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Old 21st February 2019, 04:15 PM   #81
Myriad
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Originally Posted by baron View Post
The other prisoner is doing cartwheels.

Thus proving she was the guilty one all along!
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Old 21st February 2019, 04:18 PM   #82
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Originally Posted by Myriad View Post
Thus proving she was the guilty one all along!
You understood that reference.
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Old 21st February 2019, 04:55 PM   #83
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Originally Posted by baron View Post
The easiest way to grasp the prisoner one is to imagine 100 prisoners from which only 1 will be freed; the rest will be executed. A's chance of being freed is therefore 1 in 100.

The guard, being a nice guy, agrees with A that he will shoot 98 of the prisoners who are destined for death right there and then, but he will leave A and one other alive. This he does.

It's much easier to see that A's chances are not suddenly 1 in 2 that he'll be set free, they're still 1 in 100.

The other prisoner is doing cartwheels.
Haha, indeed.
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Old 21st February 2019, 07:22 PM   #84
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Wow. 99% of posts for the answer wrong. Uncanny.
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Old 21st February 2019, 07:34 PM   #85
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Originally Posted by a_unique_person View Post
Wow. 99% of posts for the answer wrong. Uncanny.
Well, as has been pointed out, the problem as given is unsolvable.
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Old 21st February 2019, 07:42 PM   #86
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I didn't think I had to add smileys for Australians, only for Americans.
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Old 21st February 2019, 09:07 PM   #87
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Originally Posted by a_unique_person View Post
I didn't think I had to add smileys for Australians, only for Americans.
Sorry, we don't have humour here, and I sometimes fail to notice unless I'm concentrating.
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Old 22nd February 2019, 05:59 AM   #88
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Originally Posted by JoeMorgue View Post
Because "Tell me which of us are going to die" and "Tell me which us are going to die but you can't tell me specifically if I am going to die" are not the same question, so the "missing" probability that on the surface looks like a jump from 1/3 to 1/2 is hidden in the fact that the guard can't tell Ted directly if he is going to live so one possibility has been taken off the table, so it has to be assumed.

Look at it this way. In all scenarios the guard will not actually lie about anything and in all scenarios he is forbidden from directly telling Ted whether or not he will live or die. He is only allowed to give a definitive statement about 1 of the 2 prisoners who isn't Ted.

Scenario 1: Ted is the one who has been selected to be released in the morning. That means Harry or Bill are going to die. Here's all the possible scenarios.

1. Bill and Harry are going to die, the guard tells Ted that Harry will be executed. Bill and Harry are executed. Ted goes free.
1. Bill and Harry are going to die, the guard tells Ted that Bill will be executed. Bill and Harry are executed. Ted goes free.

Scenario 2. Ted is one of the two scheduled to be executed. That means either Harry or Bill is going to go free. Here's all the possible scenarios.

1. Bill and Ted are going to die, the guard tells Ted that Bill is going to die. Harry goes free.
2. Ted and Harry are going to die, the guard tells Ted that Harry is going to die. Bill goes free.

That's it. That's all possibilities. Any other combination of executions/information given by the guard would require the guard to either lie, directly tell Ted if he is going to live or die, or give information about both the other two prisoners, known of which he does.

If I'm wrong walk me through scenario with a different outcome that doesn't break the rules.
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.
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Old 22nd February 2019, 06:16 AM   #89
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Originally Posted by Thermal View Post
But it has. Bill is no longer possible to be the one going free.
But there was always going to be at least one other person than him who was going to die. The guard's information changes nothing.
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Old 22nd February 2019, 06:20 AM   #90
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I think the mistake people are making is thinking that because there are two options, they both have the same chance of occurring. The look at the statement

* Bill will be executed OR
* Ted will be executed

and conclude that the chances of each are 50/50. This is not necessarily so.
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Old 22nd February 2019, 06:30 AM   #91
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Originally Posted by Belz... View Post
But there was always going to be at least one other person than him who was going to die. The guard's information changes nothing.
It does. It positively eliminates a variable as one going free. Yes, one was going to anyway. But as each variable is eliminated, the odds change in the real world.

