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Old 22nd February 2019, 10:30 AM   #121
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Originally Posted by MEequalsIxR View Post
Saw this on another forum and the debate on the correct answer was interesting to me. No idea where it was originally was from. I thought I'd bring it here to see what the smart crowd does with it. Two possible answers - 15 or 60 which is correct.

My take on which answer is correct:

I believe 15 is correct based on PEMDAS but the way I was taught in grade school the answer would be 60 but then in those days printing was the new thing.
I got an exact answer of:

15 + 21/2

Or an estimated answer of:

16.4142...

Can anyone figure out how I got this answer?
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Old 22nd February 2019, 11:22 AM   #122
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The problem as given is not unsolvable.

For the same reason that we assume xx implies multiplication and thus = x squared, we can assume that two shocks next to each other with no indication of the appropriate operation is assumed to be multiplication.

I will use x for the symbol of each spring. The third level of the problem then reduces to:
5 + xx + xx = 9
x = plus square root 2 & minus square root

The final level has 2 possible solutions:
5 + (square root 2) X 10 = 5 + 10(square roots of 2) = approx. 19.14
5 - (square root 2) X 10 = 5 + 10(square roots of 2) = approx -9.14
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Old 22nd February 2019, 11:23 AM   #123
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Originally Posted by Crossbow View Post
I got an exact answer of:

15 + 21/2

Or an estimated answer of:

16.4142...

Can anyone figure out how I got this answer?
Very similarly to how I solved this problem, except I used PEMDAS.
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Old 22nd February 2019, 11:34 AM   #124
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Originally Posted by Thermal View Post
OK. Consider your two scenarios, and their possible outcomes, which you say are all the possibilities. They are exactly 50/50 that Ted goes free. Endgame, by your reasoning.

The problem with these things are that they are perspective shifting, not actual probabilities. The probability of Ted going free, as you so aptly demonstrated, is .5.

Put another way: 100 prisoners, one will be set free. Jim walks up after the first 98 prisoners have been executed, but he knows nothing about them. He is only faced with the fates of the last two. Jim's a betting man. He wants to place a bet on which goes free.

Under your reasoning, Ted has only a one in a hundred chance. Presumably, so does the other prisoner, from his perspective. You can be the House, and we'll tilt the odds in your favor, of course. You pay out only 50 to one on correct guesses. You in on this? I'll be Jim, and have my Paypal at the ready.

That isn't a valid extrapolation of the original prisoner problem.

In the original problem, the guard tells Ted that Bill will be executed. That doesn't change Ted's 1/3 chance of survival. If Jim walks up at that point and wants to make a fair bet that Ted will survive, the bet should reflect those odds. It should pay 2 to 1.

If Jim walks up after someone has been executed, though, and Ted is still alive, then his fair bet on Ted's survival should be even money.

What's changed? The fact that Ted isn't already dead at that point. After all, the guard never promised Ted that Bill would be executed first. Only that Bill would be one of the prisoners executed. So, Ted being alive after the first execution eliminates some previous possibilities.

Specifically, Ted has a 2/3 chance of being executed, and since the chance of being executed first out of two people is half the chance of being executed overall, there's a 1/3 chance he'll be executed first. Similarly there's a 1/6 chance of Harry being executed first, and a 1/2 chance Bill will be executed first.

If Ted is alive when Jim walks up after one execution (which happens in 4 chances out of 6), 1/4 of the time (1 chance out of the original 6) Bill will also be alive, and 3/4 of the time (3 chances out of the original 6) Harry will also be alive.

In the former case, Ted knows he'll survive, and Bill will definitely be executed.

In the latter case, nothing has changed for Ted; he knew Bill was going to die and his chances of survival remain 1/3 to Harry's 2/3.

So Ted's overall chance to survive if he's alive when Jim walks up is 1/4 * 1 (for the first case) + 3/4 * 1/3 (for the second case), or 1/4 + 1/4, or 1/2.

So Jim can place a fair bet at even odds on Ted's survival, if he doesn't know what the guard told Ted and/or he doesn't know the men and can't tell Bill from Harry.


