@oih82w8 said:
Does the "heads" of the coin portray a cat? Don't cats always land right side up...on their feet...unless they have buttered toast on their back...then it's up for grabs?
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Okay this has to be one of the strangest things I have heard in awhile "unless they have buttered toast on their backs" omg that's odd.
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@Cakes said:
Okay this has to be one of the strangest things I have heard in awhile "unless they have buttered toast on their backs" omg that's odd.
LOL....It's my Wife's favorite "scientific" experiment she is planning!
"If a cat always lands on it's feet....and buttered toast always lands buttered side down....what would happen if you strapped buttered toast on the back of a cat, and dropped it?"
She's pretty sure the cat/toast complex would float in mid-air.
@Cakes said:
Okay this has to be one of the strangest things I have heard in awhile "unless they have buttered toast on their backs" omg that's odd.
LOL....It's my Wife's favorite "scientific" experiment she is planning!
"If a cat always lands on it's feet....and buttered toast always lands buttered side down....what would happen if you strapped buttered toast on the back of a cat, and dropped it?"
She's pretty sure the cat/toast complex would float in mid-air.
Science suggests the cat will only float if the cat forgets it's falling in which case it would be both dead and alive.
It would seem that logic applied to luck/randomness circumstances can be quite the mind bender. Many years ago we were at the horse race track (harness) and the #1 horse won the first 9 races out of a total of 12 races. After each #1 horse winning most said it wouldn't happen again. Finally on race 10 the streak ended. What a parlay that could of been.
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@TommyType said:
I don't believe anything I see on video clips now-a-days. I've seen very convincing ones where people jump out of computer screens, disappear behind sheets, fly over houses, etc. The editors are just too good....
(I'm getting cranky in my old age....)
True, but on my google search there were many to chose from. Logic says at least one if not most of them are real.
First off, the OP gave only 2 options H or T.
AFAIK there is no finite known probability that a coin will land on its edge.
If it does then one would ignore it and keep going or start over.
Also there are 3 probabilities to contend with if one wants to get picky.
The probability that 10 coin flips will produce all H or T. That's fixed at 1/1024.
The probability that the string will continue as each coin is flipped. This changes as you go.
The probability that a single flip will give H or T is always 1/2.
probability 10 flips produce all heads is 1/1024 all tails is also 1/1024 so all heads or all tails would be 1/512.
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
It is simple fact that each flip (honest coin, honest flip) has a 50/50 chance of being either heads or tails. It does not matter what happened in previous flips. Barring the extremely rare occurrence of an edge landing (it has happened), the odds are even. What most are thinking of, when considering the original premise, is 'probability'.. The 50/50 still applies to each flip, but probability changes. Cheers, RickO
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.
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
It's got to be yes as you increase your odds from 1 in 3 to 1 in 2 (the host knows the correct door).
@david3142 said:
Yes is correct but the odds are 1/3 for staying and 2/3 for switching. This is known as the Monty Hall problem.
Uhhhmmm...no. Once you know the goat is gone, your odds are 1/2 for doing either
The assumption is that the host knows what is behind the doors and he will only open a non-prize door. I'm not going to try to convince you here as you can Google the problem yourself.
@david3142 said:
Yes is correct but the odds are 1/3 for staying and 2/3 for switching. This is known as the Monty Hall problem.
Uhhhmmm...no. Once you know the goat is gone, your odds are 1/2 for doing either
Quick logic sometimes is wrong when you play in the probability realm.
As noted, it is a well known problem covered in any probability class....and as much as I tried to fight it using the same logic you did, the answer is your probability increases if you switch!!! It all has to do with "what you know, and when you know it".
When you make your original pick, you obviously have 1/3 chance of picking the "big prize". And there is a 2/3 chance the "big prize" is behind the other two doors. (Good so far).
But when Monty reveals the "goat" behind one of the two unpicked doors, (and we KNOW he's not going to reveal the "big prize"), the probability of the "big prize" being behind the two unpicked doors doesn't change. The probability is just assigned to the ONE door that remains of those two. Your door is still 1/3, but the remaining unpicked door is now 2/3!
So, the act of Monty removing ONLY a "goat prize", actually doubles your chance of winning by SWITCHING doors.
