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Old 23rd February 2019, 06:01 AM   #161
p0lka
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Originally Posted by Thermal View Post
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.
Ah sorry about that, yeah you are correct it was from the same post
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Old 23rd February 2019, 09:04 AM   #162
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Pigeons and Monty Hall

"Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans (Herbranson and Schroeder, 2010)."
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Old 23rd February 2019, 09:49 AM   #163
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Originally Posted by Myriad View Post
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.
Yeah, I get all that, especially #4 being equivalent to #5, as the turning of the jack was superfluous. Where we differ is what we are calling the probability. Continuing your card analogy, the queen was always determined in advance, just like Ted's freedom or not was determined in advanced. The probability sought, as I see it, is forecasting the odds of which will be the queen or freedom. With one option eliminated by any means, it goes to even odds. In the prisoner problem, one jack is revealed by the guard. That positively alters the odds as I see them.

Other posters insist that you knew one would go anyway, so the guard does not give new information. I think it does, as the possibility of that particular one of being the queen has been positively eliminated, as in your variants one through three. This is different from 4 & 5, where you only know that the queen is in there somewhere, and statistically two will be jacks. Identifying one certain jack switches the bet from 1 in three to even money, as you showed.

If I am right, the prisoners problem is not a paradox, but just a change in perspective. Ted's odds of being already destined for freedom were always one in three, but the new information of Bill's fate changes the probability for impartial observer Jim. And this is the disconnect: I think Jim and Ted should see the odds as the same.
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Old 23rd February 2019, 10:21 AM   #164
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Originally Posted by Thermal View Post
If I am right, the prisoners problem is not a paradox, but just a change in perspective. Ted's odds of being already destined for freedom were always one in three, but the new information of Bill's fate changes the probability for impartial observer Jim. And this is the disconnect: I think Jim and Ted should see the odds as the same.
The new info changes the odds for the third prisoner too, it makes it twice that of Teds.
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Old 23rd February 2019, 10:29 AM   #165
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Originally Posted by p0lka View Post
The new info changes the odds for the third prisoner too, it makes it twice that of Teds.
See, I get the math and the manipulation of odds, conceptually. What I don't get is the logical soundness of applying them.

Been avoiding diving into Wiki or wherever till I thought it through, but maybe its time. Can't keep driving and texting about Ted's fate
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Old 23rd February 2019, 12:43 PM   #166
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Originally Posted by Thermal View Post
If I am right, the prisoners problem is not a paradox, but just a change in perspective. Ted's odds of being already destined for freedom were always one in three, but the new information of Bill's fate changes the probability for impartial observer Jim. And this is the disconnect: I think Jim and Ted should see the odds as the same.

Jim and Ted do see the odds the same, if they have the same information.

(Of course if they don't have the same information they won't see the odds the same. If the guard told Bill he'd definitely be reprieved, but Jim didn't know that, they'd see different odds.)

The information of Bill's fate does change the probability of Bill's survival (from 1/3 to zero) and of Harry's (from 1/3 to 2/3). For Jim as well as for Ted. It just doesn't happen to change the probability of Ted's survival.

Other new information could. For instance, morning comes and Ted sees Harry get executed. He (and Jim, if Jim has all the same information) now know that Ted is guaranteed to survive. (Unfortunately for Ted, if the executions are in random order, that course of events has only a 1/6 chance of happening.)
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Old 23rd February 2019, 01:30 PM   #167
p0lka
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Originally Posted by Thermal View Post
See, I get the math and the manipulation of odds, conceptually. What I don't get is the logical soundness of applying them.

Been avoiding diving into Wiki or wherever till I thought it through, but maybe its time. Can't keep driving and texting about Ted's fate
If I was actually in the position re: the monty hall problem, I would apply it and swap, because it is logically sound. Which makes it 2/3 instead of 1/3.

The three prisoner malarky,

If I was A or Ted, and magically I could swap with the third prisoner that isn't mentioned, I would apply it as it's logically sound, their chances are twice mine in this context. Which makes it 2/3 instead of 1/3.

