Bill V
Well-Known Member
Sorry grumpy, that's incorrect.
Anyone else?
Anyone else?
CAN YOU FIGURE OUT THIS RIDDLE?????
- Thread starter Peter Vincent
- Start date
Help Support Ford SportTrac Forum:
Jake Levin
Well-Known Member
Bill, why did I pick a curtain instead of a door???
Bill V
Well-Known Member
Sorry, my mistake. 
I was crossing up two different game show formats in my head. I've corrected it--if you ignored that inconsistency, the problem still remains valid.
I was crossing up two different game show formats in my head. I've corrected it--if you ignored that inconsistency, the problem still remains valid.Jeff C
Well-Known Member
Your probability of losing on your first guess is 2/3. You only win if your first guess was wrong and you switch. Since you're most likely to have gotten the first guess wrong, you should switch, in which case, your odds are 2/3 that you win.
Bill V
Well-Known Member
Ding ding ding! We have a winner! You have a 2-in-3 chance of winning if you switch. You have a 1-in-3 chance of winning if you don't.
If your original choice was the door with the money (1-in-3 chance), and you stick with it throughout, those 1-in-3 odds never change.
If your original choice was the door without the money (2-in-3 chance), then the money is behind one of the other two doors. If you change, you effectively get both of those doors to be yours, and thus have a 2-in-3 chance of winning.
Don't believe it? Then try it yourself! Get three playing cards--two kings and an ace. Have a friend shuffle those cards and place them face down in front of you--and have that friend know which one is the ace. Then you pick a card. Your friend then turns over one of the two cards--he HAS to always turn over a card that he KNOWS is a king. Then decide if you want to switch or not.
Play it many times--20 to 30 should be good--where you decide to stay with your original card. You'll discover that you end up with the ace about one-third of the time. Play it anotehr 20 to 30 times where you decide to switch to the other card. You'll discover that you end up with the ace about two-thirds of the time.
If your original choice was the door with the money (1-in-3 chance), and you stick with it throughout, those 1-in-3 odds never change.
If your original choice was the door without the money (2-in-3 chance), then the money is behind one of the other two doors. If you change, you effectively get both of those doors to be yours, and thus have a 2-in-3 chance of winning.
Don't believe it? Then try it yourself! Get three playing cards--two kings and an ace. Have a friend shuffle those cards and place them face down in front of you--and have that friend know which one is the ace. Then you pick a card. Your friend then turns over one of the two cards--he HAS to always turn over a card that he KNOWS is a king. Then decide if you want to switch or not.
Play it many times--20 to 30 should be good--where you decide to stay with your original card. You'll discover that you end up with the ace about one-third of the time. Play it anotehr 20 to 30 times where you decide to switch to the other card. You'll discover that you end up with the ace about two-thirds of the time.
Last edited by a moderator:
Daeron is correct.
This riddle is amost as old as I am perhaps even older, because I heard it in grade school. except then it had 3 guys renting the room. In either case the story includes the incorrect method of solving the problem and thus the mystery of the other dollar is not a mystery.
They paid $30, and later got a $5 refund. That means each paid $12.50. They each got $1 back ($2) and the bellhop kept $3 for himself. Deducting the $2 refund from their bill is where they are forcing you to use incorrect logic, although it sounds like it should be correct.
Years ago it was 3 guys pay $30 for a room. The innkeeper gives the bellhop $5 to return. The bellhop thinks $5 will be hard to divide so he gives each guy $1 and keeps the remaining $2 for himself
if the guys paid $30, that's $10 each. If the bellhop gave them each $1 back, that means they paid $9 each. 3 X 9 =27 plus the $2 the bellhop kept is only $29. What happened to the other $1 ??? Exactly the same thing. The riddle includes a solution that is mathmaticaly correct but logicaly incorrect.
...Rich
This riddle is amost as old as I am perhaps even older, because I heard it in grade school. except then it had 3 guys renting the room. In either case the story includes the incorrect method of solving the problem and thus the mystery of the other dollar is not a mystery.