(On the other hand, if we assume Monty was only randomly picking doors as well, the probabilities become 1/2, and 1/2, when he reveals the goat.....but if he reveals the "Big Prize" at random, your probability changes to 0/2 and 0/2, and you CAN'T win.)
It's all about what you know, and when you know it....And Monty's prior knowledge of where the prize is has a huge influence.
Given everything is equal and the coin is a regular, un-tampered coin, it can flip 1000 times or more and land on the same side every time. theoretically, it can land on the same side until the cows come home. and then the x-flip will change it and it lands 10 times or one time on the other side and then reverses again.
as I said, everything being equal. so why does it do that. I believe it is gravity , how it falls, how the high points of the coin are attracted to gravity, if perhaps someone disturbed the airflow around it.. it all matters.
There is absolutely NO guarantee. However, with every flip the unlimited possibility increases by 1 in favor of the other side.
"The probability that a single flip will give H or T is always 1/2" is not exactly true, but the closest you can come to. It is actually 1/3. you forgot the possibility of it landing on the rim and staying there. However I would not bet on it.
@MICHAELDIXON said:
Why did I waste 10 minutes reading this thread? I guess maybe I landed on my head one too many times!
Without those 10 minutes you would not have read the cat and buttered toast theory:
"LOL....It's my Wife's favorite "scientific" experiment she is planning!
"If a cat always lands on it's feet....and buttered toast always lands buttered side down....what would happen if you strapped buttered toast on the back of a cat, and dropped it?"
She's pretty sure the cat/toast complex would float in mid-air. "
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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.
@FranklinHalfAddict said:
For a single flip the odds are 50/50. But what are the odds that in 10 flips you'd get heads 10x? Just for arguments sake let's make the numbers easy and say 1 in 100, or 1%. So after 9 flips and 9 heads is there still only a 1% chance that that last flip is heads? If so, shouldn't there be a 99% chance the last flip will end up tails? Or am I way off here?
The problem is assuming the coin knows what the last 9, (or 15, or 100), flips were, and will act accordingly based on that history.
As noted, a fair flip of a fair coin will be 50/50, no matter what happened in the last 10 minutes, or the last 100 years.
(I still say your proper response to hearing that there were 9 consecutive flips of Heads is to stop trusting either the coin, or the person who told you that).
There was some scenario in the movie 21, with Kevin Spacey, where he asked a student to pick which door of three held some prize, and what the odds were he chose the correct door. Obviously the beginning odds are roughly 33%. Spacey's character reveals what's behind one door and there's nothing. So the prize is still behind one of the two remaining doors, one of which the student chose. Before opening another door the teacher asks the student what his odds are now that he chose the door with the prize behind it. One would assume it's now 50%. But for some reason it wasn't. The odds were actually greater than 50%. The teacher asked why and the student correctly explained, but it still didn't make sense to me.
Anyone remember that scene?
@FranklinHalfAddict said:
But for some reason it wasn't. The odds were actually greater than 50%. The teacher asked why and the student correctly explained, but it still didn't make sense to me.
Anyone remember that scene?
See Ikeigwin's post directly above yours. Same problem, with explanation.
@YQQ said:
Given everything is equal and the coin is a regular, un-tampered coin, it can flip 1000 times or more and land on the same side every time. theoretically, it can land on the same side until the cows come home. and then the x-flip will change it and it lands 10 times or one time on the other side and then reverses again.
as I said, everything being equal. so why does it do that. I believe it is gravity , how it falls, how the high points of the coin are attracted to gravity, if perhaps someone disturbed the airflow around it.. it all matters.
There is absolutely NO guarantee. However, with every flip the unlimited possibility increases by 1 in favor of the other side.
"The probability that a single flip will give H or T is always 1/2" is not exactly true, but the closest you can come to. It is actually 1/3. you forgot the possibility of it landing on the rim and staying there. However I would not bet on it.
So how close to random does a flip have to be in order to be truly random? If you flip it with your thumb and first finger to varying heights and it drops down onto the same table top is that random enough? The more uniform you try to make the flip, the more likely that you are approaching a situation where a uniform outcome results. I would think that the H/T differences for a Lincoln cent would be inconsequential but one could always get a planchet and mark each side with a magic marker.
I stated earlier that AFAIK there are no known viable odds of a flipped coin landing on its edge and staying there, so for practical purposes it can be ignored.