In the 100 people (99 to be executed case), if Jim walks up and sees Ted and one other person remaining, he can indeed make a fair even money bet on Ted's survival, for the same overall reasons. Ted's originally small chance of survival has been increased by the happenstance that he hasn't been executed yet. That's true whether or not a guard the night before told him the names of 98 of the other prisoners who were going to die.

(If he'd been given 98 names, he'd be focusing his attention on the one prisoner whose name he wasn't given. Let's call that prisoner Aloysius. If Aloysius had been executed by the time only Ted and one other prisoner is still alive, then Ted knows for certain he'll live. In fact he's know that since Aloysius died. But if it's down to him and Aloysius, Ted knows he's almost certain to die. Assuming the order of the executions has been random throughout, this situation where Ted is one of the last two survivors is unlikely to come about, but if it does, Aloysius is close to equally likely to be dead or alive, giving Jim his fair 50-50 bet even though by then Ted knows much better what the outcome is going to be.)
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Old 22nd February 2019, 11:55 AM   #125
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Originally Posted by Thermal View Post
I am confident I am remembering correctly as it was posed in my intro class. What you are showing is a variant where 'other side/same color' has meaning. In the example posed in my class, it did not. We used GG, GS, and SS to represent the coins, and only determined probabilities for G and S.

eta: your 'other side/same color' wouldn't make sense as an additional variable if you are only determining gold or silver.

I an confident you are either remembering wrong, or correctly remembering a class that got the wrong answer. The common denominator is "wrong."

It doesn't matter what words or letters you use to represent the coins or what you call the two sides. The coins have two sides, right?

Before we formulate this into a wager, I urge you to simply try it yourself. Make three cardboard chips and write G on both sides of one, S on both sides of another, and G and S on either side of the third. Get a pencil and paper to keep score. Draw the chips out of a hat at random and look at a random side. If you see "S" then return it to the hat and draw again. If you see "G" then turn it over and see if it's also "G" on the other side. If it is, write a check mark on the paper. If it's not, write an X on the paper. Return it to the hat and draw again.
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Old 22nd February 2019, 11:58 AM   #126
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Originally Posted by Thermal View Post
Monte is a tricky variant. I am sticking with JoeMorgue's version for now, where he says the odds of the prisoners are not 50/50 when down to the fates of two.
right, after a beer I know where i'm going now. I (think) i can show why they're not 50/50.

lets look at the the possibilities
prisoners A,B and C

prisoner A talks to guard.

1. A lives, B and C die. guard says B dies.
2. A lives, B and C die. guard says C dies.

3. B lives, A and C die. guard says C dies

4. C lives, A and B die. guard says B dies

as it's random, then 1/3 of the time A lives, 1/3 of the time B lives and 1/3 of the time C lives.

so A = 1/3, B = 1/3, C = 1/3,

but half the time that A asks when he is the survivor, the guard answers B and half the time the guard answers C. Giving 2 equally possible answers, 1/2 chance B, 1/2 chance C.

so possibility 1. above happens 1/3 x 1/2 = 1/6 of the time,
possibility 2. happens 1/3 x 1/2 = 1/6 of the time
possibility 3. happens 1/3 of the time
possibility 4. happens 1/3 of the time.

When the guard answers B, that wipes out 2. and 3. leaving 1. and 4.

but possibility 1.(A lives) happens 1/6 of the time and
possibility 4.(C lives) happens 1/3 of the time

so when the guard says B, C is twice as likely to survive as A when A is the one asking.

That's why the odds aren't 50/50 when A is told by the guard that B dies.

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Old 22nd February 2019, 12:45 PM   #127
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Originally Posted by baron View Post
Iain Banks wrote of a similar weapon although he steered clear of religious associations.
One of my favourite writers when he has the M in the middle.
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Old 22nd February 2019, 12:59 PM   #128
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Originally Posted by BowlOfRed View Post
We could always try to resurrect the downwind faster than the wind instead
Ooh, ooh, can we please!!!
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Old 22nd February 2019, 02:21 PM   #129
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Originally Posted by ServiceSoon View Post
The problem as given is not unsolvable.

For the same reason that we assume xx implies multiplication and thus = x squared, we can assume that two shocks next to each other with no indication of the appropriate operation is assumed to be multiplication.

