomad Camp
« Reply #87 on: April 18, 2012, 02:25:04 am »
0

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

Thank you Donald, this is now my FB status.
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Axxle

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Re: Math request: Nomad Camp
« Reply #88 on: April 18, 2012, 02:47:08 am »
0

Also: You are standing on one side of a river, with a fox, a goat, and a basket of cabbages. One of them always lies, one always tells the truth, and one is a basket of cabbages. You have to determine which is which using only one weighing on a scale.

Thank you Donald, this is now my FB status.
Mine too, hahaha ;D
I added a bit to the end though. "Do you testify against your partner?"
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Kirian

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Re: Math request: Nomad Camp
« Reply #89 on: April 18, 2012, 02:51:04 am »
+1

The pop quiz is actually a different problem...it's about game theory or logic more than probability, and the reasoning behind it is quite different.  The most common set up for it is as the Paradox of the Unexpected Hanging.

What's awesome is that 5 clicks away from that paradox on Wikipedia I got to analysis paralysis, an experience many of us have had in Dominion as well.  I love this thread, but the answer is 40%.
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Axxle

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Re: Math request: Nomad Camp
« Reply #90 on: April 18, 2012, 04:27:15 am »
0

The pop quiz is actually a different problem...it's about game theory or logic more than probability, and the reasoning behind it is quite different.  The most common set up for it is as the Paradox of the Unexpected Hanging.

What's awesome is that 5 clicks away from that paradox on Wikipedia I got to analysis paralysis, an experience many of us have had in Dominion as well.  I love this thread, but the answer is 40%.
5 clicks is a lot of clicks: It's also 5 clicks away from Hitler (Paradox > Impossible object > M.C._Escher > Dutch people > Hitler).
Thank Godwin.
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qmech

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Re: Math request: Nomad Camp
« Reply #91 on: April 18, 2012, 04:49:43 am »
+1

But the clever proof he's remembering shows that you can't actually take advantage of this extra information to win more than 50% of the time, and the reasoning is, because you need the bottom card to be a particular color to win.

The extra small observation you need is that while in the original game you were betting on the the next card being black, it's entirely equivalent to be betting on the bottom card being black, as it's just one of the remaining cards chosen uniformly at random.  Formally it's a simple coupling argument.

It's slightly embarrassing that this is the topic that gets me to register for such excellent forums.

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BadAssMutha

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Re: Math request: Nomad Camp
« Reply #92 on: April 18, 2012, 09:45:55 am »
+1

One comment about the Lightning Town problem - while the most likely day for the next lightning storm is indeed tomorrow, we can still say that there will probably not be a lightning storm until next week. By this I mean that  the cumulative probability of there being a storm won't reach 50% until a week has passed. Each day has a decreasing probability of being the first day that it rains, but we need to add these all up to get the intuitive answer that storms happen, on average, once a week. Hope I didn't open another can of worms here.
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Davio

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Re: Math request: Nomad Camp
« Reply #93 on: April 18, 2012, 09:57:07 am »
0

Just sneakily look at the bottom card through a glass table and be done with it.
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toaster

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Re: Math request: Nomad Camp
« Reply #94 on: April 18, 2012, 11:47:30 am »
0

One comment about the Lightning Town problem - while the most likely day for the next lightning storm is indeed tomorrow, we can still say that there will probably not be a lightning storm until next week. By this I mean that  the cumulative probability of there being a storm won't reach 50% until a week has passed. Each day has a decreasing probability of being the first day that it rains, but we need to add these all up to get the intuitive answer that storms happen, on average, once a week. Hope I didn't open another can of worms here.

Actually, that's not correct.  If storms happen once per week, the median time between storms is actually about 4.5 days.  For some very inexact intuition, think about that fact that at a minimum, the time to the next storm is a day, but there is no maximum time to the next storm...and thus for a mean interval of a week, there are more intervals less than a week than greater than a week.  See http://en.wikipedia.org/wiki/Geometric_distribution
« Last Edit: April 18, 2012, 11:50:28 am by toaster »
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BadAssMutha

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Re: Math request: Nomad Camp
« Reply #95 on: April 18, 2012, 12:05:44 pm »
0

One comment about the Lightning Town problem - while the most likely day for the next lightning storm is indeed tomorrow, we can still say that there will probably not be a lightning storm until next week. By this I mean that  the cumulative probability of there being a storm won't reach 50% until a week has passed. Each day has a decreasing probability of being the first day that it rains, but we need to add these all up to get the intuitive answer that storms happen, on average, once a week. Hope I didn't open another can of worms here.

Actually, that's not correct.  If storms happen once per week, the median time between storms is actually about 4.5 days.  For some very inexact intuition, think about that fact that at a minimum, the time to the next storm is a day, but there is no maximum time to the next storm...and thus for a mean interval of a week, there are more intervals less than a week than greater than a week.  See http://en.wikipedia.org/wiki/Geometric_distribution

True, but my main point was that even though tomorrow's the most likely day for the first storm, it still probably won't rain for a few days. I was pretty inexact with my definition of "average" time between storms (mean, median, etc.), plus I didn't read close enough to see if there is on average, "one storm per week" or "a week between storms", which are indeed be different. Gotta triple check the details before post here, I suppose.
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Avin

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Re: Math request: Nomad Camp
« Reply #96 on: April 18, 2012, 04:59:35 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 canít 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 donít 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.