Do you think that the device used to pick the white balls for PowerBall does so in a manner that is truly random or is as random as is practical to achieve?
There is absolutely no way that choosing one blind door out of two leads to a 2/3 chance of being correct. Sorry
The falsehood is in not adjusting the odds of the initial door to 50:50 once the goat is revealed. It's like Schrödinger's cat - once you open the box to reveal the goat you already disturbed the odds.
I could do a fault tree analysis but I'm just not up to it...
The premise is that the two unchosen blind doors have a 2/3 chance of holding the car. The falsehood is stating that once the goat is revealed from those two that the other door has the entire 2/3 odds still. Those odds included the 1/3 for the door now revealed.
It's a nice parlor trick with math but that's all it is.
@tradedollarnut said:
There is absolutely no way that choosing one blind door out of two leads to a 2/3 chance of being correct. Sorry
The falsehood is in not adjusting the odds of the initial door to 50:50 once the goat is revealed. It's like Schrödinger's cat - once you open the box to reveal the goat you already disturbed the odds.
I could do a fault tree analysis but I'm just not up to it...
Believe me, I understand your issues. I made the same argument when I first heard the problem. But this is a classic problem that you'll see in any probability course, and it's selected exactly BECAUSE it doesn't seem reasonable. (And, it introduces the concept of prior knowledge).
Read closely the explanation, (with all possible situations), laid out in lkeigwin's post. The key is that Monty has intentionally removed one goat based on what he knew, but you didn't. As a result, you gain new knowledge.
Imagine you’re a contestant on a game show and given the choice of three doors. Behind one door is a car; behind the other two doors, goats. You pick a door–let’s say No. 1–and the host, who knows what’s behind the doors, opens another door, let’s say No. 3, which has a goat. He then says, “Do you want to change your pick?”
The correct answer, which seems quite counterintuitive, is Yes, you should switch. The odds of the car being behind door No. 3 are 2/3 vs 1/3 for door No. 1.
The reason is that the odds have changed now that we know an additional piece of information. The chance the car is behind door No.1 is still 1/3 but since we can rule out door No. 2, the chance of the car being behind door No. 3 is the combined probability of doors No. 2 and No. 3.
Here’s another way to think about it to drive home the point. Imagine there are 100 doors with 99 goats and 1 car behind them. You pick a door, let’s assume it’s No. 1. The host then opens 98 other doors revealing goats. What are the odds of the car being behind your initial pick, door No. 1, or the remaining unopened door? The answer is you have a 1/100 chance of getting it right with door No. 1 but a 99/100 chance with the remaining door. So you should always switch to the remaining door.
The basic rule of thumb with any conditional probability question is that given new information, you should always change your pick!
TDN, this is a well known problem, but you have to accept the rules of the game for 2/3 to be the correct odds after switching. The choice of the door reveal is not random so the original odds of your 1/3 guess do not change. Just for fun, how much are you willing to bet that you are correct in this discussion if I offer you even odds?
TDN, it's not a parlor trick. It's a word problem.
The key is that the host KNOWS what is behind each door. He will never show the prize behind a door he opens. His choice of door is not by chance.
Look at it this way:
You are the contestant and get to choose a door. I am the host and I get both doors. Off the bat your chance to win is 1/3 and mine is 2/3. I peak behind my two doors and then open one of the losing doors for you to see. Has that act magically changed your odds or my odds? Nope.
Where people get hung up on that problem is that they are now looking at it as a new problem/probability, rather than a conditional one. If a 3rd person came into the room and saw two people, two closed doors and one open door and was asked to choose a door, they would have a 50% chance to pick the right door. But this person does not have all the information that I do as the host, or that you do as the contestant knowing the rules of the game.
Imagine we were playing a game of blackjack with one deck. In the first hand we see three of the four aces get dealt out. On the second hand you get dealt a face card first. What are the chances you'll get dealt the last ace and make a blackjack? Much lower.
Now a new player sits down at the table for the 2nd hand (having not seen the cards from the first hand). He is dealt a face card. What does he think his chances of getting an ace to make a blackjack are?
The Monty Hall Problem is easily googled. If there's dissension among the ranks of mathematicians, it should show up there. Marilyn Vos Savant wrote about this in Parade Magazine circ 1990. And there was quite a furor over it.
