Suppose that I have two urns, with one magnetic dollar in each urn. I begin randomly throwing more magnetic dollars in the general vicinity of the urns, and the coins fall in the urns by a simple rule:

Let's say that Urn A has \(x\) coins, and Urn B has \(y\) coins. The probability that the next coin falls into Urn A is \(\dfrac{x}{x+y}\), and the probability that the next coin falls into Urn B is \(\dfrac{y}{x+y}\).

You keep throwing magnetic dollars until there is a total of \(1,000,000\) magnetic dollars in total.

What would you bet would be the price on average, of the urn with the smaller amount of coins?

\(\$ 10\), perhaps? Maybe \(\$ 100\) or \(\$ 500\)? Post your bet or write it down on a piece of paper before looking at the next section.

The less-priced urn, on average, is actually worth a grand total of a quarter of a million dollars! Don't worry if you guessed wrong; many professional mathematicians also guessed much lower than this. In fact, when a group of mathematicians were asked this question and were asked to bet, most people only bet \(\$ 10\) and only one person bet over \(\$ 100\).

But why does the lower-priced urn price so high? You may want to try the problem out yourself before I go over a very nice and elegant solution. See if you can find it!

Tried it out yet? In the case that you have, let's see how this problem can be so elegantly solved, as I claimed.

Suppose that you have a deck of cards; one red, and \(999,998\) white. Currently, you just have a red card. Now every turn, you place a white card in any available slot. For example, in the first move, you have \(2\) available slots: one above the red card, and one below. In the second move, you have \(3\) available slots, and so on.

But wait! Let's say that the empty slots above the red card are the magnetic dollars in Urn A, and the empty slots below the red card are the magnetic dollars in Urn B. Notice that if you had \(x\) empty slots above the red card and \(y\) empty slots below the red card, then the probability that the next card will be above the red card is \(\dfrac{x}{x+y}\), and the probability that the next card will be below the red card is \(\dfrac{y}{x+y}\)! We've found a one-to-one correspondence between the original magnetic dollar problem and this new card problem!

Finally, we know that in a random placements of white cards in this fashion will result in a uniform distribution of where the red card is, every single final position is of equal probability. This means that in the original problem, the probability of \(42\) magnetic dollars being in Urn A is the same as the probability of \(314,159\) magnetic dollars in Urn A. Therefore, the average price of the lower priced urn is clearly \(250,000\). \(\Box\)

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## Comments

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TopNewestI know some of you have been waiting for me to post for a long time. Sorry for the inactivity, I had a bunch of homework and stuff.

Well, here is my next #CosinesGroup posts. The problem in my opinion is a very cool problem because of the awesome elegant solution.

This problem has also been called "Polya's Urns", but I looked that up and "Polya's Urns" actually covers a lot of random complicated stuff, so I kept it at magnetic dollars.

Hope you enjoy! Feedback is appreciated.

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The result is certain not intuitive, nor obvious.

You can look at this problem John's Red And Blue Balls, which is based off the same idea.

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I'm curious to know why only one mathematician guessed over $100. I personally guessed around $150,000 becuase while the probability might be pretty low at some point, in A MILLION tries, it would be bound to happen quite a bit.

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Well, their reasoning would probably be then assuming a sort of "reverse" normal distribution curve. The average of all the games would probably be pretty low.

They were asked to bet immediately, depriving them of the chance to think it over. Their first instinct told them that the more coins that get in a urn, the higher probability of getting more coins in, giving a feedback loop. This means that the smaller urn probably won't really have any coins at all.

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How do you do this?

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This is the solution in Peter Winkler ' Mathematical Puzzles. An excellent book.

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