I can prove this (D2).

What I will show is that the odds of winning this game when there are n cards left in the deck (in other words 52-n cards have been flipped over) in an arbitrary deck are 50%.

This can be expressed as the sum from b=0 to n of the probability of there being b black cards left multiplied by the odds of winning with b black cards left in the deck.

Here's the interesting part. Consider all permutations of 52 cards and divide them up into two groups, ones which have "black" as the final card on one side and ones with "red" as the final card on the other side. You can form a one to one correspondence between these two groups because if you take each card in one group and swap each black card with a red card and each red card with a black card (in other words, inverting the color of each card), you will get a unique permutation from the other group.

Now you can see that the probability of there being b=k black cards left has to be the same as the probability of there being b=(n-k) black cards left, because for each permutation that has k black cards left, we can use that inverting function to obtain a corresponding permutation that has n-k black cards left.

Also, the odds of winning when there are b black cards left out of n cards is b/n (see part c, only I am using n instead of r+b), and the odds of winning when there are n-b black cards left is (n-b)/n. If we suppose we were playing on both situations simultaneously, the odds then of winning one of them is the sum of the odds of winning either one, since the decks are opposite each other:

(b/n) + (n-b)/n = (n-b+b) / n = n/n = 1

So now we can reduce our big summation above by pairing up those inverted scenarios:

sum from b=0 to n of the probability of there being b black cards left multiplied by the odds of winning with b black cards left in the deck
= sum from b=0 to n/2 rounded down of the probability of there being b black cards left multiplied by the odds of winning with either b black cards or n-b black cards left in the deck *
= sum from b=0 to n/2 rounded down of the probability of there being b black cards left multiplied by 1
= 1 * sum from b=0 to n/2 rounded down of the probability of there being b black cards left
= 1/2 * sum from b=0 to n of the probability of there being b black cards left
= 1/2 * 1, since the sum of the probability of every possibility has to be 1
= 1/2

* edit to note the case in the above step that if n is even, then the case where b=n/2 is also 1/2 because then n-b=b, meaning there are just as many reds as blacks left.

P.S. 40% to the OP
« Last Edit: April 18, 2012, 05:21:06 pm by Avin »
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qmech

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Re: Math request: Nomad Camp
« Reply #97 on: April 18, 2012, 05:18:37 pm »
0

Avin, that works for strategies of the form "wait until there are n cards left, then guess", but there are more complicated strategies that you could follow (you can take into account the number of black cards seen so far, or make a choice randomly if you like).  The easiest way to sidestep these problems is to observe that the problem doesn't change if you switch to betting on the last card instead of the next card, as mentioned above.

The hardest thing about the question as presented is that it comes after three basic calculations, which primes you to calculate rather than look for a more elegant solution.
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Avin

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Re: Math request: Nomad Camp
« Reply #98 on: April 18, 2012, 05:29:25 pm »
0

Avin, that works for strategies of the form "wait until there are n cards left, then guess", but there are more complicated strategies that you could follow (you can take into account the number of black cards seen so far, or make a choice randomly if you like).  The easiest way to sidestep these problems is to observe that the problem doesn't change if you switch to betting on the last card instead of the next card, as mentioned above.

The hardest thing about the question as presented is that it comes after three basic calculations, which primes you to calculate rather than look for a more elegant solution.

You can't "guess" because the requirement is that the card has to be black - you don't get to say what color it is. If you got to pick what color then there WOULD be a strategy that would win 100% of the time - wait until you've seen 26 cards of either color, then guess the opposite color for the next card.

And I believe the proof above is sufficient for all possible strategies because it doesn't take into account what strategy you're using, it just takes into account which turn the final declaration was on. So you could randomly pick a turn by rolling a 52 sided die prior to the game begins and guess that way, or you could attempt to wait until you've seen more reds than blacks and declare then - it wouldn't matter.
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blueblimp

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Re: Math request: Nomad Camp
« Reply #99 on: April 18, 2012, 05:56:50 pm »
0

Since this same problem came up in the Venture thread, I've thought a bit about solutions.

My favourite so far is:

Since you don't know the order of the remaining cards, it's irrelevant whether you draw the next card from the top of the deck or the bottom of the deck. (This can easily be proven formally using a bijection argument.) So, imagine that every non-bet card is drawn from the bottom of the deck, whereas once you bet the next card is black, then the next card is drawn from the top of the deck. This game is the same as the original one, from a probability standpoint.

Now play the new game with a fixed shuffling of the deck. No matter when you choose to bet, it's obviously not going to affect the top card of the deck, since the timing of your bet only affects how many cards you first draw from the bottom. Therefore, your strategy does no different than 50%.
« Last Edit: April 18, 2012, 06:01:40 pm by blueblimp »
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