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vegasrenter
Participantstockstradr, Thanks for your input. My experience with trading options has been FAR more profitable (nice multi-X gains on small at-risk $ amounts) and less worrying than when I used to practice “buy and hold” (gonad-shriveling losses) with stocks & mutual funds. However, I’ve found that 80% of put options that I’ve sold for your stated reasons went on to end way more in the money than my exit point.
Still looking for information on my original question.
Participantstockstradr, Thanks for your input. My experience with trading options has been FAR more profitable (nice multi-X gains on small at-risk $ amounts) and less worrying than when I used to practice “buy and hold” (gonad-shriveling losses) with stocks & mutual funds. However, I’ve found that 80% of put options that I’ve sold for your stated reasons went on to end way more in the money than my exit point.
Still looking for information on my original question.
Participantstockstradr, Thanks for your input. My experience with trading options has been FAR more profitable (nice multi-X gains on small at-risk $ amounts) and less worrying than when I used to practice “buy and hold” (gonad-shriveling losses) with stocks & mutual funds. However, I’ve found that 80% of put options that I’ve sold for your stated reasons went on to end way more in the money than my exit point.
Still looking for information on my original question.
Participantstockstradr, Thanks for your input. My experience with trading options has been FAR more profitable (nice multi-X gains on small at-risk $ amounts) and less worrying than when I used to practice “buy and hold” (gonad-shriveling losses) with stocks & mutual funds. However, I’ve found that 80% of put options that I’ve sold for your stated reasons went on to end way more in the money than my exit point.
Still looking for information on my original question.
Participant[quote=Neveov][quote=vegasrenter]”it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”
In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.
I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.
[/quote]I am going to have to agree with this poster. I believe the oddity comes, not from the elimination in all cases of the sort, but only in the case of the knowledgeable “host” entity performing the elimination process.
Suppose there is no host eliminating the doors, and after coming upon 1000 closed doors, you pick one at random, knowing there is a car behind one. Then before you open that one, you are granted the capacity to open 998 before you must settle on your 999th decision. So you, yourself, open the 998 other doors, and frustratingly you find no car. Would we then say that the door you haven’t opened yet has a better chance than the door you didn’t open? I have a hard time believing this to be the case in this scenario.[/quote]
OK, I was wrong & now I finally get it. There’s only three possible conditions before the game begins, all equally likely: The car is behind door 1, 2, or 3. If I always pick door 1 to start and then always switch, I win when the car is behind 2 or 3, and lose if it was behind 1, so I win 2/3 of the time. Typical case where a technically educated person (me) couldn’t get something simple until beat over the head with it. Lots of good explanations on this thread, by the way – cool to see different POV’s on the same puzzle.
Participant[quote=Neveov][quote=vegasrenter]”it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”
In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.
I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.
[/quote]I am going to have to agree with this poster. I believe the oddity comes, not from the elimination in all cases of the sort, but only in the case of the knowledgeable “host” entity performing the elimination process.
Suppose there is no host eliminating the doors, and after coming upon 1000 closed doors, you pick one at random, knowing there is a car behind one. Then before you open that one, you are granted the capacity to open 998 before you must settle on your 999th decision. So you, yourself, open the 998 other doors, and frustratingly you find no car. Would we then say that the door you haven’t opened yet has a better chance than the door you didn’t open? I have a hard time believing this to be the case in this scenario.[/quote]
OK, I was wrong & now I finally get it. There’s only three possible conditions before the game begins, all equally likely: The car is behind door 1, 2, or 3. If I always pick door 1 to start and then always switch, I win when the car is behind 2 or 3, and lose if it was behind 1, so I win 2/3 of the time. Typical case where a technically educated person (me) couldn’t get something simple until beat over the head with it. Lots of good explanations on this thread, by the way – cool to see different POV’s on the same puzzle.
Participant[quote=Neveov][quote=vegasrenter]”it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”
In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.
I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.
