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Monty Hall, we have a PROBLEM
- Thread starter owslachief
- Start date
Ralph Baer
Let's Go 'Tute!
Re: Monty Hall, we have a PROBLEM
KenKen is both mathematical and logical while Soduku is purely logical. The numerals in Soduku could be replaced by any nine symbols.
KenKen is both mathematical and logical while Soduku is purely logical. The numerals in Soduku could be replaced by any nine symbols.
Re: Monty Hall, we have a PROBLEM
Could you qualify to become a US citizen?
http://civicseducationinitiative.com/take-the-test
Could you qualify to become a US citizen?
http://civicseducationinitiative.com/take-the-test
Re: Monty Hall, we have a PROBLEM
Two years ago, one of my relatives celebrated a birthday: his/her age was 3 times a prime number.
Last year, his/her age was twice a prime number.
This year, his/her age is a prime number.
How old is s/he?
I found 3 possible solutions for regular human ages, and 4 additional possible solutions for "Biblical" ages (i.e., less than 1,000)
Two years ago, one of my relatives celebrated a birthday: his/her age was 3 times a prime number.
Last year, his/her age was twice a prime number.
This year, his/her age is a prime number.
How old is s/he?
I found 3 possible solutions for regular human ages, and 4 additional possible solutions for "Biblical" ages (i.e., less than 1,000)
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5 - but was s/he vaccinated?
Re: Monty Hall, we have a PROBLEM
"1" is NOT a prime number, it is the only number that is neither prime nor composite.
"1" is NOT a prime number, it is the only number that is neither prime nor composite.
aparch
Well-known member
11, and scheduled for booster shots before high school...
Ralph Baer
Let's Go 'Tute!
owslachief
occupe toi de tes oignons
Re: Monty Hall, we have a PROBLEM
A large box contains five cone-shaped hats, three of which are colored red and two blue. After having been told about the colors of all the hats, three logically gifted but blind-folded people are each asked to take one hat out of the box and put it on his or her head.
Now each in turn is allowed to take the his/her own blindfold off and make a determination as to what color hat they have on.
Subject #1 removes his blindfold, and after a moment says "I have no idea what color hat I'm wearing."
Subject #2 then removes her blindfold, and after a moment admits the same about her own hat.
Subject #3 thinks for a moment and, without having to remove her blindfold, declares the color of hat she's wearing ... and is correct.
What color hat is #3 wearing, and how did she come to the correct conclusion?
A large box contains five cone-shaped hats, three of which are colored red and two blue. After having been told about the colors of all the hats, three logically gifted but blind-folded people are each asked to take one hat out of the box and put it on his or her head.
Now each in turn is allowed to take the his/her own blindfold off and make a determination as to what color hat they have on.
Subject #1 removes his blindfold, and after a moment says "I have no idea what color hat I'm wearing."
Subject #2 then removes her blindfold, and after a moment admits the same about her own hat.
Subject #3 thinks for a moment and, without having to remove her blindfold, declares the color of hat she's wearing ... and is correct.
What color hat is #3 wearing, and how did she come to the correct conclusion?
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Re: Monty Hall, we have a PROBLEM
Lady in red
Is dancing with me
She's wearing a red hat, the other two were each wearing blue hats, leaving red as the only available color. The first two would have seen that the other participants were evenly split between red and blue, therefore not knowing at a 100% CI what was on top of his or her own head.
Lady in red
Is dancing with me
She's wearing a red hat, the other two were each wearing blue hats, leaving red as the only available color. The first two would have seen that the other participants were evenly split between red and blue, therefore not knowing at a 100% CI what was on top of his or her own head.
unofan
Well-known member
3 is wearing red.
1 doesn't know, which means at least one, but possibly both, of 2 & 3 is wearing red. It also means they are not both wearing blue, because 1 would know he had red if they were.
2 then says they don't know. However, if 3 had blue then 2 would know they had red since they could not both be blue. Since 2 doesn't know, that means 3 must be red.
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unofan
Well-known member
Right answer but wrong reasoning. 1 or 2 could each have either color.
Re: Monty Hall, we have a PROBLEM
Correct, I wasn't taking into account the fact that 3 would've been thinking about the reasoning 1 and 2 would need to use in order to make their answers.
Correct, I wasn't taking into account the fact that 3 would've been thinking about the reasoning 1 and 2 would need to use in order to make their answers.
owslachief
occupe toi de tes oignons
Re: Monty Hall, we have a PROBLEM
<strike>Sort of</strike> Yes - there are <strike>3</strike> oopsee 4 scenarios where the story comes to the same conclusion ... all of them ending with #3 wearing red:
#1 #2 #3
RED RED RED
RED BLUE RED
BLUE BLUE RED
RED BLUE RED
<strike>Sort of</strike> Yes - there are <strike>3</strike> oopsee 4 scenarios where the story comes to the same conclusion ... all of them ending with #3 wearing red:
#1 #2 #3
RED RED RED
RED BLUE RED
BLUE BLUE RED
RED BLUE RED
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mookie1995
there's a good buck in that racket.
Re: Monty Hall, we have a PROBLEM
#1 can look and see 3 in blue and 2 in red.
the #2 can look and see 3 in blue and 1 in red.
there are five hats. one blue and one red are still on a table.
#1 can look and see 3 in blue and 2 in red.
the #2 can look and see 3 in blue and 1 in red.
there are five hats. one blue and one red are still on a table.
