@lkeigwin said:
Sorry for starting a tangent on this thread with the Monty Hall problem. I'll try to end it here. (Explanation below not my words.)
Lance.
Rules:
1) The host must always open a door that was not picked by the contestant.
2) The host must always open a door to reveal a goat and never the car.
3) The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
There are three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.
An intuitive explanation is that, if the contestant initially picks a goat (2 of 3 doors), the contestant will win the car by switching because the other goat can no longer be picked, whereas if the contestant initially picks the car (1 of 3 doors), the contestant will not win the car by switching. The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.
Fun fact: A study showed that pigeons repeatedly exposed to the problem rapidly learn always to switch, unlike humans.
This problem only works if in every scenario the host pics the goat. In reality it does not work.
There you go....I guess it was a dirty trick to swap which boxes you control. If I had it to do over again, I would have kept the same situation...my bad.
Or, let's really go extreme.
There are 1 million boxes. One box has $1million dollars.
You choose one box, and give it to me. You keep the other 999,999.
I think we agree that there is a VERY small likelihood that you gave me the $1million. Very small.
No matter how many of the remaining 999,999 boxes I open, (knowing that I know which contained the $1mil), the chances of the $1mil being in those boxes...and NOT the one you gave me....is still very high.
Even if we get down to ONE box of those 999,999, it is MUCH more likely to contain the $1mil than the box we set aside for me. No way it's a 50/50 choice.
All these theater of the mind tricks bamboozle so many, yet if you just think it through.
It's like coin grading. Don't think about it, just get your coins graded with pcgs!
Now I am off my high merry go-around horse and will step off of my soap box.
@FranklinHalfAddict said:
This problem only works if in every scenario the host pics the goat. In reality it does not work.
Absolutely correct.
But that's part of the setup for the question. See "2)" of the rules in Ikeigwin's post. The important part is that the host DOES know where the prize is.
If he doesn't, the entire problem changes.
(By the way....have I revealed yet how much I like logical puzzles? I'm completely unbearable sometimes. My apologies. )
It doesn't matter that Monte knows where the goat is - what matters is the probabilities were assigned from a set of three unknowns. Then, when one unknown is revealed it changes the probabilities of all the events originally in that set - not just in the subset of two doors not chosen.
There is no way the odds are 2/3 that the one door contains the car - simply because people are mixing up open and closed sets
@tradedollarnut said:
Open sets and closed sets. When you mix them up, you can prove 2=1
I don't think you're giving the problem fair consideration.
In the million box scenario, whittled down to two boxes, you would be completely justified to assume a 50/50 chance if, and only if, you had no knowledge of how we got to 2 boxes.
If you DID know how we got to 2 boxes, there is an obvious superior choice to be made.
Added:
And I hasten to point out that this isn't just a mental game, or a trick. This applies to the REAL WORLD. If you did a similar number of tests, that 1/3 vs. 2/3 probability, (or 1/1,000,000 vs. 999,999/1,000,000), would really reveal itself.
Probability has no purpose if it didn't apply to the real world.
The player picks one door out of three, their odds of having the car is 1 in 3.
Monty's odds of having the car in the remaining two doors is 2 in 3.
Just because Monty reveals a goat doesn't mean his odds suddenly drop to 1 in 3, especially since he knew where the goat was.
@AllCoinsRule said:
Here's a little bit tougher puzzle. (I'm sure some of you have seen in various other non coin forms!).
You have a roll of 50 1982 Lincoln Cents. One is bronze, the other 49 are zinc. Let's assume you can't tell them apart without weighing them in some manner. You only have a balance scale. What is the minimum number of uses of the balance scale needed, in order to determine which coin is the bronze one.
@AllCoinsRule said:
Here's a little bit tougher puzzle. (I'm sure some of you have seen in various other non coin forms!).
You have a roll of 50 1982 Lincoln Cents. One is bronze, the other 49 are zinc. Let's assume you can't tell them apart without weighing them in some manner. You only have a balance scale. What is the minimum number of uses of the balance scale needed, in order to determine which coin is the bronze one.