1/3 of the time I would be wrong, 2/3 of the time I would be right. I'm pretty sure there is a poker term for that

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Old 24th February 2019, 08:46 AM   #168
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Originally Posted by p0lka View Post
If I was actually in the position re: the monty hall problem, I would apply it and swap, because it is logically sound. Which makes it 2/3 instead of 1/3.

The three prisoner malarky,

If I was A or Ted, and magically I could swap with the third prisoner that isn't mentioned, I would apply it as it's logically sound, their chances are twice mine in this context. Which makes it 2/3 instead of 1/3.

1/3 of the time I would be wrong, 2/3 of the time I would be right. I'm pretty sure there is a poker term for that
Ok. I Wiki'd it.

The gist seems to be that from Ted's perspective, he receives no new information about his fate, but learns more about Harry's odds.

Ted sees it as he has a 1/3 chance of being the random freed prisoner. As he doesn't care about the other guys much, he just sees 'somebody else' as having a 2/3 chance. Learning Bill's fate, Wiki says that Ted's fate remains at 1/3, but Harry (third prisoner), now has 2/3 odds of being freed, from Ted's perspective (Ted doesn't care who specifically wins if it is not him, and he gets no other chances to mathematically average it out).

Mathematically, this is true. But it does seem to be a perspective twisting, not a probability paradox.

Let's say Harry (you selfish bastards never consider Harry) also hears that they're gonna Kill Bill. From his perspective, his odds also remain at 1/3 for freedom, and Ted's went up to 2/3. Meanwhile, impartial observer Jim only cares about the betting odds. Three prisoners, one was randomly selected for freedom, and one is eliminated as a possibility. That's even odds from Jim's neutral perspective.

So: the real odds are zero or one for any of the prisoners to be freed, but we are trying to figure the 'betting odds'. All four know that the guard is going to Kill Bill. Ted thinks the odds are 1/3 for him, and 2/3 for Harry. Harry thinks they are 1/3 for him, and 2/3 for Ted. Jim thinks they are 1/2 for either. Bill is in the corner frantically calculating Pascal's Wager.

Three betters, three different sets of odds given the same information. I think only Jim's are relevant. Why do others think only Ted's trump them?
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Old 24th February 2019, 05:12 PM   #169
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Originally Posted by ServiceSoon View Post
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.
As I said previously, that is an unwarranted assumption. The two-shocks symbol is a single symbol, distinct from the one-shock symbol. If it were two one-shock symbols, you would be correct. But it isn't. Algebraically, the two-shocks and the one shock symbols are as distinct from each other as x and y.
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Old 25th February 2019, 12:34 AM   #170
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Originally Posted by arthwollipot View Post
As I said previously, that is an unwarranted assumption. The two-shocks symbol is a single symbol, distinct from the one-shock symbol. If it were two one-shock symbols, you would be correct. But it isn't. Algebraically, the two-shocks and the one shock symbols are as distinct from each other as x and y.
In which case the setter of the puzzle would be playing stupid games and wasting everybody's time. I'd say it's reasonable to suppose they're not; that the puzzle is intended as a bit of lightweight but meaningful fun.
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Old 25th February 2019, 12:38 AM   #171
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Originally Posted by GlennB View Post
In which case the setter of the puzzle would be playing stupid games and wasting everybody's time. I'd say it's reasonable to suppose they're not; that the puzzle is intended as a bit of lightweight but meaningful fun.
I disagree. Given the second example that was posted, I think it's very much a case of the puzzle setter playing stupid games and wasting everyone's time.
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Old 25th February 2019, 06:11 AM   #172
p0lka
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Originally Posted by Thermal View Post
Ok. I Wiki'd it.

The gist seems to be that from Ted's perspective, he receives no new information about his fate, but learns more about Harry's odds.