Fwiw I've never flipped a coin and had it land on its edge. But, one time I was tossing a key back on to its storage shelf, and it did land on its edge. The only time it's ever happened to me.
You know.....The difference between 1/3 chance of winning vs. 2/3 chance of winning is fairly significant, and should reveal itself quite quickly in any real-world test.
Get a friend to play Monty Hall, 3 plastic cups, and a Lincoln cent. Do the random picking, have Monty reveal an empty cup between his two, and keep track of where the cent ends up, (your cup, or the remaining Monty cup).
@tradedollarnut said:
There is absolutely no way that choosing one blind door out of two leads to a 2/3 chance of being correct. Sorry
The falsehood is in not adjusting the odds of the initial door to 50:50 once the goat is revealed. It's like Schrödinger's cat - once you open the box to reveal the goat you already disturbed the odds.
I could do a fault tree analysis but I'm just not up to it...
You could run a simulation that follows the rules. I'll give you the source code for one. Schrödinger's goat will end up in the curry and not walking home with the contestant 2 times out of 3. The odds of the contestant choosing to switch doors if switching is based on a coin toss are even, but that's not the odds of winning should they do so.
TDN has a lot of company. Here's more from wikipedia:
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.
Sometimes, the best way to get a mental grip on a problem is to take it to a logical extreme. Consider this:
There are 100 boxes. One box contains $1million dollars.
You are asked to choose one box, and give it to me. YOU get the other 99 boxes.
At this point, you are very excited, because you have a 99% chance of being a millionaire!!
I know which box has the $1mil.
I start opening your boxes that do NOT have the $1mil, one at a time.
After a period of time, there are two boxes left. The original box you gave me, (which had a 1% chance of having $1mil), and the last remaining box of 99 that you had.
Which box do you want NOW?
The smart answer in this case is to KEEP the box you have! The only reason that box remains is either because the 1% box had the $1mil....or because of my insider knowledge, YOUR box has the $1mil. Your chance is still 99%....it didn’t change just because I opened the ones I KNEW didn’t contain $1mil.
@TommyType said:
Regarding the “Monty Hall Problem”:
Sometimes, the best way to get a mental grip on a problem is to take it to a logical extreme. Consider this:
There are 100 boxes. One box contains $1million dollars.
You are asked to choose one box, and give it to me. YOU get the other 99 boxes.
At this point, you are very excited, because you have a 99% chance of being a millionaire!!
I know which box has the $1mil.
I start opening your boxes that do NOT have the $1mil, one at a time.
After a period of time, there are two boxes left. The original box you gave me, (which had a 1% chance of having $1mil), and the last remaining box of 99 that you had.
Which box do you want NOW?
The smart answer in this case is to KEEP the box you have! The only reason that box remains is either because the 1% box had the $1mil....or because of my insider knowledge, YOUR box has the $1mil. Your chance is still 99%....it didn’t change just because I opened the ones I KNEW didn’t contain $1mil.
That's dumb. Better to switch boxes as the 98 opened don't have the cash and neither does the one you picked originally. Now, if it is random and no one knows, including the box opener you would be right. That's not the case. The box opener knows which of the 99 boxes that haven't been picked has the million and proceeds to open all the other 98.
@roadrunner said:
The Monty Hall Problem is easily googled. If there's dissension among the ranks of mathematicians, it should show up there. Marilyn Vos Savant wrote about this in Parade Magazine circ 1990. And there was quite a furor over it.
Fwiw I've never flipped a coin and had it land on its edge. But, one time I was tossing a key back on to its storage shelf, and it did land on its edge. The only time it's ever happened to me.
AHA, my earlier comment that not one of us has ever been involved in a coin flip that resulted in it landing on it's edge holds true. Keys, smeys, lol.
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Comments
edit -
Vince Lombardi said there are three things that can happen on an offensive play throwing the ball and two of them are bad.
"Three things can happen when you throw the ball, and two of them are bad."
The probability that 10 coin flips will produce all H or T. That's fixed at 1/1024. Bad
The probability that the string will continue as each coin is flipped. This changes as you go. Bad
The probability that a single flip will give H or T is always 1/2. Good
Okay this has to be one of the strangest things I have heard in awhile "unless they have buttered toast on their backs" omg that's odd.
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I have never seen this happen in my lifetime. Has anyone?