25/25
12/12
6/6
3/3
?
That would be a lot of weighings since you need to weigh each stack. Assume the bronze cent is heavier. Take the heavier stack of 25 and add a coin from the other lighter stack to make 26. Divide into 2 stacks of 13 and discard the lighter stack. Add a coin to the heavier stack to get 14. Divide into 2 stacks of 7 and repeat. Add a coin to the heavier stack of 7 to get 8 and divide into 2 stacks of 4. Take the heavier stack of 4 and divide into 2 stacks of 2. Divide the heavier stack of 2 in half. The heavier one is the bronze cent.
No need to add 1 coin back into the stack.
Take the heavier stack of 25 and split into two groups of 12.
If they are equal in weight, then the bronze coin is one the excluded.
Otherwise split the heavier group of 12 in two groups of 6.
Weigh those and take the heavier group of 6 and split into groups of 3.
Weigh those and take the heavier group of 3 and set one coin aside and weigh the other two.
If one is heavier, then you found the bronze one.
If they weigh the same then it is the one you set aside.
So the most uses of the scale you would need is 5.
And it may be as few as two times if the two groups of 12 coins weighed the same and the bronze coin was the one you set aside at that point.
@10000lakes said:
No need to add 1 coin back into the stack.
Take the heavier stack of 25 and split into two groups of 12.
If they are equal in weight, then the bronze coin is one the excluded.
Otherwise split the heavier group of 12 in two groups of 6.
Weigh those and take the heavier group of 6 and split into groups of 3.
Weigh those and take the heavier group of 3 and set one coin aside and weigh the other two.
If one is heavier, then you found the bronze one.
If they weigh the same then it is the one you set aside.
So the most uses of the scale you would need is 5.
And it may be as few as two times if the two groups of 12 coins weighed the same and the bronze coin was the one you set aside at that point.
You guys are all missing the actual issue.
we are talking about a COIN, not mathematics and probabilities of square circles, goats, cars or monkeys.
A round , any size coin , a good, not doctored in any way coin, any metal , landing on one side or the other.
(Lets leave the faint probability out of this experiment for now.)
NO mathematics needed here or applicable.
IMO:
Facts are this: the coin is attracted by gravity. The structure, relief and other properties of the coin together with effects and influences (and there are a ton)and strengths of gravity at your specific flipping location will determine on which side this coin will land. Not if you flip it with your fingers, hand, left toe or a flipping machine.
You only have ONE item, your coin, not a multiple of 10 different coins where you know 100% for sure that ever 11th coin must be a different coin. You can extend that into multiple types of coins. it will only multiply your count accordingly.
But there is a mathematical result.
With our coin, a mathematical or theoretical probability result is impossible. what formula would you base it on?
@tradedollarnut said:
It doesn't matter that Monte knows where the goat is - what matters is the probabilities were assigned from a set of three unknowns. Then, when one unknown is revealed it changes the probabilities of all the events originally in that set - not just in the subset of two doors not chosen.
There is no way the odds are 2/3 that the one door contains the car - simply because people are mixing up open and closed sets
Incorrect. Given the way the problem is laid out, they are what is known as "conditional probabilities". Google that and you'll find a large amount of reading material to keep you busy a long time.
Another way to think about the Monte Hall problem that may help:
You get to pick one door. I as the host gets to pick two doors. I get to look behind both doors, and then show you an empty one. Do you still want your door that you picked initially? Or do you want to trade me for my last door?
There is a reason why it took mathematicians so long to "solve" probability formulas and theorems. It is not intuitive.
@neildrobertson said:
This would be a good time to remind people of the Dunning-Kruger effect.
This pattern of over-estimating competence was seen in studies of skills as diverse as reading comprehension, practicing medicine, operating a motor vehicle, and playing games such as chess or tennis. Dunning and Kruger proposed that, for a given skill, incompetent people will:[4]
fail to recognize their own lack of skill
fail to recognize the extent of their inadequacy
fail to accurately gauge skill in others
recognize and acknowledge their own lack of skill only after they are exposed to training for that skill
Wut, you are pulling my lariat.
