08 Aug 08
Originally posted by ZahlanziAh, but then again, what if you're playing russian roulette with two bullets, and the bullets are in successive chambers. If there are 3 players to play, who has the best odds: the person who goes first? second? third? Who has the worst odds?
this is not how the problem is formulated. the problem asks if you have more chances of winning if you change your choice. the dude picked the first door with a 1/3 chance of success. but now he makes a new pick from 2 choices.
if you play russian rullette and you pull the trigger 4 times what are your chances on the 5th pull? things would not be 50% ...[text shortened]... ld have asked before the 1st pull of the trigger what are your chances of reaching the 5th pull.
08 Aug 08
Originally posted by ZahlanziWhat I don't understand is why you still consider you have a 50/50 chance when the link ATY provided will plainly show you that it isn't. Play his game as often as you like, making sure you do the switch, and you will win more than fifty percent of the time. Theoretically you could lose but, like the ten coin tosses I mentioned, you won't.
this is not how the problem is formulated. the problem asks if you have more chances of winning if you change your choice. the dude picked the first door with a 1/3 chance of success. but now he makes a new pick from 2 choices.
if you play russian rullette and you pull the trigger 4 times what are your chances on the 5th pull? things would not be 50% ...[text shortened]... ld have asked before the 1st pull of the trigger what are your chances of reaching the 5th pull.
Maybe this will help. Given the presenter will show you the other salt door you get 100% certainty of winning if you pick the wrong door the first time and then switch. I guess you're happy to admit that the odds of picking the wrong door the first time are, in fact 2/3.
08 Aug 08
Originally posted by forkedknightyes, but the odds of the second and third person change depending on what the first guy gets. so to calculate the chances you need both events. so in the second event you could have bullet bullet, nothing bullet or bullet nothing.
Ah, but then again, what if you're playing russian roulette with two bullets, and the bullets are in successive chambers. If there are 3 players to play, who has the best odds: the person who goes first? second? third? Who has the worst odds?
in our example, we know that the first door(the eliminated one) is salt. we know what the final pair of possibilities is.
08 Aug 08
Originally posted by Wheelythat doesn't prove anything. probability isn't something you can verify in a finite series of tries.
What I don't understand is why you still consider you have a 50/50 chance when the link ATY provided will plainly show you that it isn't. Play his game as often as you like, making sure you do the switch, and you will win more than fifty percent of the time. Theoretically you could lose but, like the ten coin tosses I mentioned, you won't.
Maybe this w ...[text shortened]... u're happy to admit that the odds of picking the wrong door the first time are, in fact 2/3.
a coin has 50% chance of landing on heads. try flipping the coin 2 times. if it lands both times on heads, what does it say about the calculated odds? if you flip it 100 times, it can land on heads from 0 to 100. some ranges of values have better statistical possibility. so in 100 series of 100 tosses each, heads will be around the value 50.
did you play that game 100 times 100?
Maybe this will help. Given the presenter will show you the other salt door you get 100% certainty of winning if you pick the wrong door the first time and then switch. I guess you're happy to admit that the odds of picking the wrong door the first time are, in fact 2/3.
but you don't know. you don't know that you picked wrong the first time. you just choose to switch or choose not to switch. the fact that you chose a certain door the first time doesn't give you any extra information.
if the presenter would have opened the door with the porche, what would have been your chances of picking salt? 2*salt / 2 possibilities gives 1 and i reached this conclusion without caring which door did i choose the first time.
08 Aug 08
Originally posted by ZahlanziIndeed. You can never verify a probability exactly. However, if you buy a meter of plastic piping from the local hardware shop you're not going to get an exact meter either. We deal in reality here, not theory.
that doesn't prove anything. probability isn't something you can verify in a finite series of tries.
a coin has 50% chance of landing on heads. try flipping the coin 2 times. if it lands both times on heads, what does it say about the calculated odds? if you flip it 100 times, it can land on heads from 0 to 100. some ranges of values have better statistical possibility. so in 100 series of 100 tosses each, heads will be around the value 50.
In the real world in which we all have to exist, you WILL win your porche more times if you do the switch than if you don't. In the theoretical world from which you seem to be arguing, it is true that it is slightly possible that if you played this game to the end of time, you might have lost the porche more than you won it. This, however, is extremely unlikely.
08 Aug 08
Originally posted by Wheelythats not probability, it is statistics. you say "if you change you will win more times". but it is at most statistics. the same statistic that describes the results of a coin toss series. why is this any different? because before getting to choose between two options you made a choice that had absolutely no result?