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LOL....It's my Wife's favorite "scientific" experiment she is planning!
"If a cat always lands on it's feet....and buttered toast always lands buttered side down....what would happen if you strapped buttered toast on the back of a cat, and dropped it?"
She's pretty sure the cat/toast complex would float in mid-air.
https://youtu.be/hI-j9TItxtU
He was likely talking about in person and not a video. Then again, I can't remember the last time I flipped a coin and had it land on a flat surface.
https://en.m.wikipedia.org/wiki/Gambler%27s_fallacy
Science suggests the cat will only float if the cat forgets it's falling in which case it would be both dead and alive.
Maybe the coin being used is a uniface piece.
It would seem that logic applied to luck/randomness circumstances can be quite the mind bender. Many years ago we were at the horse race track (harness) and the #1 horse won the first 9 races out of a total of 12 races. After each #1 horse winning most said it wouldn't happen again. Finally on race 10 the streak ended. What a parlay that could of been.
Anyone remember Jeff da flipper?
This is going back a long time.
probability 10 flips produce all heads is 1/1024 all tails is also 1/1024 so all heads or all tails would be 1/512.
A coin can stand on it's edge...
...in the Twilight Zone.
(The Twilight Zone episode "A Penny for Your Thoughts")
The Mysterious Egyptian Magic Coin
Coins in Movies
Coins on Television
Here's a classic:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
It is simple fact that each flip (honest coin, honest flip) has a 50/50 chance of being either heads or tails. It does not matter what happened in previous flips. Barring the extremely rare occurrence of an edge landing (it has happened), the odds are even. What most are thinking of, when considering the original premise, is 'probability'.. The 50/50 still applies to each flip, but probability changes. Cheers, RickO
Why did I waste 10 minutes reading this thread? I guess maybe I landed on my head one too many times!
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.
It's got to be yes as you increase your odds from 1 in 3 to 1 in 2 (the host knows the correct door).
Yes is correct but the odds are 1/3 for staying and 2/3 for switching. This is known as the Monty Hall problem.
Common sense dictates it'd be AU, if it started out MS, for the tenth time.
``https://ebay.us/m/KxolR5
Uhhhmmm...no. Once you know the goat is gone, your odds are 1/2 for doing either
Well, if the person tossing the coin floats, and a duck floats, and a duck weighs the same as wood, then the coin tosser is a witch!
Keeper of the VAM Catalog • Professional Coin Imaging • Prime Number Set • World Coins in Early America • British Trade Dollars • Variety Attribution
The assumption is that the host knows what is behind the doors and he will only open a non-prize door. I'm not going to try to convince you here as you can Google the problem yourself.
Quick logic sometimes is wrong when you play in the probability realm.
As noted, it is a well known problem covered in any probability class....and as much as I tried to fight it using the same logic you did, the answer is your probability increases if you switch!!! It all has to do with "what you know, and when you know it".
I'll try to find the answer why, if I have time.
See if this makes sense:
When you make your original pick, you obviously have 1/3 chance of picking the "big prize". And there is a 2/3 chance the "big prize" is behind the other two doors. (Good so far).
But when Monty reveals the "goat" behind one of the two unpicked doors, (and we KNOW he's not going to reveal the "big prize"), the probability of the "big prize" being behind the two unpicked doors doesn't change. The probability is just assigned to the ONE door that remains of those two. Your door is still 1/3, but the remaining unpicked door is now 2/3!
So, the act of Monty removing ONLY a "goat prize", actually doubles your chance of winning by SWITCHING doors.
(On the other hand, if we assume Monty was only randomly picking doors as well, the probabilities become 1/2, and 1/2, when he reveals the goat.....but if he reveals the "Big Prize" at random, your probability changes to 0/2 and 0/2, and you CAN'T win.)
It's all about what you know, and when you know it....And Monty's prior knowledge of where the prize is has a huge influence.
Given everything is equal and the coin is a regular, un-tampered coin, it can flip 1000 times or more and land on the same side every time. theoretically, it can land on the same side until the cows come home. and then the x-flip will change it and it lands 10 times or one time on the other side and then reverses again.
as I said, everything being equal. so why does it do that. I believe it is gravity , how it falls, how the high points of the coin are attracted to gravity, if perhaps someone disturbed the airflow around it.. it all matters.
