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This was the best explanation for me instead of the pictures:

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
 
Sammich117 said:
This was the best explanation for me instead of the pictures:

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
Click to expand...

👍👍
 
lstone13 said:
Do share!
Click to expand...

okay! here goes:

There are 100 gnomes being oppressed by an evil, gun-wielding dictator. The dictator tells the gnomes that tomorrow he is going to line them up so that all the gnomes are facing one direction (each gnome faces the back of a fellow gnome’s head, except for the front gnome who sees nothing). The gnomes will be outfitted with either a black hat or a white hat, and they will not be able to see which hat they have personally received nor any of the hats behind them. There is no predetermined distribution of black and white hats – it could be 50/50, it could be 100/0…the evil, gun-wielding dictator is very, very evil so he will not tell the gnomes how many hats of each color there will be.

The dictator plans to start from the back of the line, asking each gnome what color hat he (or she) is wearing. If the gnome answers correctly, his (or her) life is spared. If the gnome answers incorrectly, the dictator guns down the gnome on the spot, with all the remaining gnomes hearing the shot and becoming very, very sad over the death of a comrade.

We can’t have sad gnomes…so how many gnomes can be guaranteed safety if the gnomes come up with a clever plan? And what is their plan?
 
Sammich117 said:
This was the best explanation for me instead of the pictures:

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
Click to expand...

You know, supposedly, Marilyn vos Savant has the highest IQ in the world...
 
jla314 said:
okay! here goes:

There are 100 gnomes being oppressed by an evil, gun-wielding dictator. The dictator tells the gnomes that tomorrow he is going to line them up so that all the gnomes are facing one direction (each gnome faces the back of a fellow gnome’s head, except for the front gnome who sees nothing). The gnomes will be outfitted with either a black hat or a white hat, and they will not be able to see which hat they have personally received nor any of the hats behind them. There is no predetermined distribution of black and white hats – it could be 50/50, it could be 100/0…the evil, gun-wielding dictator is very, very evil so he will not tell the gnomes how many hats of each color there will be.

The dictator plans to start from the back of the line, asking each gnome what color hat he (or she) is wearing. If the gnome answers correctly, his (or her) life is spared. If the gnome answers incorrectly, the dictator guns down the gnome on the spot, with all the remaining gnomes hearing the shot and becoming very, very sad over the death of a comrade.

We can’t have sad gnomes…so how many gnomes can be guaranteed safety if the gnomes come up with a clever plan? And what is their plan?
Click to expand...
Say the color of the hat in front of them? That way they at least know there are some of them...OH OH! Then the person in front would know what hat to say. So only the last gnome would die..
 
Wait.. That wouldn't work... If the last gnome said the color in front of him, then that one would know what to say, but that may not necessarily be the color of the hat in front of HIM..

Hmmm.
 
Ok.

The last gnome could say the color of the hat in front of him.
That gnome would say what the guy behind him said, who may or may not live.
If that color is ALSO the color of the hat in front of him, he can cough.

They all live. Except that last one in line may or may not.
 
Sammich117 said:
Wait.. That wouldn't work... If the last gnome said the color in front of him, then that one would know what to say, but that may not necessarily be the color of the hat in front of HIM..

Hmmm.
Click to expand...

yeah its tricksy, too tricksy...

do you guys want hints?
 
jla314 said:
okay! here goes:

There are 100 gnomes being oppressed by an evil, gun-wielding dictator. The dictator tells the gnomes that tomorrow he is going to line them up so that all the gnomes are facing one direction (each gnome faces the back of a fellow gnome’s head, except for the front gnome who sees nothing). The gnomes will be outfitted with either a black hat or a white hat, and they will not be able to see which hat they have personally received nor any of the hats behind them. There is no predetermined distribution of black and white hats – it could be 50/50, it could be 100/0…the evil, gun-wielding dictator is very, very evil so he will not tell the gnomes how many hats of each color there will be.

The dictator plans to start from the back of the line, asking each gnome what color hat he (or she) is wearing. If the gnome answers correctly, his (or her) life is spared. If the gnome answers incorrectly, the dictator guns down the gnome on the spot, with all the remaining gnomes hearing the shot and becoming very, very sad over the death of a comrade.

We can’t have sad gnomes…so how many gnomes can be guaranteed safety if the gnomes come up with a clever plan? And what is their plan?
Click to expand...
99.... you say the color of the hat in front of you. if the dude in front of you is wearing white and you're wearing black (you know this because of the guy behind you who called it out) you stutter your own color. if the gnome in front is the same color as you just say it normally...
 
Sammich117 said:
This was the best explanation for me instead of the pictures:

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
Click to expand...