Indeed. You can never verify a probability exactly. However, if you buy a meter of plastic piping from the local hardware shop you're not going to get an exact meter either. We deal in reality here, not theory.
In the real world in which we all have to exist, you WILL win your porche more times if you do the switch than if you don't. In the theoretica ...[text shortened]... e, you might have lost the porche more than you won it. This, however, is extremely unlikely.
08 Aug 08
Originally posted by ZahlanziIt did have a result. It determined which of the doors the presenter opened which is key. I'm pretty sure we have all agreed that the presenter knows what's behind the doors. Even if we haven't agreed that, in the one case that Shav stated in the original post in what seems like several years ago now, the presenter did show us a salt door which is equivalent to him knowing what's behind each door and choosing to show the the salt. As has been stated rather regularly now, as he has shown us a salt door and as we are most likely to have chosen a salt door originally ourselves and as there are only two salt doors the porche is most likely to be in the one remaining door.
thats not probability, it is statistics. you say "if you change you will win more times". but it is at most statistics. the same statistic that describes the results of a coin toss series. why is this any different? because before getting to choose between two options you made a choice that had absolutely no result?
08 Aug 08
4 edits
Originally posted by ZahlanziYou just hit the nail exactly on the head, and then came to the wrong conclusion.
Maybe this will help. Given the presenter will show you the other salt door you get 100% certainty of winning if you pick the wrong door the first time and then switch. I guess you're happy to admit that the odds of picking the wrong door the first time are, in fact 2/3.
but you don't know. you don't know that you picked wrong the first time. you just ...[text shortened]... e fact that you chose a certain door the first time doesn't give you any extra information.
"you get 100% certainty of winning if you pick the wrong door the first time and then switch."
"the odds of picking the wrong door the first time are, in fact 2/3"
100% * 2/3 = 2/3 -- your chances of winning if you switch!
08 Aug 08
Originally posted by forkedknightExcept he didn´t say that, I did 🙂 He has quoted what I said within his text itself.
You just hit the nail exactly on the head, and then came to the wrong conclusion.
[b]"you get 100% certainty of winning if you pick the wrong door the first time and then switch."
"the odds of picking the wrong door the first time are, in fact 2/3"
100% * 2/3 = 2/3 -- your chances of winning if you switch![/b]
09 Aug 08
ok she explained it better why one should switch. (scroll down to the example of the alien landing on the show)
it is the fact that we know stuff prior to the second experiment that changes the odds from 50/50. to my deffence, the phd's initially thought as i did.
and even if she is wrong and the odds stay at 50 percent, then it wouldn't matter if you do switch since the odds are equal. so i guess the answer is "always switch"
http://www.marilynvossavant.com/articles/gameshow.html
09 Aug 08
Originally posted by shavixmirTry it in practice with a friend. My class did this, and that helped.
Right. My head is about to implode... I need some help.
Say there are three doors. Behind two of the doors there are sacks of salt and behind the third door there is a Porche 911 GTI turbo 10.6 liter 1.000.000 horse power... convertable...
The door I eventually pick will yield me the prize behind it.
So, I pick a door. Before it's opened, the pre ...[text shortened]... on why I was discussing odds at a lap dancing bar, stick to the question at hand...
Thanks.
10 Aug 08
Originally posted by ZahlanziYou are misunderstanding the purpose of probability. Odds are simply a mathematical representation
a coin has 50% chance of landing on heads. try flipping the coin 2 times. if it lands both times on heads, what does it say about the calculated odds? if you flip it 100 times, it can land on heads from 0 to 100. some ranges of values have better statistical possibility. so in 100 series of 100 tosses each, heads will be around the value 50.
of the likelihood of some event to obtain. Heads will obtain 50% of the time on a fair coin. If
you flip the coin five times, and it lands heads all five times, the odds before the toss were
still 50%, and the odds for the next toss is also 50%. That it landed heads five times, or a hundred
times in succession has nothing to 'say about the calculated odds.'
Look at the problem this way: You have a 2/3 chance of being wrong on your first choice. So,
if you pick 'A,' then it's 2/3 likely that the the car will be behind 'B' or 'C.' Monty Hall then
opens either 'B' or 'C' showing a goat, and allows you to switch. Do you see how it's as if
Monty Hall is letting you pick two doors, a door you know to be wrong (because he showed you)
and another one? If you were 2/3 likely to be wrong before, now you must be 2/3 likely to be
right now.
Make sense?
Nemesio