Re: Monty Hall, we have a PROBLEM
:shiva six-handed face palm:
I saw an episode of Jeopardy in which all three contestants ended with $0!
The one in the lead committed the inexcusable error of betting everything. The cardinal rule if you are in the lead is that you only bet $1 more than the highest total that the second-place contestant can possibly reach if they bet everything. How can anyone "smart enough" to get on the show in the first place not do that??
This rule, in turn, opens up some really interesting game theory approaches for second and third.
The game theory approaches for # 2 are quite interesting and probably are too varied to discuss, since they depend upon how much # 2 has relative to # 1, and also depend upon whether # 2 expects # 3 to use optimal game theory or not.
In other words, if # 1 has $15,000, # 2 has $10,000, then # 1 bets $5,001. Ignoring # 3 strategy for now, # 2 bets $0, since if # 1 is wrong, $10,000 > $9,999. # 1 "has to" always follow the cardinal rule, or else loses if # 2 bets everything and is right. # 1 cannot take that risk.
Generally, if I am in third, I have to hope that 1 and 2 are both wrong. And so I bet $1 more than what # 1 would end up with assuming s/he followed the rule just cited, and then was wrong.
If # 3 has $6,000, say, and # 1 bets $5,001, # 3 bets $4,000...unless # 3 anticipates # 2 to bet $0, and so bets $4,001.
It is annoying to me the number of times when it is "obvious" that the optimal bet for one of the contestants is $0 and then they don't do it.
:shiva six-handed face palm:
I saw an episode of Jeopardy in which all three contestants ended with $0!
The one in the lead committed the inexcusable error of betting everything. The cardinal rule if you are in the lead is that you only bet $1 more than the highest total that the second-place contestant can possibly reach if they bet everything. How can anyone "smart enough" to get on the show in the first place not do that??
This rule, in turn, opens up some really interesting game theory approaches for second and third.
The game theory approaches for # 2 are quite interesting and probably are too varied to discuss, since they depend upon how much # 2 has relative to # 1, and also depend upon whether # 2 expects # 3 to use optimal game theory or not.
In other words, if # 1 has $15,000, # 2 has $10,000, then # 1 bets $5,001. Ignoring # 3 strategy for now, # 2 bets $0, since if # 1 is wrong, $10,000 > $9,999. # 1 "has to" always follow the cardinal rule, or else loses if # 2 bets everything and is right. # 1 cannot take that risk.
Generally, if I am in third, I have to hope that 1 and 2 are both wrong. And so I bet $1 more than what # 1 would end up with assuming s/he followed the rule just cited, and then was wrong.
If # 3 has $6,000, say, and # 1 bets $5,001, # 3 bets $4,000...unless # 3 anticipates # 2 to bet $0, and so bets $4,001.
It is annoying to me the number of times when it is "obvious" that the optimal bet for one of the contestants is $0 and then they don't do it.
FlagDUDE08
Banned
Re: Monty Hall, we have a PROBLEM
Was that the second episode of the current run? That ended all with $0. It's called pulling a Claven. Named after Cliff Claven, the character from Cheers, who went on Jeopardy, got a big lead going into Final, and got cocky and bet it all.
Was that the second episode of the current run? That ended all with $0. It's called pulling a Claven. Named after Cliff Claven, the character from Cheers, who went on Jeopardy, got a big lead going into Final, and got cocky and bet it all.
Twitch Boy
Defacing this tagline is a MAX foul
Re: Monty Hall, we have a PROBLEM
The kind folks at J! Archive have you covered.
http://www.j-archive.com/wageringcalculator.php
Be sure to find your way to the glossary for explanations of some of the various wagering scenarios.
http://www.j-archive.com/help.php#glossary
:shiva six-handed face palm:
I saw an episode of Jeopardy in which all three contestants ended with $0!
The one in the lead committed the inexcusable error of betting everything. The cardinal rule if you are in the lead is that you only bet $1 more than the highest total that the second-place contestant can possibly reach if they bet everything. How can anyone "smart enough" to get on the show in the first place not do that??
This rule, in turn, opens up some really interesting game theory approaches for second and third.
The game theory approaches for # 2 are quite interesting and probably are too varied to discuss, since they depend upon how much # 2 has relative to # 1, and also depend upon whether # 2 expects # 3 to use optimal game theory or not.
In other words, if # 1 has $15,000, # 2 has $10,000, then # 1 bets $5,001. Ignoring # 3 strategy for now, # 2 bets $0, since if # 1 is wrong, $10,000 > $9,999. # 1 "has to" always follow the cardinal rule, or else loses if # 2 bets everything and is right. # 1 cannot take that risk.
Generally, if I am in third, I have to hope that 1 and 2 are both wrong. And so I bet $1 more than what # 1 would end up with assuming s/he followed the rule just cited, and then was wrong.
If # 3 has $6,000, say, and # 1 bets $5,001, # 3 bets $4,000...unless # 3 anticipates # 2 to bet $0, and so bets $4,001.
It is annoying to me the number of times when it is "obvious" that the optimal bet for one of the contestants is $0 and then they don't do it.
The kind folks at J! Archive have you covered.
http://www.j-archive.com/wageringcalculator.php
Be sure to find your way to the glossary for explanations of some of the various wagering scenarios.
http://www.j-archive.com/help.php#glossary
P
Priceless
Guest
Re: Monty Hall, we have a PROBLEM
Who are three people who have never been in my kitchen?
Who are three people who have never been in my kitchen?