I will use x for the symbol of each spring. The third level of the problem then reduces to:
5 + xx + xx = 9
x = plus square root 2 & minus square root

The final level has 2 possible solutions:
5 + (square root 2) X 10 = 5 + 10(square roots of 2) = approx. 19.14
5 - (square root 2) X 10 = 5 + 10(square roots of 2) = approx -9.14
If we're allowed to make extra assumptions we can make the sum be anything we want. We could assume that in the problem as posted the symbols some take to be numeric base 10 digits are actually base 12, or we could assume the operators do completely different things than we are conventionally used to. And that's just for starters, there's no limit to what we can assume. The problem therefore doesn't become suddenly solveable when introducing your own assumptions, your solution just becomes gibberish to everyone not making the same assumptions.

And of course the problem as originally stated is just intended to troll people.
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Old 22nd February 2019, 02:27 PM   #130
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Originally Posted by Crossbow View Post
I got an exact answer of:

15 + 21/2

Or an estimated answer of:

16.4142...

Can anyone figure out how I got this answer?

Using the single shock as the square root of 2 it would seem.
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Old 22nd February 2019, 02:54 PM   #131
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Originally Posted by Dr. Keith View Post
Ooh, ooh, can we please!!!
Not going to happen unless you can come up with a Humber replacement. No thread can survive long if everyone is in agreement.
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Old 22nd February 2019, 02:56 PM   #132
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Originally Posted by BowlOfRed View Post
Not going to happen unless you can come up with a Humber replacement. No thread can survive long if everyone is in agreement.
Ok, give me a few minutes . . . [leaves room in a hurry]
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Old 22nd February 2019, 02:57 PM   #133
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Lightbulb

Originally Posted by p0lka View Post
right, after a beer I know where i'm going now. I (think) i can show why they're not 50/50.

lets look at the the possibilities
prisoners A,B and C

prisoner A talks to guard.

1. A lives, B and C die. guard says B dies.
2. A lives, B and C die. guard says C dies.

3. B lives, A and C die. guard says C dies

4. C lives, A and B die. guard says B dies

as it's random, then 1/3 of the time A lives, 1/3 of the time B lives and 1/3 of the time C lives.

so A = 1/3, B = 1/3, C = 1/3,
Yeah, a refreshing beverage does put things in perspective.

I think this part shows the disconnect. The structure of the dilemma does not allow the guard to tell A he gets the ax. That restricts the possibilities to the four you indicate, in which A lives in half. So right out of the gate, his odds of a surviving outcome are 50/50. But then we slip back to unrestricted odds, where there is a one in three of being freed.

This perspective shifting is what clouds the odds, so I turn to my home boy Jim again. Jim says that, knowing none of the background, if one of the three random prisoners are to be freed, and we know the fate of one, the odds are 50/50 for either of the other two to be the chosen one. He is still taking bets.

Quote:
but half the time that A asks when he is the survivor, the guard answers B and half the time the guard answers C. Giving 2 equally possible answers, 1/2 chance B, 1/2 chance C.

so possibility 1. above happens 1/3 x 1/2 = 1/6 of the time,
possibility 2. happens 1/3 x 1/2 = 1/6 of the time
possibility 3. happens 1/3 of the time
possibility 4. happens 1/3 of the time.

When the guard answers B, that wipes out 2. and 3. leaving 1. and 4.

but possibility 1.(A lives) happens 1/6 of the time and
possibility 4.(C lives) happens 1/3 of the time

so when the guard says B, C is twice as likely to survive as A when A is the one asking.

That's why the odds aren't 50/50 when A is told by the guard that B dies.
I remember from I think it was calc that when you arrive at a nonsensical answer, you approach the problem differently. That's why I keep consulting with Jimmy. The tail end odds remain, as you first posited, 50/50. I see these odds as indisputable, although I get the reasoning that the guard doesn't actually provide new information since either B or C was going down anyway, so naming the name should make no difference. But the endgame odds are fiddy fiddy, and me and Jimbo gots money on the table
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Old 22nd February 2019, 03:08 PM   #134
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Here's a restatement of the problem. Some folks like it better this way:

You have 100 marbles (with names on them). Put them in vase 1.

Executor comes in and picks one marble (presumably at random) to "save". Puts that one in vase 2.