Would you take the bet I proposed a couple posts up? You can be the House. Hell, your payout can be only 10 to one. You in?
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Old 22nd February 2019, 06:34 AM   #92
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Originally Posted by baron View Post
I think the mistake people are making is thinking that because there are two options, they both have the same chance of occurring. The look at the statement

* Bill will be executed OR
* Ted will be executed

and conclude that the chances of each are 50/50. This is not necessarily so.
If the choice of who was executed is random, that would exactly be the odds. No?
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Old 22nd February 2019, 06:41 AM   #93
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Originally Posted by JoeMorgue View Post
Hey here's a good one along these lines, to step away from the Monty Hall archetype for a moment.

You are standing before the King, facing the possibility of execution. (All great philosophical math problems have to include threat of death.)

Two jars sit in front of you. Each jar contains 50 white marbles and 50 black marbles. You have to reach, sight unseen, into one of the two jars and pick out a marble. White marble, you go free. Black marble, you meet the ax man.

You ask the king if you can rearrange the marbles to increase your odds. The King, on the advice of his scheming Vicar, agrees. The number of jars and marbles is fixed, you cannot increase your odds.

You dump all the marbles into one jar so it has 100 black and 100 white. You then take one white marble and put it in the empty jar.

Boom. You have now increased your odds from 50% to 75% (well 74.99% or whatever if you want to get technical.)

You now have a 50/50 chance of picking a jar with only white marbles, and if you by chance pick the jar with both marbles you have a (nearly) 50/50 chance of picking a white marble.
Yes, but this is gaming a system where you can influence the odds. Whole different structure, that. The Monte and Prisoner problems have no means of restructuring the possibilities to influence your odds
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Old 22nd February 2019, 06:51 AM   #94
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Originally Posted by Thermal View Post
It does. It positively eliminates a variable as one going free. Yes, one was going to anyway. But as each variable is eliminated, the odds change in the real world.
Let's look at it this way:

Either Ted and another prisoner dies, or another prisoner and another prisoner die. In both cases (Ted lives or dies) at least another prisoner dies. This is 100% of the time, so clarifying that doesn't change the odds.
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Old 22nd February 2019, 06:51 AM   #95
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Originally Posted by Thermal View Post
It does. It positively eliminates a variable as one going free. Yes, one was going to anyway. But as each variable is eliminated, the odds change in the real world.

Would you take the bet I proposed a couple posts up? You can be the House. Hell, your payout can be only 10 to one. You in?
I think your bet proposition would only relate if jim was one of the prisoners.

100 prisoners, 98 are killed leaving jim and 1 other prisoner,
the odds that jim is one of the 99 killed are not equal to the odds of jim being the 1 in 100 that goes free,

jim should not bet on himself, he should switch his bet to the other prisoner.

99 times out of a hundred jim will be in the executed group, and if he bets on the other prisoner he would win,
only 1 time in a hundred will he lose as he was picked to go free that 1 time.

It's equivalent to the way the monty hall problem works.

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Old 22nd February 2019, 07:07 AM   #96
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Originally Posted by Belz... View Post
Let's look at it this way:

Either Ted and another prisoner dies, or another prisoner and another prisoner die. In both cases (Ted lives or dies) at least another prisoner dies. This is 100% of the time, so clarifying that doesn't change the odds.
I had a similar problem in Stats or Philosophy back in school:

You have three coins. One is gold both sides, one silver both sides, and one gold on one side and silver on the other. If you pick one from a bag, and it has gold on one side, what are the odds that it is gold on the other?

Intuitively, one might think that the odds were one in three. but by showing gold on one side, you have eliminated the two silver sided coin from possibility. That means your odds are 50/50 of having gold on the other side.

The prisoner problem is the same, plus or minus: by eliminating one possibility positively, he is no longer a variable, and the odds are recalculated with the remaining variables, at .5

I get the 'one was guaranteed to die anyway, so the guard gives no new information' approach. But the guard does- he eliminates one variable of going free. That changes Ted from being one possibility in three to go free to one in two.
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Old 22nd February 2019, 07:09 AM   #97
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Originally Posted by Thermal View Post
I had a similar problem in Stats or Philosophy back in school:

You have three coins. One is gold both sides, one silver both sides, and one gold on one side and silver on the other. If you pick one from a bag, and it has gold on one side, what are the odds that it is gold on the other?