They paid $30, and later got a $5 refund. That means each paid $12.50. They each got $1 back ($2) and the bellhop kept $3 for himself. Deducting the $2 refund from their bill is where they are forcing you to use incorrect logic, although it sounds like it should be correct.
Years ago it was 3 guys pay $30 for a room. The innkeeper gives the bellhop $5 to return. The bellhop thinks $5 will be hard to divide so he gives each guy $1 and keeps the remaining $2 for himself
if the guys paid $30, that's $10 each. If the bellhop gave them each $1 back, that means they paid $9 each. 3 X 9 =27 plus the $2 the bellhop kept is only $29. What happened to the other $1 ??? Exactly the same thing. The riddle includes a solution that is mathmaticaly correct but logicaly incorrect.
...Rich
Bill V
Well-Known Member
Sorry Rich, but Daeron is not correct.
He claims that each of the two guys (in the version of the problem that started this thread) paid $13.50, for a total of $27. But that's not true--it's based on the same flawed math/logic that resulted in the original problem's $31 total.
They each paid $15, they each got $1 back, therefore they are each out $14; combined they are out $28. Of that $28, the innkeeper has $25, the bellhop has the other $3. End of story.
He claims that each of the two guys (in the version of the problem that started this thread) paid $13.50, for a total of $27. But that's not true--it's based on the same flawed math/logic that resulted in the original problem's $31 total.
They each paid $15, they each got $1 back, therefore they are each out $14; combined they are out $28. Of that $28, the innkeeper has $25, the bellhop has the other $3. End of story.
Robert Sayre
Active Member
You don't add 14 and 14 AND 3, that's meaningless. You add 14 and 14 to get 28, and then SUBTRACT 3 to get 25, the true cost of the room. The bellhop's 3 bucks is part of the two 14's. The true cost of the rooms was 12.50 each, it's 14 bucks because the bellhop stole 3 dollars.
--Bob
--Bob
Thomas Rogers
Well-Known Member
I heard a similar riddle, and that one and the one posed trick you into thinking there is a "footing" problem, but the reality is the math presented isn't accurate.
The way I heard it was that there were THREE guys, the room was $30, they each paid $10, and then the clerk realized his overcharge of $5 and tried to split the $5 between the men. The clerk gave them back each $1 since it split easily. The rub: Each man paid $9, the clerk kept $2. And as we know, $9 x 3 = $27, add the $2 and you get $29....Where did the other dollar go?
The way I heard it was that there were THREE guys, the room was $30, they each paid $10, and then the clerk realized his overcharge of $5 and tried to split the $5 between the men. The clerk gave them back each $1 since it split easily. The rub: Each man paid $9, the clerk kept $2. And as we know, $9 x 3 = $27, add the $2 and you get $29....Where did the other dollar go?
TJR,
That's what I just said !! When I heard it about 50 years ago it was three guys and $30.
Bill V.
I may have misread Daeron's post but in the end $13.50 each is what they paid. He is also correct that the problem presents a mathmatical illogical solution that leads the into thinking there is a descrepancy of $1.
In the end, they guys only paid $27 ($13.50 each) for the room and the bellhop pocketed $3 which totals $30. You cannot deduct the $2 refund from what they each paid since you can only deduct what the bellhop kept which was $3
...Rich
That's what I just said !! When I heard it about 50 years ago it was three guys and $30.
Bill V.
I may have misread Daeron's post but in the end $13.50 each is what they paid. He is also correct that the problem presents a mathmatical illogical solution that leads the into thinking there is a descrepancy of $1.
In the end, they guys only paid $27 ($13.50 each) for the room and the bellhop pocketed $3 which totals $30. You cannot deduct the $2 refund from what they each paid since you can only deduct what the bellhop kept which was $3
...Rich
Bill V
Well-Known Member
RichardL, no, they didn't pay $13.50--you're still not getting the math right, nor is Daeron.