What does Dunning-Kruger say about coin grading skills?
@YQQ said:
You guys are all missing the actual issue.
we are talking about a COIN, not mathematics and probabilities of square circles, goats, cars or monkeys.
A round , any size coin , a good, not doctored in any way coin, any metal , landing on one side or the other.
(Lets leave the faint probability out of this experiment for now.)
NO mathematics needed here or applicable.
IMO:
Facts are this: the coin is attracted by gravity. The structure, relief and other properties of the coin together with effects and influences (and there are a ton)and strengths of gravity at your specific flipping location will determine on which side this coin will land. Not if you flip it with your fingers, hand, left toe or a flipping machine.
You only have ONE item, your coin, not a multiple of 10 different coins where you know 100% for sure that ever 11th coin must be a different coin. You can extend that into multiple types of coins. it will only multiply your count accordingly.
But there is a mathematical result.
With our coin, a mathematical or theoretical probability result is impossible. what formula would you base it on?
So what would the expected H/T ratio be for flipping a cent planchet at the top of Mount Everest? 5.00001 to 4.99999?
It may be easier to appreciate the solution by considering the same problem with 100 doors instead of just three. In this case there are 99 doors without the prize behind them and one door with the prize. The contestant picks a door. Monty Hall then has to reveal 98 of the other doors without the prize leaving just the contestant's door and one other door.
Unless the contestant chose the one door with the prize (a 1/100 probability) the only door left unopen by Monty must be the one with the prize behind it. So there is 99/100 probability that the other door will contain the prize, as 99 out of 100 times the player first picked a door without the prize. The contestant's probability of winning is therefore 99/100 if he switches compared to 1/100 if he does not. What if you had 10,000,000 doors? Would you switch after it was done to the last 2 doors?
In the original 3 door question....you increase your odds from 1/3 to 2/3 by switching doors after a losing door is revealed. 2/3 chance from the start that the prize is not behind the door you picked.
The answer is that it can be done with just 4 uses of the scale. I'll give it a couple more days before explaining in case someone wants to still try to figure out how.
Ok, I like a good mental challenge.
Here is one solution that may not be the simplest, but it does do it with 4 or less uses of the scale.
Hopefully it's not to convoluted to follow
Take the 50 coins and split into groups of 22,22,6.
Step #1 weigh 22 vs 22 setting 6 coins aside.
If the 22 vs 22 are equal then proceed to step #3b with the 6 coins that you set aside.
Otherwise take the heavier group of 22 and split into groups of 8,8,6 and go to step #2.
Step #2 weigh 8 vs 8 setting aside 6.
If the 8 vs 8 are equal then proceed to Step #3b with the 6 coins that you set aside.
Otherwise take the heavier group of 8 and split into groups of 3,3,2 and go to step #3a
Step #3a weigh 3 vs 3 setting aside 2 coins.
If the 3 vs 3 are not equal then go to step #4a with the heavier group of 3 coins.
If they are equal then take the 2 coins you set aside and go to step #4b.
Step #3b weigh 3 vs 3 with no coins set aside.
Take the heavier group of 3 coins and go to step #4a.
Step #4a weigh 1 vs 1 from the group of 3 that is heavier, setting 1 coin aside.
If they are equal, it is the coin that was set aside, otherwise it is the heavier coin.
Step #4b weigh 1 vs 1 whichever is heavier is the bronze cent.
@AllCoinsRule said:
You got it. Essentially for a set of N candidate coins you can just always split it into groups of 3, leaving at most the ceil of N/3 candidates left. The number of uses of the scale is then just the ceil of the log base 3 of N.
What you just said is Greek to me, but thanks for the confirmation that I got it right.
