Subject:
|
Re: math question (or pattern... whatever...)
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Tue, 4 Mar 2003 12:40:42 GMT
|
Viewed:
|
541 times
|
| |
|
|
In lugnet.off-topic.geek, Bruce Schlickbernd writes:
> Inasmuch as you get to pick a second time, the odds are 50/50. If you did
> not get to pick a second time, the odds would be 1/3, not 1/2. Of course,
> this is leaving out the poker aspect: if your initial pick was incorrect,
> would they even offer the second chance? In fact, they did, but what
> percentage of the time I couldn't tell you, but I would suspect that they
> decided on showing a wrong door no matter what as part of the script (take X
> long to make a presentation).
See the link I posted and follow some of the links it has. You are in good
company, lots of mathemeticians came to the same conclusion. (and roasted
Marilyn Vos Savant about it at the time it first got a lot of publicity)
That was before they rigorously defined the problem, and ran computer
simulations. Once they did that, they realised that there IS a 2/3 chance
that you're right if you switch, given the parameters of the problem.
I like Monty Hall's explanation best, actually.
|
|
Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
|
| (...) Inasmuch as you get to pick a second time, the odds are 50/50. If you did not get to pick a second time, the odds would be 1/3, not 1/2. Of course, this is leaving out the poker aspect: if your initial pick was incorrect, would they even offer (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
61 Messages in This Thread:
 
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