There is absolutely NO guarantee. However, with every flip the unlimited possibility increases by 1 in favor of the other side.
"The probability that a single flip will give H or T is always 1/2" is not exactly true, but the closest you can come to. It is actually 1/3. you forgot the possibility of it landing on the rim and staying there. However I would not bet on it.
Without those 10 minutes you would not have read the cat and buttered toast theory:
"LOL....It's my Wife's favorite "scientific" experiment she is planning!
"If a cat always lands on it's feet....and buttered toast always lands buttered side down....what would happen if you strapped buttered toast on the back of a cat, and dropped it?"
She's pretty sure the cat/toast complex would float in mid-air.
"
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So much poor understanding of probability and math in this thread
Haven't read other responses yet, but if it's a fair coin, 10th flip ( and all flips) are 50:50, regardless of past results.
However, if i flip 9 heads in a row, I'd suspect coin is not fair (is loaded ) and would expect another head on next flip.
Liberty: Parent of Science & Industry
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.
There was some scenario in the movie 21, with Kevin Spacey, where he asked a student to pick which door of three held some prize, and what the odds were he chose the correct door. Obviously the beginning odds are roughly 33%. Spacey's character reveals what's behind one door and there's nothing. So the prize is still behind one of the two remaining doors, one of which the student chose. Before opening another door the teacher asks the student what his odds are now that he chose the door with the prize behind it. One would assume it's now 50%. But for some reason it wasn't. The odds were actually greater than 50%. The teacher asked why and the student correctly explained, but it still didn't make sense to me.
Anyone remember that scene?
See Ikeigwin's post directly above yours. Same problem, with explanation.
Why do I have a craving for rogan josh right now?
Keeper of the VAM Catalog • Professional Coin Imaging • Prime Number Set • World Coins in Early America • British Trade Dollars • Variety Attribution
So how close to random does a flip have to be in order to be truly random? If you flip it with your thumb and first finger to varying heights and it drops down onto the same table top is that random enough? The more uniform you try to make the flip, the more likely that you are approaching a situation where a uniform outcome results. I would think that the H/T differences for a Lincoln cent would be inconsequential but one could always get a planchet and mark each side with a magic marker.
I stated earlier that AFAIK there are no known viable odds of a flipped coin landing on its edge and staying there, so for practical purposes it can be ignored.
Do you think that the device used to pick the white balls for PowerBall does so in a manner that is truly random or is as random as is practical to achieve?
There is absolutely no way that choosing one blind door out of two leads to a 2/3 chance of being correct. Sorry
The falsehood is in not adjusting the odds of the initial door to 50:50 once the goat is revealed. It's like Schrödinger's cat - once you open the box to reveal the goat you already disturbed the odds.
I could do a fault tree analysis but I'm just not up to it...
The premise is that the two unchosen blind doors have a 2/3 chance of holding the car. The falsehood is stating that once the goat is revealed from those two that the other door has the entire 2/3 odds still. Those odds included the 1/3 for the door now revealed.
It's a nice parlor trick with math but that's all it is.
Believe me, I understand your issues. I made the same argument when I first heard the problem. But this is a classic problem that you'll see in any probability course, and it's selected exactly BECAUSE it doesn't seem reasonable. (And, it introduces the concept of prior knowledge).
Read closely the explanation, (with all possible situations), laid out in lkeigwin's post. The key is that Monty has intentionally removed one goat based on what he knew, but you didn't. As a result, you gain new knowledge.
I feel your pain...honestly.
Imagine you’re a contestant on a game show and given the choice of three doors. Behind one door is a car; behind the other two doors, goats. You pick a door–let’s say No. 1–and the host, who knows what’s behind the doors, opens another door, let’s say No. 3, which has a goat. He then says, “Do you want to change your pick?”
The correct answer, which seems quite counterintuitive, is Yes, you should switch. The odds of the car being behind door No. 3 are 2/3 vs 1/3 for door No. 1.
The reason is that the odds have changed now that we know an additional piece of information. The chance the car is behind door No.1 is still 1/3 but since we can rule out door No. 2, the chance of the car being behind door No. 3 is the combined probability of doors No. 2 and No. 3.