Guard comes in and is told: "Go through vase 1 and pull out the first 98 marbles you can find that don't have "A"s name on it. Throw them in the 'dead' vase"

There are now two unidentified marbles in two vases. Because of the convoluted procedure, do you bet 50:50 that "A" is in vase1 vs vase2? No one has touched vase2 since the executor picked the 1-in-a-hundred marble.
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Old 22nd February 2019, 03:10 PM   #135
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Thermal - never be tempted to play scratchcards.
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Old 22nd February 2019, 03:18 PM   #136
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Originally Posted by baron View Post
Thermal - never be tempted to play scratchcards.
Never would. Odds are stacked against me.

Jim and I have money on the table. You in?
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Old 22nd February 2019, 03:27 PM   #137
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Originally Posted by Thermal View Post
Never would. Odds are stacked against me.
That's good, because I had a vision of you playing a scratch card, one of those where you have nine blobs that you 'scratch off' and if you reveal three matching numbers you win a prize, and you scratch off eight of them and you've got three pairs already and the odds of getting a match are really great and you scratch off that last one and.... damn, no match, but you were so near, so you buy another...
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Old 22nd February 2019, 03:30 PM   #138
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Originally Posted by BowlOfRed View Post
Here's a restatement of the problem. Some folks like it better this way:

You have 100 marbles (with names on them). Put them in vase 1.

Executor comes in and picks one marble (presumably at random) to "save". Puts that one in vase 2.

Guard comes in and is told: "Go through vase 1 and pull out the first 98 marbles you can find that don't have "A"s name on it. Throw them in the 'dead' vase"

There are now two unidentified marbles in two vases. Because of the convoluted procedure, do you bet 50:50 that "A" is in vase1 vs vase2? No one has touched vase2 since the executor picked the 1-in-a-hundred marble.
Wait: how many names? Three, one hundred different, or what? How many times does A appear? And there are three vases, 1&2& a dead one?
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Old 22nd February 2019, 03:39 PM   #139
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Originally Posted by BowlOfRed View Post
Not going to happen unless you can come up with a Humber replacement. No thread can survive long if everyone is in agreement.
Damn, he's been gone a long time. Last post in 2011. I feel old now. Get off my lawn!
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Old 22nd February 2019, 03:42 PM   #140
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Originally Posted by Thermal View Post
Wait: how many names? Three, one hundred different, or what? How many times does A appear? And there are three vases, 1&2& a dead one?
Why are you trying to complicate the problem like this? It is a restatement of the original Monty Hall problem with 100 unique entities instead of 3, with still only one of those entities having the attribute you want (a car, staying alive, whatever).

This restatement was meant as a way to make everybody who has a problem seeing how the initial odds keep playing a role after extra information was revealed realize their mistake. But you introduced some extra crud onto it which can only serve to cloud your understanding of the problem.
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Old 22nd February 2019, 03:44 PM   #141
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Originally Posted by Thermal View Post
Yeah, a refreshing beverage does put things in perspective.

I think this part shows the disconnect. The structure of the dilemma does not allow the guard to tell A he gets the ax. That restricts the possibilities to the four you indicate, in which A lives in half.
A doesn't live in half of the 4 possibilities.

A lives in 2 out of the 4 possibilities but the possibilities are not weighted equally because of the guard having the freedom to pick between two names.
It's right there in my post.

so possibility 1. above happens 1/3 x 1/2 = 1/6 of the time,
possibility 2. happens 1/3 x 1/2 = 1/6 of the time
possibility 3. happens 1/3 of the time
possibility 4. happens 1/3 of the time.

lets add it up,

A lives in possibility 1. and possibility 2.

that's 1/6 + 1/6 which equals 1/3.

That's A's chance.

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Old 22nd February 2019, 03:48 PM   #142
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Originally Posted by baron View Post
That's good, because I had a vision of you playing a scratch card, one of those where you have nine blobs that you 'scratch off' and if you reveal three matching numbers you win a prize, and you scratch off eight of them and you've got three pairs already and the odds of getting a match are really great and you scratch off that last one and.... damn, no match, but you were so near, so you buy another...
Jimbo is still drumming his fingers, waiting for someone to take him up. Good money, as he calculates it.

