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Maths riddle - where has the penny gone
Someone at uni told me this today. Funny but can drive you loopy. 
3 friends go for a meal, the meal comes to 30p. They each pay 10p. The waiter then says the meal actually costs 25p and gives them each a penny back and keeps the extra 2p for himself.
Each of the 3 people have paid 9p. The waiter has kept 2p.
3 x 9p + 2p = 29p
Where has the extra penny gone?
3 friends go for a meal, the meal comes to 30p. They each pay 10p. The waiter then says the meal actually costs 25p and gives them each a penny back and keeps the extra 2p for himself.
Each of the 3 people have paid 9p. The waiter has kept 2p.
3 x 9p + 2p = 29p
Where has the extra penny gone?
Scroll to see replies
the friends have 1p each, the till has 25p in, the waiter has 2p
(1x3)+25+2 = 30.
in the sum you posted, you are counting the money which is not in the possession of the friends, and that which is in the possession of the waiter. tha's just rubbish, since some which the friends do not have, the waiter has, not the till, and some money in the till is not counted.
(1x3)+25+2 = 30.
in the sum you posted, you are counting the money which is not in the possession of the friends, and that which is in the possession of the waiter. tha's just rubbish, since some which the friends do not have, the waiter has, not the till, and some money in the till is not counted.
manps
Someone at uni told me this today. Funny but can drive you loopy.
3 friends go for a meal, the meal comes to 30p. They each pay 10p. The waiter then says the meal actually costs 25p and gives them each a penny back and keeps the extra 2p for himself.
Each of the 3 people have paid 9p. The waiter has kept 2p.
3 x 9p + 2p = 29p
Where has the extra penny gone?
3 friends go for a meal, the meal comes to 30p. They each pay 10p. The waiter then says the meal actually costs 25p and gives them each a penny back and keeps the extra 2p for himself.
Each of the 3 people have paid 9p. The waiter has kept 2p.
3 x 9p + 2p = 29p
Where has the extra penny gone?
I used to use this years ago to teach numbers in Secondary School French!
Trois copains mangent dans un petit restaurant. Ils paient....
The listener is so mesmerised by the fact that he has to do mental arithmetic, that he takes the implied logic for granted! Once you add the linguistic complication, even the brightest have no chance!
Aitch
More importantly...where can 3 people eat for 30p? lol
Ok then, here's another conundrum/mathematical fallacy, this time with probability.
You and a friend have one envelope each, one with £5 in it, the other with £10, but they're sealed and you don't know which is which. Would you profit by swapping with your friend? Well, half the time you'd gain £5, and half the time lose £5, so on average no profit is to be made by swapping envelopes.
Suppose now that you don't know the quantities in the envelopes but do know that there's twice in one what there is in the other. Call the amount in your envelope £x. Half the time then there is £2x in the other envelope, and swapping would gain you £x; half the time there would be £(x/2) in the other envelope and swapping would lose you (£x/2). So the average gain from swapping is 1/2(£x - £(x/2)) = £(x/4) - a clear profit! Reapply the above argument and swap again to make still more money!
Why is this all BS?
You and a friend have one envelope each, one with £5 in it, the other with £10, but they're sealed and you don't know which is which. Would you profit by swapping with your friend? Well, half the time you'd gain £5, and half the time lose £5, so on average no profit is to be made by swapping envelopes.
Suppose now that you don't know the quantities in the envelopes but do know that there's twice in one what there is in the other. Call the amount in your envelope £x. Half the time then there is £2x in the other envelope, and swapping would gain you £x; half the time there would be £(x/2) in the other envelope and swapping would lose you (£x/2). So the average gain from swapping is 1/2(£x - £(x/2)) = £(x/4) - a clear profit! Reapply the above argument and swap again to make still more money!
Why is this all BS?
This "Monty Hall" problem is quite famous so I'm guessing someone is going to answer it correctly straight away but here goes (I'm paraphrasing it by the way.)
You are on a gameshow - the presenter gives shows you 3 boxes. Inside 1 of these is the prize. You pick one of the boxes; the presenter then opens another box, revealing it to be empty. You are then given the choice to stick with your original box or switch to another before the location of the prize is revealed by the presenter.
Do you stick or switch to the other, unopened box, or do you not care which is chosen? Most importantly, why? Give an explanation.
You are on a gameshow - the presenter gives shows you 3 boxes. Inside 1 of these is the prize. You pick one of the boxes; the presenter then opens another box, revealing it to be empty. You are then given the choice to stick with your original box or switch to another before the location of the prize is revealed by the presenter.
Do you stick or switch to the other, unopened box, or do you not care which is chosen? Most importantly, why? Give an explanation.
SsEe
The value of x doubles from one half of the argument to the next. Then you put them into the same sum to get a nonsense profit.