Intuitively, one might think that the odds were one in three. but by showing gold on one side, you have eliminated the two silver sided coin from possibility. That means your odds are 50/50 of having gold on the other side.

The prisoner problem is the same, plus or minus: by eliminating one possibility positively, he is no longer a variable, and the odds are recalculated with the remaining variables, at .5

I get the 'one was guaranteed to die anyway, so the guard gives no new information' approach. But the guard does- he eliminates one variable of going free. That changes Ted from being one possibility in three to go free to one in two.
That's where we don't connect: you always knew that at least another prisoner was going to die. The guard tells you nothing you didn't know beforehand.
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Old 22nd February 2019, 07:12 AM   #98
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Originally Posted by Myriad View Post
In the Monty Hall problem, switching changes your odds exactly because NOT switching does NOT change your odds. That is, the revelation of one goat doesn't alter your odds of winning with the door you initially chose, which remain at 1/3. Which means switching improves your odds to 2/3.

In the Prisoner problem, you have no option to switch, so your odds of survival don't change. They remain at 1/3. (The odds of survival of the third person, who is not you and not the doomed person the guard names, improves to 2/3.)
Okay, got you now. No matter what, one of the other prisoners would die. That you know which one it was doesn't make a difference. That makes more sense; I was thinking as if you asked about a specific prisoner. It's analogous to not switching in Monty Hall.
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Old 22nd February 2019, 07:13 AM   #99
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Originally Posted by p0lka View Post
I think your bet proposition would only relate if jim was one of the prisoners.

100 prisoners, 98 are killed leaving jim and 1 other prisoner,
the odds that jim is one of the 99 killed are not equal to the odds of jim being the 1 in 100 that goes free,

jim should not bet on himself, he should switch his bet to the other prisoner.

99 times out of a hundred jim will be in the executed group, and if he bets on the other prisoner he would win,
only 1 time in a hundred will he lose as he was picked to go free that 1 time.
Disagreed. Jim is a constant, removed from the perspective shifting that is clouding the odds.

Jim walks up with only two prisoners left. What are the odds of his bet? They should be the same as Bill's, Ted's, the guard's, or ours.

And my wager remains on the table for those who think the odds are other than 50/50.

Quote:
It's equivalent to the way the monty hall problem works.
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.
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Old 22nd February 2019, 07:19 AM   #100
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Like I said I prefer the 3 Prisoner Problem because it's mathematically identical but it doesn't have the two things that people get hung up on with the Monte Hall version, the motivation of Monte and the person being able to change their decision after the new information is presented to them.

One because those two variables sort of cancel each other out if you think about, in this version the Monte equivalent, the guard, isn't specifically trying to trick you into a specific outcome but since you have no control over the outcome it makes no difference but more so because neither of them actually changes the underlying mathematics of the problem.

Both of them trick you into directly comparing two similar but not equivalent scenarios that leads to a false inequality when in reality what you are calculating the probability of hasn't changed between the first half and second half of the problem by giving you information that sounds like new information (and yes is on a technical level) but doesn't give you any new information that actually affects your POV probability of the future event.

The Three Prison Problem is alike a 'double blind' version of the Monte Hall Problem, since the Ted the Prisoner and the Guard neither have any control over the final events, Ted's fate has been decided before the "problem" starts and isn't affected by anything that happens in it.
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Old 22nd February 2019, 07:21 AM   #101
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Originally Posted by Belz... View Post
That's where we don't connect: you always knew that at least another prisoner was going to die. The guard tells you nothing you didn't know beforehand.
Right. Ted's odds were always 50/50, then. JoeMorgue posed in post #47 that his odds were one in three. That's what I am arguing.

eta: When the guard says Bill is doomed, he has picked up a gold sided coin. Even though we know that it was one in three knowing no variables, the guard has indeed narrowed the odds.
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Old 22nd February 2019, 07:31 AM   #102
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Originally Posted by JoeMorgue View Post
Like I said I prefer the 3 Prisoner Problem because it's mathematically identical but it doesn't have the two things that people get hung up on with the Monte Hall version, the motivation of Monte and the person being able to change their decision after the new information is presented to them.