Say you have three quarters in your pocket. No, wait...you're from over there. You have three Euros.... ah, damn it. You have three one Pound notes.

You lose one on a bet with Jimmy. Your down by one third.

You bet with Harry, who too often gets overlooked in this, and win it back. You're now up 50%.

That's what I see happening with the prisoner odds. The reference point for determining odds gets shifted without comment, and it seems like a counterintuitive little paradox. That's why Jim offers to take anyone's money... er, place a fair wager.

The board consensus seems to be that poor Ted has a one in three chance of freedom, given Bill's known fate. I say its one in two. We can mock this up easily enough.

Any takers?
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Old 22nd February 2019, 03:59 PM   #143
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Originally Posted by Thermal View Post
Jimbo is still drumming his fingers, waiting for someone to take him up. Good money, as he calculates it.

Say you have three quarters in your pocket. No, wait...you're from over there. You have three Euros.... ah, damn it. You have three one Pound notes.

You lose one on a bet with Jimmy. Your down by one third.

You bet with Harry, who too often gets overlooked in this, and win it back. You're now up 50%.

That's what I see happening with the prisoner odds. The reference point for determining odds gets shifted without comment, and it seems like a counterintuitive little paradox. That's why Jim offers to take anyone's money... er, place a fair wager.

The board consensus seems to be that poor Ted has a one in three chance of freedom, given Bill's known fate. I say its one in two. We can mock this up easily enough.

Any takers?
No that would be like taking candy from a baby. Or taking a bet from someone over a math problem that was solved decades ago.
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Old 22nd February 2019, 04:03 PM   #144
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Originally Posted by Turgor View Post
Why are you trying to complicate the problem like this? It is a restatement of the original Monty Hall problem with 100 unique entities instead of 3, with still only one of those entities having the attribute you want (a car, staying alive, whatever).

This restatement was meant as a way to make everybody who has a problem seeing how the initial odds keep playing a role after extra information was revealed realize their mistake. But you introduced some extra crud onto it which can only serve to cloud your understanding of the problem.
Asking for you to be clear in your terms of the problem is not 'introducing extra crud'. Its requesting you to be clear.

Most of these dilemmas like to use three names. Hint: since they rely on binary odds, three is reeeeaaally convenient to slip the changing paradigm.

So I'll ask again, strictly for the clarity you did not provide: you have three vases, and one hundred different names, not repeated, yes? And one is 'A', correct?

While we're at it, you throwing down against Jimmy-J?
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Old 22nd February 2019, 04:04 PM   #145
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Originally Posted by Thermal View Post
Jimbo is still drumming his fingers, waiting for someone to take him up. Good money, as he calculates it.

Say you have three quarters in your pocket. No, wait...you're from over there. You have three Euros.... ah, damn it. You have three one Pound notes.

You lose one on a bet with Jimmy. Your down by one third.

You bet with Harry, who too often gets overlooked in this, and win it back. You're now up 50%.

That's what I see happening with the prisoner odds. The reference point for determining odds gets shifted without comment, and it seems like a counterintuitive little paradox. That's why Jim offers to take anyone's money... er, place a fair wager.

The board consensus seems to be that poor Ted has a one in three chance of freedom, given Bill's known fate. I say its one in two. We can mock this up easily enough.

Any takers?
Well, obviously I'd be a taker if it was a real bet, because I already know the solution.

If the examples don't convince you then I genuinely recommend you mock it up. Just write a quick piece of code to simulate the problem and set it running for a a few thousand iterations. I'd do it now if I had PHP on here but for some reason I can't find it right now.
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Old 22nd February 2019, 04:09 PM   #146
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Originally Posted by Turgor View Post
No that would be like taking candy from a baby. Or taking a bet from someone over a math problem that was solved decades ago.
You can prove Jim's POV wrong then, yes? To refresh you: Jim knows nothing about discussions with guards or anything else. He is betting on the scenario he walks into, that out of three prisoners, one randomly goes free. He knows the fate of one. He determines the odds of the remaining two to be a .5 chance of freedom. Other posters put their odds as less, one in three.

You putting cash against my home boy?
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Old 22nd February 2019, 04:14 PM   #147
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Originally Posted by Thermal View Post
Asking for you to be clear in your terms of the problem is not 'introducing extra crud'. Its requesting you to be clear.