Each man paid $15. Each man got $1 back. Therefore, each man paid $14, not $13.50. End of story.
Taking $30 and subtracting $3 does not give you the amount paid. It only gives you the total of the amount that ended up in the hands of the hotel ($25) and the amount that ended up in the hands of the guests ($2), which is rather meaningless. And there is no way that you can include that $2 in the amount "paid", as it ended up back in the hands of the guests. The amount paid is simply the original amount paid ($30) minus the amount refunded ($2), which is $28, or $14 a piece--not $13.50.
Here's another way of looking at it--pretend you're one of the hotel guests. You paid $15, under the belief that the room was $30. Later you got $1 back. You have no idea that the room was actually only $25--that was only discussed by the hotel clerk and the bellhop. Similarly, you have no idea that the bellhop pocketed $3. All you know is that you handed over $15, and later you got $1 back. Now, how much have you paid? $13.50? Hell no. You've paid $15, and received $1 back, for a net of $14. End of story.
Each man paid $15. Each man got $1 back. Therefore, each man paid $14, not $13.50. End of story.
Taking $30 and subtracting $3 does not give you the amount paid. It only gives you the total of the amount that ended up in the hands of the hotel ($25) and the amount that ended up in the hands of the guests ($2), which is rather meaningless. And there is no way that you can include that $2 in the amount "paid", as it ended up back in the hands of the guests. The amount paid is simply the original amount paid ($30) minus the amount refunded ($2), which is $28, or $14 a piece--not $13.50.
Here's another way of looking at it--pretend you're one of the hotel guests. You paid $15, under the belief that the room was $30. Later you got $1 back. You have no idea that the room was actually only $25--that was only discussed by the hotel clerk and the bellhop. Similarly, you have no idea that the bellhop pocketed $3. All you know is that you handed over $15, and later you got $1 back. Now, how much have you paid? $13.50? Hell no. You've paid $15, and received $1 back, for a net of $14. End of story.
Chet 02
Well-Known Member
It similar to when I was young and stupid and worked in a shoe store. A customer bought a pair of shoes which with tax came to $30.02. He didn't have any change so he gave me $31.00 and I gave him .98 in change. He then said he had the 2 cents now from the change and gave me 2 pennies in exchange for that extra dollar. .....Yes I fell for it then[Broken External Image]:
Bill V
you are falling for the proposed solution by deducting the $2 they got back is your math error. The $2 has no bearing on the correct calculation which is $30 minus the $3 the bellhop kept is $27 or $13.50 each. Surely the simple fact that following the math presented in the riddle gives you a total of $31 which is impossible, would tell any rational person that there is an error in the solution offered by the riddle.
The riddle relies on the correct sounding logic that if they paid $30 that's $15 each and with a $1 refund each they only paid $14 each. That is misleading and simply incorrect, which is the real hook of the riddle.
The correct way to calculate this riddle is: They paid $30 they were refunded $5, but only got $2 which made the room $27 and the bellhop kept the remaining $3.
How can you say that is incorrect when the math and logic are correct. You are stuck on the riddles premis that they paid $15 each and got a $1 back so that falsely implies that they only paid $14 each which leads you to $31 which you already know is wrong. It is confusing the proper handling of debits and credits with a solution that sounds correct.
...Rich
Remember, the first sign of insanity is doing the same thing over and over again and expecting the results will be different.
you are falling for the proposed solution by deducting the $2 they got back is your math error. The $2 has no bearing on the correct calculation which is $30 minus the $3 the bellhop kept is $27 or $13.50 each. Surely the simple fact that following the math presented in the riddle gives you a total of $31 which is impossible, would tell any rational person that there is an error in the solution offered by the riddle.
The riddle relies on the correct sounding logic that if they paid $30 that's $15 each and with a $1 refund each they only paid $14 each. That is misleading and simply incorrect, which is the real hook of the riddle.