Ted sees it as he has a 1/3 chance of being the random freed prisoner. As he doesn't care about the other guys much, he just sees 'somebody else' as having a 2/3 chance. Learning Bill's fate, Wiki says that Ted's fate remains at 1/3, but Harry (third prisoner), now has 2/3 odds of being freed, from Ted's perspective (Ted doesn't care who specifically wins if it is not him, and he gets no other chances to mathematically average it out).

Mathematically, this is true. But it does seem to be a perspective twisting, not a probability paradox.

Let's say Harry (you selfish bastards never consider Harry) also hears that they're gonna Kill Bill. From his perspective, his odds also remain at 1/3 for freedom, and Ted's went up to 2/3. Meanwhile, impartial observer Jim only cares about the betting odds. Three prisoners, one was randomly selected for freedom, and one is eliminated as a possibility. That's even odds from Jim's neutral perspective.

So: the real odds are zero or one for any of the prisoners to be freed, but we are trying to figure the 'betting odds'. All four know that the guard is going to Kill Bill. Ted thinks the odds are 1/3 for him, and 2/3 for Harry. Harry thinks they are 1/3 for him, and 2/3 for Ted. Jim thinks they are 1/2 for either. Bill is in the corner frantically calculating Pascal's Wager.

Three betters, three different sets of odds given the same information. I think only Jim's are relevant. Why do others think only Ted's trump them?
If Harry was the one that asked the guard then it would be correct that harry would be 1/3 and then ted would be 2/3.

If Harry just overheard the guard and teds conversation, then it would still be 1/3 Ted and 2/3 Harry.

The guard being constrained by the rule of not being able to tell the asker his own fate is the difference.
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Old 25th February 2019, 06:42 AM   #173
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Originally Posted by p0lka View Post
If Harry was the one that asked the guard then it would be correct that harry would be 1/3 and then ted would be 2/3.

If Harry just overheard the guard and teds conversation, then it would still be 1/3 Ted and 2/3 Harry.

The guard being constrained by the rule of not being able to tell the asker his own fate is the difference.
I've been thinking about this. What you are saying makes it a relativity kind of problem, rather than probability, which is what I've been thinking it is. Like considering whether you are 33 1/3% down or 50% up when losing then winning back the same dollar in a bet. It's just relative to your POV.

But you make a correct point- the information is relative to Ted, not randomly revealed (I think Myriad was trying to explain this earlier). My last scenario assumes it was announced to all, and I'm still not sure it doesn't apply. Harry still hears they're gong to Kill Bill, and I think he would still interpret the information the same way, as would Jim.

I think if we added one more factor to the original proposition, it would work: that in the case of Ted being freed, the guard had to choose to reveal which by fair coin flip. That would up Harry's odds of not being named by predictable probability, rather than random.

eta: I think the slip I'm having is determining if the news of Bill's fate has to be interpreted as relative to Ted. Also, apologies to all for the derail. I think I'll stop now.

eta II: I've also been obstinate in thinking that by eliminating Bill as a variable by any means, 1/3 were no longer viable relative odds.
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Old 25th February 2019, 07:30 AM   #174
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Thermal:

I was missing it as well for a bit. You have to consider there are two different sets of interacting probabilities.

At the start, you have three possible outcomes:
1. A and B die
2. A and C die
3. B and C die

Each outcome has a 1 in 3 probability.

Now, A asks the guard. So, we have to consider what answer the guard gives under each of the above three conditions.

For 1, the guard will say "B dies" 100% of the time. Multiplying by the original 1 in 3 probability, that's a 1 in 3 answer.

For 2, the guard will say "B dies" 100% of the time. Multiplying by the original 1 in 3 probability, that's a 1 in 3 answer.

For 3, 50% of the time the guard will say "B dies" and 50% of the time the guard will say "C dies". Multiplying that by the original probability gives a 1 in 6 chance for each, or still 1 in 3 for both together.

The difference is that the guard has the option of a choice in the case where A survives. It's two sets of probabilities interacting, rather than a single probability.

Does that help?
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Old 25th February 2019, 07:39 AM   #175
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Originally Posted by Hellbound View Post
Thermal:

I was missing it as well for a bit. You have to consider there are two different sets of interacting probabilities.