Most of these dilemmas like to use three names. Hint: since they rely on binary odds, three is reeeeaaally convenient to slip the changing paradigm.

So I'll ask again, strictly for the clarity you did not provide: you have three vases, and one hundred different names, not repeated, yes? And one is 'A', correct?

While we're at it, you throwing down against Jimmy-J?
Let's generalize it even, since the exact amount of names (entities) is not that relevant:

There are 2 Vases, X and Y. There are Z (with Z being 3 or greater) names/marbles/doors, all different. Only one of those is 'A'.

All names are in Vase X. I pull out one name at random and put it in Vase Y.
What is the chance that Vase Y contains 'A'?

I now show you every name in Vase X, except for one. None of the names shown are 'A'.

What is the chance that Vase Y contains contains 'A'?

If you get two different answers for both times this question is asked, please show your work.
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Old 22nd February 2019, 04:19 PM   #148
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Ah, don't want to confuse things... Deleted for now. Go with Turgor's.

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Old 22nd February 2019, 04:20 PM   #149
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Originally Posted by Turgor View Post
Let's generalize it even, since the exact amount of names (entities) is not that relevant:

There are 2 Vases, X and Y. There are Z (with Z being 3 or greater) names/marbles/doors, all different. Only one of those is 'A'.

All names are in Vase X. I pull out one name at random and put it in Vase Y.
What is the chance that Vase Y contains 'A'?

I now show you every name in Vase X, except for one. None of the names shown are 'A', even if 'A'is still in vase X.

What is the chance that Vase Y contains contains 'A'?

If you get two different answers for both times this question is asked, please show your work.
ETA: added something that i highlighted to make it more clear
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Old 22nd February 2019, 04:28 PM   #150
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What are we betting?

A forty dollar bill or a golden fiddle?

Nothing's ever fifty-fifty

If that's your answer it's wrong 90% of the time.
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Old 22nd February 2019, 04:36 PM   #151
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Originally Posted by Thermal View Post
You can prove Jim's POV wrong then, yes? To refresh you: Jim knows nothing about discussions with guards or anything else. He is betting on the scenario he walks into, that out of three prisoners, one randomly goes free. He knows the fate of one. He determines the odds of the remaining two to be a .5 chance of freedom. Other posters put their odds as less, one in three.

You putting cash against my home boy?

How he knows the fate of one matters.

If he knows the fate of one because one prisoner has already been executed when he arrives, his fair bet on either of the remaining two is even money.

If he knows the fate of one because he asked the guards about the fate of that specific one and they told him truthfully that the one he asked about would die, his fair bet on either of the other two is even money.

If he knows the fate of one because before placing his bet he asked the guards to point out one of the two prisoners who was going to be executed, and they did so honestly, his fair bet on either of the remaining two is even money.

HOWEVER - If he knows the fate of one because he asked the guards whether a specific one "Ted" would live or die, and they refused to tell him, but instead pointed out that a certain one of the other two ("Bill") has been chosen to die, then his fair bet on Ted's survival is two to one (1/3 chance to win), and his fair bet on the third prisoner ("Harry") is one to two (2/3 chance to win). And that is the scenario that matches the original terms. None of the previous ones do.
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Old 22nd February 2019, 04:37 PM   #152
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Originally Posted by Turgor View Post
ETA: added something that i highlighted to make it more clear
For the first scenario, it is a 1/Z chance of it being the sole A, assuming all the perfect fair stuff.

For the second scenario, it is 50/50, as all other variables have been eliminated except two.
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Old 22nd February 2019, 04:42 PM   #153
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Originally Posted by Thermal View Post
For the first scenario, it is a 1/Z chance of it being the sole A, assuming all the perfect fair stuff.

For the second scenario, it is 50/50, as all other variables have been eliminated except two.
There is no second scenario, they are the same scenario continued. Where did the Z go in your second answer? Show me the calculation you did that made it irrelevant in the second answer.
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Old 22nd February 2019, 05:02 PM   #154
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Originally Posted by Turgor View Post
There is no second scenario, they are the same scenario continued. Where did the Z go in your second answer? Show me the calculation you did that made it irrelevant in the second answer.
Sorry, scenario was the wrong word. Second calculation of odds was what I meant. But you knew that, based on asking about the second answer. I deduce that you are being cranky by this faux confusion that you damn right well understood.