One because those two variables sort of cancel each other out if you think about, in this version the Monte equivalent, the guard, isn't specifically trying to trick you into a specific outcome but since you have no control over the outcome it makes no difference but more so because neither of them actually changes the underlying mathematics of the problem.

Both of them trick you into directly comparing two similar but not equivalent scenarios that leads to a false inequality when in reality what you are calculating the probability of hasn't changed between the first half and second half of the problem by giving you information that sounds like new information (and yes is on a technical level) but doesn't give you any new information that actually affects your POV probability of the future event.

The Three Prison Problem is alike a 'double blind' version of the Monte Hall Problem, since the Ted the Prisoner and the Guard neither have any control over the final events, Ted's fate has been decided before the "problem" starts and isn't affected by anything that happens in it.
Like I said, I'll be Jim. You're the house, with the payout odds stacked heavily in your favor. You in?
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Old 22nd February 2019, 08:06 AM   #103
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mistake post

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Old 22nd February 2019, 08:18 AM   #104
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Originally Posted by p0lka View Post
mistake post
Damn. I was ready for that, too

eta: (I was running the math when you edited)
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Old 22nd February 2019, 08:19 AM   #105
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Originally Posted by p0lka View Post
mistake post
What are the odds that this is the only correct post in the thread?
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Old 22nd February 2019, 08:24 AM   #106
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Originally Posted by Thermal View Post
I had a similar problem in Stats or Philosophy back in school:

You have three coins. One is gold both sides, one silver both sides, and one gold on one side and silver on the other. If you pick one from a bag, and it has gold on one side, what are the odds that it is gold on the other?

Intuitively, one might think that the odds were one in three. but by showing gold on one side, you have eliminated the two silver sided coin from possibility. That means your odds are 50/50 of having gold on the other side.

I think you're remembering that one backward.

Intuitively, one might think the odds were 50-50, because when you see a gold side you know it's one of two coins (the gold-gold or the gold-silver), and one of those two will have gold on the other side and the other one won't.

But actually, the odds of gold on the other side are 2 in 3, because of the original six equally likely possibilities for choosing a coin and a side to look at:

silver coin, silver side
silver coin, other silver side
mixed coin, silver side
mixed coin, gold side
gold coin, gold side
gold coin, other gold side

... the first three are eliminated, and of the remaining three, two will also have gold on the other side.


ETA: The above is only correct if you interpret "... and it has gold on one side" as meaning you only look at one side, and the side you look at is gold.

If you interpret it instead as meaning you examine both sides of the coin and note that it has gold on at least one side, then it's 50-50 that both sides are gold.
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Old 22nd February 2019, 08:27 AM   #107
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Originally Posted by Thermal View Post
Damn. I was ready for that, too

eta: (I was running the math when you edited)
Yeah I was running the math as I was typing, then realised I wasn't sure where I was going with it.
Doh.

I'll try again later when I get back.
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Old 22nd February 2019, 08:31 AM   #108
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Originally Posted by Thermal View Post
If the choice of who was executed is random, that would exactly be the odds. No?
The original decision may well have been random but the process by which these two remain undead is not random, it adheres to a set of rules which make one outcome twice as likely as the next.

(I haven't read subsequent replies to this may have been said).
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Old 22nd February 2019, 08:39 AM   #109
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OK, so imagine an entire city about to be attacked by aliens (it's over a dog-related dispute but never mind that).

There are one million people in the city. Bob is one of them. We don't know who the others are because that's modern city culture, nobody talks to their neighbours any more.

The aliens have chosen one person in the city to be spared. The rest will be murdered by Islamic ray guns.

Bob does a deal with the aliens stipulating that they will kill everybody but him and one other. They do so. 999,998 people are killed.

So there Bob is, and next to him some random geezer. (EDIT: I forgot to mention that in the end, one of the pair will be killed, of course)


It's no surprise that Bob is still alive because the deal did not allow him to be killed.

It is a happy surprise, however, for the geezer.

Last edited by baron; 22nd February 2019 at 08:42 AM.
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Old 22nd February 2019, 08:40 AM   #110
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Originally Posted by baron View Post
OK, so imagine an entire city about to be attacked by aliens (it's over a dog-related dispute but never mind that).