@david3142 said:
The real-world answer is that it depends on how strong your prior is that the coin is in fact 50/50. If you are given that its fairness is an absolute fact (and there is no flipping skill) then it is always 1/2. In practice this is obviously never the case. What are the chances the coin is biased? What are the chances the flipper has skill? At 9 in a row I would be a little skeptical. At 20 in a row (1 in a million), I would be pretty sure something was wrong and would definitely bet heads. The above is a basic Bayesian framework. Every new bit of information should cause a slight change in your prior belief consistent with the relative probabilities and the initial distribution.
Also, to clarify above - the binomial distribution converges toward the normal distribution as N goes to infinity.
Do you think there is a person alive who could flip with sufficient skill to produce at will all heads or all tails or some predetermined outcome?
Depending on how many turns in the air are required for you to call it a valid flip, the outcome will vary between pure skill and pure luck. I have no idea how skilled flippers may be, but I'm sure that somebody is practicing!
Anyway, all that is required for a flip not to be 50/50 is some amount of skill (or bias in the coin itself).
@abcde12345 said:
. . . wouldn't logic dictate it HAS to come up tails on the tenth flip (no cheating)? It can't still just be 50/50 even though
that what textbooks declare.
Here's another way to think of it... the coin has no memory whatsoever. It has no idea what the last flip was so the odds always remain 50/50.
@abcde12345 said:
. . . wouldn't logic dictate it HAS to come up tails on the tenth flip (no cheating)? It can't still just be 50/50 even though
that what textbooks declare.
If you and your best friend stood there and watched a LEGIT coin flipped 500 times, and it landed on heads all 500 times, and then I walked over without any information on the preceding 500 flips......why would the odds of a head on the 501st flip not be a 50/50 to me? And if it is 50/50 to me, then guess what.....its also 50/50 for you and anyone else as well.
each flip has a 50 50 chance of coming up heads or tails.
(we are eliminating the chance that it could land on its side or edge)
the probability of each previous flip and the next flips has nothing to do with the current flip.
each flip has a 50 50 chance of coming up heads or tails and thats about it.
Collector of Buffalo Nickels and other 20th century United States Coinage a.k.a "The BUFFINATOR"
Comments
This problem only works if in every scenario the host pics the goat. In reality it does not work.
There you go....I guess it was a dirty trick to swap which boxes you control.
If I had it to do over again, I would have kept the same situation...my bad.
Or, let's really go extreme.
All these theater of the mind tricks bamboozle so many, yet if you just think it through.
It's like coin grading. Don't think about it, just get your coins graded with pcgs!
Now I am off my high merry go-around horse and will step off of my soap box.
Absolutely correct.
But that's part of the setup for the question. See "2)" of the rules in Ikeigwin's post. The important part is that the host DOES know where the prize is.
If he doesn't, the entire problem changes.
(By the way....have I revealed yet how much I like logical puzzles? I'm completely unbearable sometimes.
My apologies. 
)
1 million minus 1 = 999,999
Everything else with your post is fine.
Yeah....I fixed that. I'm an engineer. What do I know about numbers?
Open sets and closed sets. When you mix them up, you can prove 2=1
It doesn't matter that Monte knows where the goat is - what matters is the probabilities were assigned from a set of three unknowns. Then, when one unknown is revealed it changes the probabilities of all the events originally in that set - not just in the subset of two doors not chosen.
There is no way the odds are 2/3 that the one door contains the car - simply because people are mixing up open and closed sets
I don't think you're giving the problem fair consideration.
In the million box scenario, whittled down to two boxes, you would be completely justified to assume a 50/50 chance if, and only if, you had no knowledge of how we got to 2 boxes.
If you DID know how we got to 2 boxes, there is an obvious superior choice to be made.
Added:
And I hasten to point out that this isn't just a mental game, or a trick. This applies to the REAL WORLD. If you did a similar number of tests, that 1/3 vs. 2/3 probability, (or 1/1,000,000 vs. 999,999/1,000,000), would really reveal itself.
Probability has no purpose if it didn't apply to the real world.
The player picks one door out of three, their odds of having the car is 1 in 3.
Monty's odds of having the car in the remaining two doors is 2 in 3.