Here’s another way to think about it to drive home the point. Imagine there are 100 doors with 99 goats and 1 car behind them. You pick a door, let’s assume it’s No. 1. The host then opens 98 other doors revealing goats. What are the odds of the car being behind your initial pick, door No. 1, or the remaining unopened door? The answer is you have a 1/100 chance of getting it right with door No. 1 but a 99/100 chance with the remaining door. So you should always switch to the remaining door.
The basic rule of thumb with any conditional probability question is that given new information, you should always change your pick!
TDN, this is a well known problem, but you have to accept the rules of the game for 2/3 to be the correct odds after switching. The choice of the door reveal is not random so the original odds of your 1/3 guess do not change. Just for fun, how much are you willing to bet that you are correct in this discussion if I offer you even odds?
TDN, it's not a parlor trick. It's a word problem.
The key is that the host KNOWS what is behind each door. He will never show the prize behind a door he opens. His choice of door is not by chance.
Look at it this way:
You are the contestant and get to choose a door. I am the host and I get both doors. Off the bat your chance to win is 1/3 and mine is 2/3. I peak behind my two doors and then open one of the losing doors for you to see. Has that act magically changed your odds or my odds? Nope.
Where people get hung up on that problem is that they are now looking at it as a new problem/probability, rather than a conditional one. If a 3rd person came into the room and saw two people, two closed doors and one open door and was asked to choose a door, they would have a 50% chance to pick the right door. But this person does not have all the information that I do as the host, or that you do as the contestant knowing the rules of the game.
Imagine we were playing a game of blackjack with one deck. In the first hand we see three of the four aces get dealt out. On the second hand you get dealt a face card first. What are the chances you'll get dealt the last ace and make a blackjack? Much lower.
Now a new player sits down at the table for the 2nd hand (having not seen the cards from the first hand). He is dealt a face card. What does he think his chances of getting an ace to make a blackjack are?
The Monty Hall Problem is easily googled. If there's dissension among the ranks of mathematicians, it should show up there. Marilyn Vos Savant wrote about this in Parade Magazine circ 1990. And there was quite a furor over it.
https://en.wikipedia.org/wiki/Monty_Hall_problem ........... plenty of info there.
Fwiw I've never flipped a coin and had it land on its edge. But, one time I was tossing a key back on to its storage shelf, and it did land on its edge. The only time it's ever happened to me.
You know.....The difference between 1/3 chance of winning vs. 2/3 chance of winning is fairly significant, and should reveal itself quite quickly in any real-world test.
Get a friend to play Monty Hall, 3 plastic cups, and a Lincoln cent. Do the random picking, have Monty reveal an empty cup between his two, and keep track of where the cent ends up, (your cup, or the remaining Monty cup).
Believe you will see it is correct.
You could run a simulation that follows the rules. I'll give you the source code for one. Schrödinger's goat will end up in the curry and not walking home with the contestant 2 times out of 3. The odds of the contestant choosing to switch doors if switching is based on a coin toss are even, but that's not the odds of winning should they do so.
Keeper of the VAM Catalog • Professional Coin Imaging • Prime Number Set • World Coins in Early America • British Trade Dollars • Variety Attribution
I guess I didn't kill the tangent. LOL.
TDN has a lot of company. Here's more from wikipedia:
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999).
The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.
Regarding the “Monty Hall Problem”:
Sometimes, the best way to get a mental grip on a problem is to take it to a logical extreme. Consider this:
The smart answer in this case is to KEEP the box you have! The only reason that box remains is either because the 1% box had the $1mil....or because of my insider knowledge, YOUR box has the $1mil. Your chance is still 99%....it didn’t change just because I opened the ones I KNEW didn’t contain $1mil.
That's dumb. Better to switch boxes as the 98 opened don't have the cash and neither does the one you picked originally. Now, if it is random and no one knows, including the box opener you would be right. That's not the case. The box opener knows which of the 99 boxes that haven't been picked has the million and proceeds to open all the other 98.
AHA, my earlier comment that not one of us has ever been involved in a coin flip that resulted in it landing on it's edge holds true. Keys, smeys, lol.
Successful card BST transactions with cbcnow, brogurt, gstarling, Bravesfan 007, and rajah 424.
abcde123456: I don't think you understand what you think you understand. Read it again.
Ahhhhh! Sorry!! I had you figured at the opposite of what you really stated. You are correct dear sir!