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23rd February 2019, 06:01 AM  #161 
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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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23rd February 2019, 09:49 AM  #163 
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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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23rd February 2019, 10:21 AM  #164 
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23rd February 2019, 10:29 AM  #165 
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23rd February 2019, 12:43 PM  #166 
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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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23rd February 2019, 01:30 PM  #167 
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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 
24th February 2019, 08:46 AM  #168 
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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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24th February 2019, 05:12 PM  #169 
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As I said previously, that is an unwarranted assumption. The twoshocks symbol is a single symbol, distinct from the oneshock symbol. If it were two oneshock symbols, you would be correct. But it isn't. Algebraically, the twoshocks and the one shock symbols are as distinct from each other as x and y.

25th February 2019, 12:34 AM  #170 
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25th February 2019, 12:38 AM  #171 
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25th February 2019, 06:11 AM  #172 
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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. 
25th February 2019, 06:42 AM  #173 
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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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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? 
25th February 2019, 07:39 AM  #175 
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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 Montyish 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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25th February 2019, 05:50 PM  #176 
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28th February 2019, 01:19 AM  #177 
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I want to add a line of clarification (bolded) to this classic explanation of the prisoner problem.
Quote:
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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4th March 2019, 12:54 PM  #178 
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4th March 2019, 01:32 PM  #179 
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4th March 2019, 02:25 PM  #180 
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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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