The correct way to calculate this riddle is: They paid $30 they were refunded $5, but only got $2 which made the room $27 and the bellhop kept the remaining $3.
How can you say that is incorrect when the math and logic are correct. You are stuck on the riddles premis that they paid $15 each and got a $1 back so that falsely implies that they only paid $14 each which leads you to $31 which you already know is wrong. It is confusing the proper handling of debits and credits with a solution that sounds correct.
...Rich
Remember, the first sign of insanity is doing the same thing over and over again and expecting the results will be different.
Thomas Rogers
Well-Known Member
What RichardL said. BillV is right in that the two people don't pay $13.50 each, but RichardL is incorrect is saying the room costs $27. It didn't. It cost $25.
The correct way to account for the money is:
$25 for the room, $5 for the refund of which $1 each goes to the two guests and $3 for the bellhop.
As RichardL said, the riddle as posed mixes up credits and debits. Another way to think what the two guests paid is $12.50 each to pay for the room (as we all agree final room cost is $25), and another $2.50 each that then gets applied as follows: $1 back to each of them, and $1.50 to the bellhop.
The mixup is when one takes that $1 back to each of the guys and tries to apply it both before and after in the two different price scenarios.
The guys paid $12.50 for the room, $1.50 to the bellhop, and $1 to themselves.
TJR
The correct way to account for the money is:
$25 for the room, $5 for the refund of which $1 each goes to the two guests and $3 for the bellhop.
As RichardL said, the riddle as posed mixes up credits and debits. Another way to think what the two guests paid is $12.50 each to pay for the room (as we all agree final room cost is $25), and another $2.50 each that then gets applied as follows: $1 back to each of them, and $1.50 to the bellhop.
The mixup is when one takes that $1 back to each of the guys and tries to apply it both before and after in the two different price scenarios.
The guys paid $12.50 for the room, $1.50 to the bellhop, and $1 to themselves.
TJR
Bill V
Well-Known Member
Rich, I'm not falling for anything in the problem--in fact, it's you who can't seem to get your mind past the "tricks" in the problem. Look at what I just quoted from you--you actually said they paid $30, were refunded $2, and that made the room $27. In what state does 30 minus 2 equal 27 ???? (I'm biting my lip to stop from pointing out that math like this is the result of what Governor Bush did to the Texas education system.
j/k!)Calculating the cost of the room to the guests is simply the amount that they are out, which is the amount they first handed over ($30), then subtracting the amount of change they actually received ($2), yielding a net of $28. It's the exact same calculation done every time you order a cheeseburger, give the cashier $3, and get $.50 change. The amount you paid is $3.00 - $.50 = $2.50. In this case, the guests handed over $30, they received $2 back, meaning they paid $28. $25 of that is with the hotel manager, and $3 is with the bellhop--but the guests do not know this. As far as they know (or care), their pockets are $28 lighter than when they checked in, and it's all in the hotel manager's hands.
Agreed--which is why I can't figure out why you keep insisting on doing the math in the same way as is illogically/erroneously proposed in the problem! Subtracting the amount that the bellhop kept from the $30 does not give you the amount that the guests paid--it just gives you a meaningless total of the amount retained by the hotel manager, and the amount returned to the guests as change.
That's just it, Rich, you math and logic are not correct--you still keep trying to claim that $30 - $2 = $27. Let me just ask, Rich--if you're so convinced that the guests paid $27, who has that $27 now? $25 is with the hotel manager. Who has the other $2? Clearly, it's not the bellhop--he has $3, not $2. And clearly, it's not the guests--when you "pay" an amount, that amount does NOT include what you get back in change. So where is it? The answer is, no one has it, because no one paid $27. They paid $28--$25 of which is with the manager, and $3 of which is with the bellhop.
I hope you can finally come to see the blatantly obvious error in your math--but if not, I guess this proves the old saying about being able to fool some of the people all of the time...
Last edited by a moderator:
Bill V
Well-Known Member
Rich, can you tell me what they typo is? Because you're still reiterating the same thing--that they paid $27, which is simply not the case.