There are one million people in the city. Bob is one of them. We don't know who the others are because that's modern city culture, nobody talks to their neighbours any more.

The aliens have chosen one person in the city to be spared. The rest will be murdered by Islamic ray guns.

Bob does a deal with the aliens stipulating that they will kill everybody but him and one other. They do so. 999,998 people are killed.

So there Bob is, and next to him some random geezer.

It's no surprise that Bob is still alive because the deal did not allow him to be killed.

It is a happy surprise, however, for the geezer.
One question: do islamic ray guns decapitate?
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Old 22nd February 2019, 08:40 AM   #111
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Originally Posted by Belz... View Post
One question: do islamic ray guns decapitate?
If you're a man, yes. If you're a woman they fire stones.
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Old 22nd February 2019, 08:41 AM   #112
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Originally Posted by Myriad View Post
I think you're remembering that one backward.

Intuitively, one might think the odds were 50-50, because when you see a gold side you know it's one of two coins (the gold-gold or the gold-silver), and one of those two will have gold on the other side and the other one won't.

But actually, the odds of gold on the other side are 2 in 3, because of the original six equally likely possibilities for choosing a coin and a side to look at:

silver coin, silver side
silver coin, other silver side
mixed coin, silver side
mixed coin, gold side
gold coin, gold side
gold coin, other gold side

... the first three are eliminated, and of the remaining three, two will also have gold on the other side.


ETA: The above is only correct if you interpret "... and it has gold on one side" as meaning you only look at one side, and the side you look at is gold.

If you interpret it instead as meaning you examine both sides of the coin and note that it has gold on at least one side, then it's 50-50 that both sides are gold.
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.
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Last edited by Thermal; 22nd February 2019 at 08:48 AM.
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Old 22nd February 2019, 08:43 AM   #113
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Originally Posted by baron View Post
If you're a man, yes. If you're a woman they fire stones.
That is conveniently selective. Definitely advanced tech.
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Old 22nd February 2019, 08:49 AM   #114
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Originally Posted by baron View Post
OK, so imagine an entire city about to be attacked by aliens (it's over a dog-related dispute but never mind that).

There are one million people in the city. Bob is one of them. We don't know who the others are because that's modern city culture, nobody talks to their neighbours any more.

The aliens have chosen one person in the city to be spared. The rest will be murdered by Islamic ray guns.

Bob does a deal with the aliens stipulating that they will kill everybody but him and one other. They do so. 999,998 people are killed.

So there Bob is, and next to him some random geezer. (EDIT: I forgot to mention that in the end, one of the pair will be killed, of course)


It's no surprise that Bob is still alive because the deal did not allow him to be killed.

It is a happy surprise, however, for the geezer.
Takeaway is that one must always negotiate a hot model as a second survivor when dealing with aliens
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Old 22nd February 2019, 08:50 AM   #115
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We would all get the (censored) beat out of us at any table in Vegas.
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Old 22nd February 2019, 08:53 AM   #116
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That's why the smart money is in rolling the winners in the parking garage
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Old 22nd February 2019, 08:58 AM   #117
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Originally Posted by Belz... View Post
That is conveniently selective. Definitely advanced tech.
Iain Banks wrote of a similar weapon although he steered clear of religious associations.

Originally Posted by Thermal View Post
Takeaway is that one must always negotiate a hot model as a second survivor when dealing with aliens
The trouble is, your partner's selection really is random, so if it's an issue I advise you to live in a city populated entirely by hot models. Me, I'm still looking.
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Old 22nd February 2019, 09:04 AM   #118
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Originally Posted by baron View Post
Iain Banks wrote of a similar weapon although he steered clear of religious associations.
Pfft. Pussy.
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Old 22nd February 2019, 09:44 AM   #119
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Dear Whoever Who Introduced the Monty Hall problem to this thread: I hate you.
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Old 22nd February 2019, 10:06 AM   #120
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Originally Posted by Trebuchet View Post
Dear Whoever Who Introduced the Monty Hall problem to this thread: I hate you.
We could always try to resurrect the downwind faster than the wind instead
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