At the start, you have three possible outcomes:
1. A and B die
2. A and C die
3. B and C die

Each outcome has a 1 in 3 probability.

Now, A asks the guard. So, we have to consider what answer the guard gives under each of the above three conditions.

For 1, the guard will say "B dies" 100% of the time. Multiplying by the original 1 in 3 probability, that's a 1 in 3 answer.

For 2, the guard will say "B dies" 100% of the time. Multiplying by the original 1 in 3 probability, that's a 1 in 3 answer.

For 3, 50% of the time the guard will say "B dies" and 50% of the time the guard will say "C dies". Multiplying that by the original probability gives a 1 in 6 chance for each, or still 1 in 3 for both together.

The difference is that the guard has the option of a choice in the case where A survives. It's two sets of probabilities interacting, rather than a single probability.

Does that help?
I have not been assuming the hilited, but considering it random/unpredictable. If we plug in fair coin odds to the original proposition, it does become Monty-ish with the odds unexpectedly leaning one way.

eta: the other thing I was reluctant to let go of was that once Bill was eliminated as a variable to live, your option #2 above was no longer a factor, leaving only 1&3, or even odds. Deceptive, that.
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Old 25th February 2019, 05:50 PM   #176
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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
The two shocks are not placed horizontally next to each other but at an angle, so the appropriate operation to assume would be tetration rather than multiplication.
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Old 28th February 2019, 01:19 AM   #177
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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.

I want to add a line of clarification (bolded) to this classic explanation of the prisoner problem.

Quote:
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.

In other words, if A is chosen for death he will be skipped until the very end until it is just two people: [A and the person who lives] or, very unlikely, [A and the real last person to die]. And also that no other chosen for death people have the luxury of being skipped.

The guard proceeds.

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 part about him being skipped if he was one of the ones destined to die really drives home to a layman that his odds at the end are still 1 in 100. baron implied it, but technically left it out.

It also helps people realize that the other person left did not have the benefit of being skipped if chosen to die.

(A perhaps even better way that I've seen is that the guard offers to read a list of the 99 names destined to die and person A's name is skipped if on the list.)
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Old 4th March 2019, 12:54 PM   #178
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Originally Posted by This is The End View Post
I want to add a line of clarification (bolded) to this classic explanation of the prisoner problem.





The part about him being skipped if he was one of the ones destined to die really drives home to a layman that his odds at the end are still 1 in 100. baron implied it, but technically left it out.

It also helps people realize that the other person left did not have the benefit of being skipped if chosen to die.

(A perhaps even better way that I've seen is that the guard offers to read a list of the 99 names destined to die and person A's name is skipped if on the list.)
I think you mean 98?
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Old 4th March 2019, 01:32 PM   #179
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Originally Posted by Turgor View Post
I think you mean 98?
I don't think they do, they're referring to the other person who isn't A (the one who asked).

I could be wrong though, 'lifts beer and says cheers'.

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Old 4th March 2019, 02:25 PM   #180
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Originally Posted by p0lka View Post
I don't think they do, they're referring to the other person who isn't A (the one who asked).

I could be wrong though, 'lifts beer and says cheers'.
If you have 100(Z) prisoners and you plan to kill 99 chosen at random, then to any subsequently randomly chosen 1 prisoner(A) list everyone who will be killed taking care not to list the A, then A will either hear a list of 98 names and know he is number 99 to be killed (99/100 chance), or he will hear 99 names and know he is safe (1/100 chance). So this is not the same problem, at the end we now have certainty.

The counterintuitivity of the problem only comes from leaving 2 entities out of a bigger amount of entities, with one entity having been in a group where the chance that the desired property of that entitiy (survival/car/whatever) is 99/100 or 2/3 and the other all by itself with still the original 1/100 or 1/3 or whatever Z you chose.

ETA oh now i see, what about this very last person that is not A. Well in that case the top half of this post still describes that scenario fully if I am not mistaken?

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