The riddle is 100% right in saying that they paid $28--there's no funny math involved with that at all. Where the funny math comes in is when they try to say that the $3 retained by the bellhop can be added to that. This is not the case--that $3 is already included in the $28, and therefore cannot be added in again. But taking the original $30 and subtracting the $3 that went to the bellhop only gives you the total of the amount kept by the hotel ($25) and the amount returned to you ($2), which is a pretty meaningless number. (It could be the answer to the question "How much of the money that changed hands did the bellhop fail to steal?", but that's about it.)
Here's another way of stating my previous analogy--Let's say you go into McDonald's and order a Big Mac. (I'm not sure what Big Macs cost these days, so if the numbers I'm about to give are way off, please ignore this fact.) You give the cashier $3.00. The cashier walks away to get your sandwich. A couple minutes later, a different cashier comes over, opens the drawer, and gives you your change--20 cents--and your Big Mac. How much did you pay for the Big Mac? The answer is simple--$3.00 minus $0.20 = $2.80.
Now, let's say that when the second cashier had the drawer open, without you seeing, he reached into the drawer, pulled out 30 cents, and stuck it in his pocket. Now how much have you paid for your sandwich? The answer is still the same--You handed over $3.00, you got $0.20 back, for a net cost of $2.80. Of that amount, $2.50 is still in the till, and $0.30 is now in the cashier's pocket. Notice that it doesn't matter whether the sandwich was supposed to be $2.80, you got the correct change, and the cashier stole the $0.30 from the till; or if the sandwich was supposed to be $2.50, the till received the correct amount, and the cashier stole the $0.30 from you. The net result is still the same--You are out $2.80, the till is $2.50 richer, and the cashier has $0.30 in his pocket.
If you take all these numbers and multiply them by ten, you now have the problem proposed here--exactly. And you still have the same answers--the hotel guests are out $28--not $27, the hotel has gained $25, the bellhop has stolen $3, and the guests received $2 change. End of story.
The riddle is 100% right in saying that they paid $28--there's no funny math involved with that at all. Where the funny math comes in is when they try to say that the $3 retained by the bellhop can be added to that. This is not the case--that $3 is already included in the $28, and therefore cannot be added in again. But taking the original $30 and subtracting the $3 that went to the bellhop only gives you the total of the amount kept by the hotel ($25) and the amount returned to you ($2), which is a pretty meaningless number. (It could be the answer to the question "How much of the money that changed hands did the bellhop fail to steal?", but that's about it.)
Here's another way of stating my previous analogy--Let's say you go into McDonald's and order a Big Mac. (I'm not sure what Big Macs cost these days, so if the numbers I'm about to give are way off, please ignore this fact.) You give the cashier $3.00. The cashier walks away to get your sandwich. A couple minutes later, a different cashier comes over, opens the drawer, and gives you your change--20 cents--and your Big Mac. How much did you pay for the Big Mac? The answer is simple--$3.00 minus $0.20 = $2.80.
Now, let's say that when the second cashier had the drawer open, without you seeing, he reached into the drawer, pulled out 30 cents, and stuck it in his pocket. Now how much have you paid for your sandwich? The answer is still the same--You handed over $3.00, you got $0.20 back, for a net cost of $2.80. Of that amount, $2.50 is still in the till, and $0.30 is now in the cashier's pocket. Notice that it doesn't matter whether the sandwich was supposed to be $2.80, you got the correct change, and the cashier stole the $0.30 from the till; or if the sandwich was supposed to be $2.50, the till received the correct amount, and the cashier stole the $0.30 from you. The net result is still the same--You are out $2.80, the till is $2.50 richer, and the cashier has $0.30 in his pocket.
If you take all these numbers and multiply them by ten, you now have the problem proposed here--exactly. And you still have the same answers--the hotel guests are out $28--not $27, the hotel has gained $25, the bellhop has stolen $3, and the guests received $2 change. End of story.