I think the logical problem that I have with it is what difference does picking a door first signify? What's the difference between picking a door first, then have the host open up the rest 999,998 doors, and the host just opening up 999,998 doors from the start without you picking a door? Mathematically obviously you should switch, but logically, why?

I think I've resolved it, just can't put it into words yet :laugh:

Sammich117 said:
Say the color of the hat in front of them? That way they at least know there are some of them...OH OH! Then the person in front would know what hat to say. So only the last gnome would die..
Click to expand...

That seems like a good answer lol
 
Sammich117 said:
Ok.

The last gnome could say the color of the hat in front of him.
That gnome would say what the guy behind him said, who may or may not live.
If that color is ALSO the color of the hat in front of him, he can cough.

They all live. Except that last one in line may or may not.
Click to expand...

no coughing allowed! they can only say black or white
 
bleargh said:
99.... you say the color of the hat in front of you. if the dude in front of you is wearing white and you're wearing black (you know this because of the guy behind you who called it out) you stutter your own color. if the gnome in front is the same color as you just say it normally...
Click to expand...

no stuttering! the dictator severely dislikes stutters and will shoot all of them
 
Sammich117 said:
Wait.. That wouldn't work... If the last gnome said the color in front of him, then that one would know what to say, but that may not necessarily be the color of the hat in front of HIM..

Hmmm.
Click to expand...

So just have every other one say the hat in front of them. So 1 says 2, 2 says 2 (lives), 3 says 4, 4 says 4 (lives)... etc.

Then all the even gnomes live. 50. Poor odd gnomes.

Edit: they are odd, why would we want them anyway??? 😛
 
i will tell you all that more than 50 gnomes live!
 
ksmi117 said:
So just have every other one say the hat in front of them. So 1 says 2, 2 says 2 (lives), 3 says 4, 4 says 4 (lives)... etc.

Then all the even gnomes live. 50. Poor odd gnomes.

Edit: they are odd, why would we want them anyway??? 😛
Click to expand...

that'll only work if they know which one in the line they will be, which won't be the case if the ONLY thing they can see is the gnome in front of them.
 
on the count of three they all attack the dictator and hopefully theres only like a 10 gnome collateral damage. so 90 live. the end.
 
jla314 said:
nope. he is a very evil dictator
Click to expand...

But wouldn't they be able to tell by counting? I mean, the first gnome asked will be the last in line, so he knows it and says the guy in front of him's color, then the second gnome will know he's second in line, and says his own color, and so on and so forth?
 
dingyibvs said:
that'll only work if they know which one in the line they will be, which won't be the case if the ONLY thing they can see is the gnome in front of them.
Click to expand...

they can see alllllllll the gnomes in front of them. so the gnome in the back can see all 99 hats in front of him
 
sdguy2008 said:
on the count of three they all attack the dictator and hopefully theres only like a 10 gnome collateral damage. so 90 live. the end.
Click to expand...

Haha.

dingyibvs said:
But wouldn't they be able to tell by counting? I mean, the first gnome asked will be the last in line, so he knows it and says the guy in front of him's color, then the second gnome will know he's second in line, and says his own color, and so on and so forth?
Click to expand...

Yeah, they could count. But apparently more would live another way.
 
The gnome in the back should say which hat he sees the most of, then they all just say that color and hope for the best. One of the colors has to be = or > 50, so at least 50 live.
 
Sammich117 said:
The gnome in the back should say which hat he sees the most of, then they all just say that color and hope for the best. One of the colors has to be = or > 50, so at least 50 live.
Click to expand...

getting warmer...
 
jla314 said:
they can see alllllllll the gnomes in front of them. so the gnome in the back can see all 99 hats in front of him
Click to expand...

Oh, that's much better. Chances are, there will be more of one color than the other, since it could be 50/50 to 100/0, so the last gnome yells out the color that is the most numerous, and every gnome answer that color. Other than a pure 50/50 distribution, this method will save the most gnomes.
 
the gnomes actually protest that black and white should be equal and without equality no one's really alive
 
Sammich117 said:
The gnome in the back should say which hat he sees the most of, then they all just say that color and hope for the best. One of the colors has to be = or > 50, so at least 50 live.
Click to expand...

jla314 said:
getting warmer...
Click to expand...

That's not right??? We've come up with some great plans for these gnomes. What more do they want?
 
dingyibvs said:
Oh, that's much better. Chances are, there will be more of one color than the other, since it could be 50/50 to 100/0, so the last gnome yells out the color that is the most numerous, and every gnome answer that color. Other than a pure 50/50 distribution, this method will save the most gnomes.
Click to expand...


ahh you are so tantalizingly close!!
 
how about even vs. odd? if the back gnome sees an even # of black in front of him, he says black. odd #, he says white. the guy in front counts. if the answer changes, he knows what color he is. does that make sense? then 99 live.
 
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