Just because Monty reveals a goat doesn't mean his odds suddenly drop to 1 in 3, especially since he knew where the goat was.
Think about it and the light bulb will go off
Well now my post makes no sense!
I hereby declare that david3142 corrected a stupid error of mine, and his post made perfect sense at the time he wrote it.
100 post and counting on a coin flip and a goat. Gotta loves these boards...
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I just came here for the goat.....
Been reading up on it. Very interesting...
25/25
12/12
6/6
3/3
?
That would be a lot of weighings since you need to weigh each stack. Assume the bronze cent is heavier. Take the heavier stack of 25 and add a coin from the other lighter stack to make 26. Divide into 2 stacks of 13 and discard the lighter stack. Add a coin to the heavier stack to get 14. Divide into 2 stacks of 7 and repeat. Add a coin to the heavier stack of 7 to get 8 and divide into 2 stacks of 4. Take the heavier stack of 4 and divide into 2 stacks of 2. Divide the heavier stack of 2 in half. The heavier one is the bronze cent.
If a coin is flipped nine times in a row and comes up heads each time. . .
The real question is which came first, the coin collector or the coin?
Or if a coin collector is talking when his wife is not around, is he still wrong?
https://www.autismforums.com/media/albums/acrylic-colors-by-rocco.291/
No need to add 1 coin back into the stack.
Take the heavier stack of 25 and split into two groups of 12.
If they are equal in weight, then the bronze coin is one the excluded.
Otherwise split the heavier group of 12 in two groups of 6.
Weigh those and take the heavier group of 6 and split into groups of 3.
Weigh those and take the heavier group of 3 and set one coin aside and weigh the other two.
If one is heavier, then you found the bronze one.
If they weigh the same then it is the one you set aside.
So the most uses of the scale you would need is 5.
And it may be as few as two times if the two groups of 12 coins weighed the same and the bronze coin was the one you set aside at that point.
RUN!!
This is what I came up with too.
Lance
You guys are all missing the actual issue.
we are talking about a COIN, not mathematics and probabilities of square circles, goats, cars or monkeys.
A round , any size coin , a good, not doctored in any way coin, any metal , landing on one side or the other.
(Lets leave the faint probability out of this experiment for now.)
NO mathematics needed here or applicable.
IMO:
Facts are this: the coin is attracted by gravity. The structure, relief and other properties of the coin together with effects and influences (and there are a ton)and strengths of gravity at your specific flipping location will determine on which side this coin will land. Not if you flip it with your fingers, hand, left toe or a flipping machine.
You only have ONE item, your coin, not a multiple of 10 different coins where you know 100% for sure that ever 11th coin must be a different coin. You can extend that into multiple types of coins. it will only multiply your count accordingly.
But there is a mathematical result.
With our coin, a mathematical or theoretical probability result is impossible. what formula would you base it on?
Incorrect. Given the way the problem is laid out, they are what is known as "conditional probabilities". Google that and you'll find a large amount of reading material to keep you busy a long time.
Another way to think about the Monte Hall problem that may help:
You get to pick one door. I as the host gets to pick two doors. I get to look behind both doors, and then show you an empty one. Do you still want your door that you picked initially? Or do you want to trade me for my last door?
There is a reason why it took mathematicians so long to "solve" probability formulas and theorems. It is not intuitive.
This would be a good time to remind people of the Dunning-Kruger effect.
https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
The more you learn, the less you know.
IG: DeCourcyCoinsEbay: neilrobertson
"Numismatic categorizations, if left unconstrained, will increase spontaneously over time." -me
This pattern of over-estimating competence was seen in studies of skills as diverse as reading comprehension, practicing medicine, operating a motor vehicle, and playing games such as chess or tennis. Dunning and Kruger proposed that, for a given skill, incompetent people will:[4]
fail to recognize their own lack of skill
fail to recognize the extent of their inadequacy
fail to accurately gauge skill in others
recognize and acknowledge their own lack of skill only after they are exposed to training for that skill
Wut, you are pulling my lariat.