We can put Z in the second answer if you like, but it is no longer relevant, as they are not variables any longer. We are calculating the odds of unknowns having a name. This is that perspective blurring I was talking about, vacillating between variables and knowns, and Z>2 variables versus binary outcomes
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Old 22nd February 2019, 06:39 PM   #155
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Originally Posted by alfaniner View Post
All I know is, I'm shocked, tired, and piston ways you can't imagine.
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Old 22nd February 2019, 06:51 PM   #156
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Originally Posted by Thermal View Post
The tail end odds remain, as you first posited, 50/50. I see these odds as indisputable, although I get the reasoning that the guard doesn't actually provide new information since either B or C was going down anyway, so naming the name should make no difference. But the endgame odds are fiddy fiddy, and me and Jimbo gots money on the table
I never posited 50/50, stop saying that.
4 possibilities,
1. is 1/6
2. is 1/6
3. is 1/3
4. is 1/3.

they're not equal possibilities,
removing 2. and 3. leaves you with possibility 1. which is 1/6 and possibility 4. which is 1/3.
possibility 4. is not the same as possibility 1. Its literally double.

It is not 50/50 and nothing in what I posted said it was,

I'm pretty sure I nailed it in my previous post to show that it wasn't 50/50. If you just follow it I can't see how you would disagree.

EDIT: or I could be completely wrong, in which case somebody slap me and show me where.

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Old 22nd February 2019, 09:47 PM   #157
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Originally Posted by p0lka View Post
I never posited 50/50, stop saying that.
I did stop saying that. You went back to our first exchange for that quote, and are replaying it now as if I have persisted? I have not.

Quote:
4 possibilities,
1. is 1/6
2. is 1/6
3. is 1/3
4. is 1/3.

they're not equal possibilities,
removing 2. and 3. leaves you with possibility 1. which is 1/6 and possibility 4. which is 1/3.
possibility 4. is not the same as possibility 1. Its literally double.

It is not 50/50 and nothing in what I posted said it was,

I'm pretty sure I nailed it in my previous post to show that it wasn't 50/50. If you just follow it I can't see how you would disagree.

EDIT: or I could be completely wrong, in which case somebody slap me and show me where.
The reason I would disagree is Jim, over there, leaning against the wall and waiting for someone to throw down. You see, Jimmy kind of understands the root of the problem. The Three Prisoner Problem is not really a paradox of probability; it's an illustration of how restricting the question and shifting the perspective will screw up the odds.

For example, when we assume the guard can't tell who is being freed, it alters the number of possible responses, thus changing the odds from that limited perspective. But Jimmy is a pragmatist. He doesn't care about the perspectives. He cares about the objective odds. They are one in three, going in cold. Were he to know that one of the prisoners other than Ted/A will be executed, that prisoner is no longer a variable. Leaving Jim with 50/50 odds for the remaining unknowns.

Jim's oddsmaking doesn't care who is who or where they heard the info from or when (assuming it is accurate). His odds are based on who will be freed: if one of the three is eliminated, it is a coin flip between the other two.

I may also of course be completely wrong on this. But I need Jim's odds explained. If I am right on his probabilities, the whole Tree Prisoner's thing is a deceptive word game, rather than an interesting paradox.
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Old 22nd February 2019, 10:21 PM   #158
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Originally Posted by Myriad View Post
How he knows the fate of one matters.

If he knows the fate of one because one prisoner has already been executed when he arrives, his fair bet on either of the remaining two is even money.

If he knows the fate of one because he asked the guards about the fate of that specific one and they told him truthfully that the one he asked about would die, his fair bet on either of the other two is even money.

If he knows the fate of one because before placing his bet he asked the guards to point out one of the two prisoners who was going to be executed, and they did so honestly, his fair bet on either of the remaining two is even money.