[/quote]I am going to have to agree with this poster. I believe the oddity comes, not from the elimination in all cases of the sort, but only in the case of the knowledgeable “host” entity performing the elimination process.
Suppose there is no host eliminating the doors, and after coming upon 1000 closed doors, you pick one at random, knowing there is a car behind one. Then before you open that one, you are granted the capacity to open 998 before you must settle on your 999th decision. So you, yourself, open the 998 other doors, and frustratingly you find no car. Would we then say that the door you haven’t opened yet has a better chance than the door you didn’t open? I have a hard time believing this to be the case in this scenario.[/quote]
OK, I was wrong & now I finally get it. There’s only three possible conditions before the game begins, all equally likely: The car is behind door 1, 2, or 3. If I always pick door 1 to start and then always switch, I win when the car is behind 2 or 3, and lose if it was behind 1, so I win 2/3 of the time. Typical case where a technically educated person (me) couldn’t get something simple until beat over the head with it. Lots of good explanations on this thread, by the way – cool to see different POV’s on the same puzzle.
Participant[quote=Neveov][quote=vegasrenter]”it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”
In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.
I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.
[/quote]I am going to have to agree with this poster. I believe the oddity comes, not from the elimination in all cases of the sort, but only in the case of the knowledgeable “host” entity performing the elimination process.
Suppose there is no host eliminating the doors, and after coming upon 1000 closed doors, you pick one at random, knowing there is a car behind one. Then before you open that one, you are granted the capacity to open 998 before you must settle on your 999th decision. So you, yourself, open the 998 other doors, and frustratingly you find no car. Would we then say that the door you haven’t opened yet has a better chance than the door you didn’t open? I have a hard time believing this to be the case in this scenario.[/quote]
OK, I was wrong & now I finally get it. There’s only three possible conditions before the game begins, all equally likely: The car is behind door 1, 2, or 3. If I always pick door 1 to start and then always switch, I win when the car is behind 2 or 3, and lose if it was behind 1, so I win 2/3 of the time. Typical case where a technically educated person (me) couldn’t get something simple until beat over the head with it. Lots of good explanations on this thread, by the way – cool to see different POV’s on the same puzzle.
Participant[quote=Neveov][quote=vegasrenter]”it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”
In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.
I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.
[/quote]I am going to have to agree with this poster. I believe the oddity comes, not from the elimination in all cases of the sort, but only in the case of the knowledgeable “host” entity performing the elimination process.
Suppose there is no host eliminating the doors, and after coming upon 1000 closed doors, you pick one at random, knowing there is a car behind one. Then before you open that one, you are granted the capacity to open 998 before you must settle on your 999th decision. So you, yourself, open the 998 other doors, and frustratingly you find no car. Would we then say that the door you haven’t opened yet has a better chance than the door you didn’t open? I have a hard time believing this to be the case in this scenario.[/quote]
OK, I was wrong & now I finally get it. There’s only three possible conditions before the game begins, all equally likely: The car is behind door 1, 2, or 3. If I always pick door 1 to start and then always switch, I win when the car is behind 2 or 3, and lose if it was behind 1, so I win 2/3 of the time. Typical case where a technically educated person (me) couldn’t get something simple until beat over the head with it. Lots of good explanations on this thread, by the way – cool to see different POV’s on the same puzzle.
ParticipantI see that the OP is enjoying new comments on some old predictions; thought I would bump this one for newer members that have not enjoyed it yet.
ParticipantI see that the OP is enjoying new comments on some old predictions; thought I would bump this one for newer members that have not enjoyed it yet.
ParticipantI see that the OP is enjoying new comments on some old predictions; thought I would bump this one for newer members that have not enjoyed it yet.
ParticipantI see that the OP is enjoying new comments on some old predictions; thought I would bump this one for newer members that have not enjoyed it yet.
ParticipantI see that the OP is enjoying new comments on some old predictions; thought I would bump this one for newer members that have not enjoyed it yet.
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