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TomasB Inner circle Sweden 1144 Posts |
A card shop owner opens some cases of mixed decks and divide them into two piles. 30 Red decks in one pile and 30 Blue decks in the other. The Blue pile he marks "3 for a dollar" and the Red one he marks "2 for a dollar".
At the end of the day he sees that he has sold them all and gotten 25 dollars in the register. The next day he again opens some boxes that has 30 Red and 30 Blue mixed, but he doesn't have the energy to divide them so he figures that putting them all in one pile and selling them with the sign "5 for 2 dollars" will be the same thing but save him some work. At the end of the day all cards are sold but in the register he only finds 24 dollars. Where is the missing dollar? Could this be the dollar Stan found in http://www.themagiccafe.com/forums/viewt......orum=101 ? /Tomas |
dlhoyt Regular user 176 Posts |
He should have sold them with the sign "12 for 5 dollars".
That is the integer no. of decks closest to the geometric mean of the previous days sale. (The geometric mean, instead of the arithmetic mean, is the appropriate one to use when averaging rates. The rates in this case are "decks per dollar". The geometric mean of 3/$1 and 2/$1 is square root of 3*2 = 2.4 decks per dollar.) The missing dollar came from his incorrect calculation. |
Steve Martin Inner circle 1119 Posts |
In the individual sales, the three-at-a-time pile runs out first, and after that, there are still five pairs (10 decks) from the other pile to sell.
When we sell them all together, we are only allowed to sell those remaining 10 decks in two sets of five (for a total of only 4 dollars) rather than in five pairs (for a total of 5 dollars). I do not know the mathematical treatment that explains it in general (for any combination of numbers). I didn't understand the "geometric mean" in the previous post. It's a great puzzle. At first glance it seems crazy, as in both cases you appear to be selling the same number of decks, for the same price, for a different total amount of money.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
TomasB Inner circle Sweden 1144 Posts |
You got it Steve!
/Tomas |
dlhoyt Regular user 176 Posts |
I've already confessed to Tomas that the Geometric mean is not the correct solution. Charge it up to a poor memory. The correct solution is to calculate the Harmonic mean of the sales prices to determine how much you should sell the single pile for. It is the Harmonic mean that is appropriate for averaging rates and ratios, not the Geometric mean.
For those who don't know these other types of averages: the harmonic mean, H, can be calculated by the following: 1/H = (Sum(1/x's))/n , where n is the number of values being averaged, and x's is my attempt at writing x subscript i, which is just saying that you should add up the reciprocal of all the values. Applied to Tomas' problem the solution is: 1/H = ((1/3) + (1/2)) / 2 = (2/6 + 3/6)/2 = (5/6)/2 = 5/12 Therefore, H = 12/5, which is 12 decks for $5. It's a coincidence that the geometric mean gives the same answer for this set of values. The other way to calculate the correct selling price is to calculate the number of decks sold per dollar from the total sales: 60 decks sold for $25, so 60/25 = 2.4 decks per $1. Now the store owner has to sell the mixed colors at that price to make the same amount of money. Since he can't sell .4 decks he converts it to the closest integer amount by multiplying both numerator and denominator by 5: 5*2.4 = 12 deck and 5*$1 = $5. So is he sells a dozen decks for $5 he won't lose any money. Thanks to Tomas for the problem and for pointing out my error. |
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