What does Dunning-Kruger say about coin grading skills?
So what would the expected H/T ratio be for flipping a cent planchet at the top of Mount Everest? 5.00001 to 4.99999?
It may be easier to appreciate the solution by considering the same problem with 100 doors instead of just three. In this case there are 99 doors without the prize behind them and one door with the prize. The contestant picks a door. Monty Hall then has to reveal 98 of the other doors without the prize leaving just the contestant's door and one other door.
Unless the contestant chose the one door with the prize (a 1/100 probability) the only door left unopen by Monty must be the one with the prize behind it. So there is 99/100 probability that the other door will contain the prize, as 99 out of 100 times the player first picked a door without the prize. The contestant's probability of winning is therefore 99/100 if he switches compared to 1/100 if he does not. What if you had 10,000,000 doors? Would you switch after it was done to the last 2 doors?
In the original 3 door question....you increase your odds from 1/3 to 2/3 by switching doors after a losing door is revealed. 2/3 chance from the start that the prize is not behind the door you picked.
Ok, I like a good mental challenge.
Here is one solution that may not be the simplest, but it does do it with 4 or less uses of the scale.
Hopefully it's not to convoluted to follow
Take the 50 coins and split into groups of 22,22,6.
Step #1 weigh 22 vs 22 setting 6 coins aside.
If the 22 vs 22 are equal then proceed to step #3b with the 6 coins that you set aside.
Otherwise take the heavier group of 22 and split into groups of 8,8,6 and go to step #2.
Step #2 weigh 8 vs 8 setting aside 6.
If the 8 vs 8 are equal then proceed to Step #3b with the 6 coins that you set aside.
Otherwise take the heavier group of 8 and split into groups of 3,3,2 and go to step #3a
Step #3a weigh 3 vs 3 setting aside 2 coins.
If the 3 vs 3 are not equal then go to step #4a with the heavier group of 3 coins.
If they are equal then take the 2 coins you set aside and go to step #4b.
Step #3b weigh 3 vs 3 with no coins set aside.
Take the heavier group of 3 coins and go to step #4a.
Step #4a weigh 1 vs 1 from the group of 3 that is heavier, setting 1 coin aside.
If they are equal, it is the coin that was set aside, otherwise it is the heavier coin.
Step #4b weigh 1 vs 1 whichever is heavier is the bronze cent.
What you just said is Greek to me, but thanks for the confirmation that I got it right.
Do you think there is a person alive who could flip with sufficient skill to produce at will all heads or all tails or some predetermined outcome?
Depending on how many turns in the air are required for you to call it a valid flip, the outcome will vary between pure skill and pure luck. I have no idea how skilled flippers may be, but I'm sure that somebody is practicing!
Anyway, all that is required for a flip not to be 50/50 is some amount of skill (or bias in the coin itself).
I wonder how the folks at Guinness Book of World Records would set it up to assure randomness and remove bias.
Ever notice how when you flip a coin sometimes it rolls away like a runaway wheel chair and disappears like a chameleon?
https://www.autismforums.com/media/albums/acrylic-colors-by-rocco.291/
Hey abcde12345 just have to ask you out of this whole big wide world you can have any avatar you want and you chose a piece of bread with eyes ?
Here's another way to think of it... the coin has no memory whatsoever. It has no idea what the last flip was so the odds always remain 50/50.
If you and your best friend stood there and watched a LEGIT coin flipped 500 times, and it landed on heads all 500 times, and then I walked over without any information on the preceding 500 flips......why would the odds of a head on the 501st flip not be a 50/50 to me? And if it is 50/50 to me, then guess what.....its also 50/50 for you and anyone else as well.
** Logic would dictate** you keep it for your next coin flip to better your odds.
each flip has a 50 50 chance of coming up heads or tails.
(we are eliminating the chance that it could land on its side or edge)
the probability of each previous flip and the next flips has nothing to do with the current flip.
each flip has a 50 50 chance of coming up heads or tails and thats about it.
a.k.a "The BUFFINATOR"