HOWEVER - If he knows the fate of one because he asked the guards whether a specific one "Ted" would live or die, and they refused to tell him, but instead pointed out that a certain one of the other two ("Bill") has been chosen to die, then his fair bet on Ted's survival is two to one (1/3 chance to win), and his fair bet on the third prisoner ("Harry") is one to two (2/3 chance to win). And that is the scenario that matches the original terms. None of the previous ones do.
I have read this several times, and am honestly trying to understand the distinction you are making with the last scenario, but I confess I don't get it. If one of the prisoners was already executed, or was truthfully fingered to be executed, or whatever, the odds would be the same, tail end. Is this some kind of Schrodinger's Pardon? You may be sitting on the answer I am looking for, but I only see the claims you are making sans the justification.
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Old 22nd February 2019, 11:13 PM   #159
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Originally Posted by Turgor View Post
There is no second scenario, they are the same scenario continued. Where did the Z go in your second answer? Show me the calculation you did that made it irrelevant in the second answer.
You know, you are right on this one (although it is a different principle than the prisoner problem). The odds would be higher of the remaining A marble in the Y vase, because there was a (Z-1)/Z chance of the A marble being in there at the onset, and the other variables were systematically removed.

That is an interesting way to look at a specific probability scenario. But it is not analogous to the prisoner problem, because of the manipulation of variables.
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Old 23rd February 2019, 12:20 AM   #160
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Originally Posted by Thermal View Post
I have read this several times, and am honestly trying to understand the distinction you are making with the last scenario, but I confess I don't get it. If one of the prisoners was already executed, or was truthfully fingered to be executed, or whatever, the odds would be the same, tail end. Is this some kind of Schrodinger's Pardon? You may be sitting on the answer I am looking for, but I only see the claims you are making sans the justification.

Okay, maybe I can explain it this way. There are five variations of a card game. All four variations have the following rules in common:
- Three cards are face down in a row, two jacks and a queen.
- The cards are fairly shuffled, so any of the face-down cards is equally likely to be the queen in any game.
- I, the dealer, always know where the queen is.
- You, the player, pay $1 to play and are paid $2.25 if you win.
- You win if you turn up the queen.

Variation #1: When setting up the cards, the dealer lays out one of the jacks face up. You pick from the remaining two cards.

Variation #2: After the cards are set up, you point to a card for the dealer to turn up. If that card is revealed as a queen, the game is a draw (you get your $1 back). Otherwise, you then make your final pick from the remaining two cards.

Variation #3: After the cards are set up, you ask the dealer to show a jack, and he turns up one of the jacks. You pick from the remaining two cards.

Variation #4: After the cards are set up, you point to a face-down card of your choice. The dealer then turns up a jack that is not the card you're pointing to. You then turn up the card you're pointing to.

Variation #5: You simply pick one of the three face-down cards and turn it up.

I think you can see that Variation #5 is a scam. Your chance to win is only 1 in 3, but it pays out less than 2 to 1. For it to be a fair bet you'd have to get $3 when you win.

Variations #1 to #4 correspond closely to the four possibilities I listed for "how he knows the fate of one" in my previous post. In all four of those variations, when you turn your card up, one card has already been turned up. So the card you're turning up is one of only two cards remaining face down, one of which has to be the queen.

This might lead you to think Variations #1 to #4 all have a 1 in 2 chance to win, making them good bets because they pay out better than even money. And that's true for the first three of them. But it's not true for Variation #4. That has the same chance of winning as Variation #5, 1 in 3, which makes it a sucker bet.

What's the difference? In Variation #4 you have to choose from three equally likely choices, just as in Variation #5. In Variations #1, #2, and #3, the dealer's actions before your final pick helps you avoid picking a jack, by removing a jack before you make your choice, leaving you only two equally likely choices to choose from. In Variation #4, the dealer's action in turning up a jack you didn't pick doesn't help you avoid picking a jack, because it's too late. You've already picked.

In short: In Variations #1 to #3, the dealer reveals information helpful to making your choice, then you make your choice. Hence, better chances. Variation #4, you make your choice, then dealer reveals information that would have been helpful to making your choice. Doesn't help! (Unless you then get to change your choice, as in the original Monty Hall riddle.)

In the Prisoner problem, Ted's choice has already been made for him at the outset. Ted can only "bet on" himself holding the queen (the reprieve) and his chances of that never shift from 1/3.
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