Half the time swapping would gain you £x; half the time swapping would lose you £x. So the average gain from swapping is 1/2(£x - £x) = £0
Half the time swapping would gain you £x; half the time swapping would lose you £x. So the average gain from swapping is 1/2(£x - £x) = £0
All I'm denoting by x is the amount of money in my envelope. That doesn't change, though I grant you it's a random variable.
JohnC
This "Monty Hall" problem is quite famous so I'm guessing someone is going to answer it correctly straight away but here goes (I'm paraphrasing it by the way.)
You are on a gameshow - the presenter gives shows you 3 boxes. Inside 1 of these is the prize. You pick one of the boxes; the presenter then opens another box, revealing it to be empty. You are then given the choice to stick with your original box or switch to another before the location of the prize is revealed by the presenter.
Do you stick or switch to the other, unopened box, or do you not care which is chosen? Most importantly, why? Give an explanation.
You are on a gameshow - the presenter gives shows you 3 boxes. Inside 1 of these is the prize. You pick one of the boxes; the presenter then opens another box, revealing it to be empty. You are then given the choice to stick with your original box or switch to another before the location of the prize is revealed by the presenter.
Do you stick or switch to the other, unopened box, or do you not care which is chosen? Most importantly, why? Give an explanation.
Yes, you should swap. You have a 1 in 3 chance of choosing the right box initially, and 2 in 3 of getting it wrong. If you picked right and he opens an empty box, you'll lose if you swap as you'll get the other empty one, but if you picked wrong and he opens an empty box, you'll get the prize if you swap. Since you're twice as likely to initially choose wrongly, you have a 2/3 chance of winning if you swap.
Nima
I don't get that problem at all. He takes 1 away, so the prize is in one of the 2 boxes. You're choosing one box by sticking or switching, so your chance is 1 in 2 either way, surely?
You'd have thought so, but it's not - because he's always removing an empty box, regardless of what you've chosen. So if you've chosen an empty one, you'll always win be switching and only if you chose the prize initially will you lose by switching, so you actually have a 2/3 chance of winning if you switch.
Nima
I don't get that problem at all. He takes 1 away, so the prize is in one of the 2 boxes. You're choosing one box by sticking or switching, so your chance is 1 in 2 either way, surely?
it's all to do with what happened before he took one box away. There are loads of ways of explaining it, some more mathematical than others, but the general gist is that you have a 2/3 probability of winning if you switch, and a 1/3 if you don't. If you want to find some explanations I suggest just googling it, there are loads around, and i would probably not expalain them very well
RichE
Ok then, here's another conundrum/mathematical fallacy, this time with probability.
You and a friend have one envelope each, one with £5 in it, the other with £10, but they're sealed and you don't know which is which. Would you profit by swapping with your friend? Well, half the time you'd gain £5, and half the time lose £5, so on average no profit is to be made by swapping envelopes.
Suppose now that you don't know the quantities in the envelopes but do know that there's twice in one what there is in the other. Call the amount in your envelope £x. Half the time then there is £2x in the other envelope, and swapping would gain you £x; half the time there would be £(x/2) in the other envelope and swapping would lose you (£x/2). So the average gain from swapping is 1/2(£x - £(x/2)) = £(x/4) - a clear profit! Reapply the above argument and swap again to make still more money!
Why is this all BS?
You and a friend have one envelope each, one with £5 in it, the other with £10, but they're sealed and you don't know which is which. Would you profit by swapping with your friend? Well, half the time you'd gain £5, and half the time lose £5, so on average no profit is to be made by swapping envelopes.
Suppose now that you don't know the quantities in the envelopes but do know that there's twice in one what there is in the other. Call the amount in your envelope £x. Half the time then there is £2x in the other envelope, and swapping would gain you £x; half the time there would be £(x/2) in the other envelope and swapping would lose you (£x/2). So the average gain from swapping is 1/2(£x - £(x/2)) = £(x/4) - a clear profit! Reapply the above argument and swap again to make still more money!
Why is this all BS?
because if there is twice in one what there is in the other, then ...
one is 2x, one is x
or one is x, one is x/2
you cannot have 2x and x/2!
http://math.ucsd.edu/~crypto/Monty/montybg.html
http://www.cut-the-knot.org/hall.shtml
Option1:
You choose a door with the prize.
Option1a:
You stick - you win.
Option1b:
You change - you lose.
Option2:
You choose the door without the prize.
Option2a:
You stick - you lose.
Option2b:
You change - you win.
Option3:
You choose the door without the prize.
Option3a:
You stick - you lose.
Option3b:
You change - you win.
Changing wins 2/3 times, sticking only 1/3.
http://www.cut-the-knot.org/hall.shtml
Option1:
You choose a door with the prize.
Option1a:
You stick - you win.
Option1b:
You change - you lose.
Option2:
You choose the door without the prize.
Option2a:
You stick - you lose.
Option2b:
You change - you win.
Option3:
You choose the door without the prize.
Option3a:
You stick - you lose.
Option3b:
You change - you win.
Changing wins 2/3 times, sticking only 1/3.
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