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A Full Monty! #1060753
03/27/16 12:46 AM
03/27/16 12:46 AM
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Mojave desert, California
CCbomber Offline OP
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One of your sisters (Jane or Mary) is taking a shower. You hear some singing.
You know that Jane always sings in the shower but that Mary only sings during
one shower out of four. What is the probability that Jane is in the shower?

easter

Re: A Full Monty! [Re: CCbomber] #1060755
03/27/16 01:42 AM
03/27/16 01:42 AM
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BobH Offline
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Assuming that they each shower daily, I would say, one chance in five. Out of the 8 showers in 4 days, 5 have someone singing. In only one of those 5 is the singer Jane. smile


Bob
As I've grown older, I've found that my memory is not as good as I used to think it was.
Re: A Full Monty! [Re: CCbomber] #1060758
03/27/16 02:50 AM
03/27/16 02:50 AM
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Mojave desert, California
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Good thinking, Bob. Except that Jane is the one that always sings.
So the answer is 4/5. Well done. praise

Re: A Full Monty! [Re: CCbomber] #1060794
03/27/16 10:57 AM
03/27/16 10:57 AM
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Ottawa Ontario Canada
CanukDenis Offline
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You were referring to the Monty Hall
famous probability problem, right CCb ? wink


I'm a man of few words, BUT I use 'em often!!
Re: A Full Monty! [Re: CCbomber] #1060800
03/27/16 12:44 PM
03/27/16 12:44 PM
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Mojave desert, California
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Yes, Denis. It was a problem I read (worded differently) in an article
about variations of the Monty Hall problem, Monty fall and Monty crawl. smile

Re: A Full Monty! [Re: CCbomber] #1060805
03/27/16 12:58 PM
03/27/16 12:58 PM
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The problem that made Marilyn Vos Savant famous...


I'm a man of few words, BUT I use 'em often!!
Re: A Full Monty! [Re: CCbomber] #1060819
03/27/16 02:13 PM
03/27/16 02:13 PM
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Coincidentally, I had just watched this Numberphile video on the Monty Hall problem (there are many to choose from!) that makes the answer much more intuitive.

The solution is not clear when there are just three doors so instead, say that there are 100 doors. You pick a door. It's clear that your odds of having picked the correct door are 1 in 100. Now Monty Hall eliminates 98 of the other 99 doors, leaving only one other door remaining. It's easier to see in this example that you would obviously want to switch to the one door of the 99 that Monty did not eliminate. The odds of that door being correct are 99 in 100.

The same holds true for 3 doors but in this case the odds of your initial pick are 1 in 3 and the odds of the other non-eliminated door are 2 in 3.


Bob
As I've grown older, I've found that my memory is not as good as I used to think it was.
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