I was extremely busy these past 2 days. Moving flat is never an easy job… And I am sorry for not posting here often… But a nice paradox just appeared in my head today:BERTRAND’S BOX PARADOX.

It is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his “Calcul des probabilités”, published in 1889.

There are three boxes:

1.a box containing two gold coins,

2.a box containing two silver coins,

3.a box containing one gold coin and one silver coin.

After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, what is the probability that the remaining coin is gold? Have a great week.

*User:*1/3 * (1+0+1/2) = 1/2?*Me:*I understand your idea, but this is not exactly the solution-
*User:*2/3 * 1/2 = 1/3, I call it “probabilità composta”*Me:*This is not exactly the solution, but not far away.*User:*Probably we’re thinking about it somewhat “badly”. It looks like a game: you are considering 3 boxes and you must pick the box with the little ball. You choose a box. After that, a man tells you that a box is empty. What do you do, do you change the box to improve you chances to win or not? The answer is “Yes “, because, in that case, you have 2/3 chances to have won. It seems a lot like the same principle.*Me:*I totally love your answer. -
*User:*1/3 or is it (1/3)*1*1+(1/3)*0*0+(1/3)*(1/2)*(1/2)*Me:*I like your second idea. But I think 1/3 is more close to the actual solution. -
*User:*(1/3)/[1/3+1/3*1/2]=2/3 ??*Me:*Good job.

As you could see in the comments, I got some interesting answers and also wonderful solutions. I want to share with you an interesting and easy solution using labels. So, we will label the coins from 1 to 6 as you can see in the image bellow:

The gold faces are 1,2 and 5, all equally likely; if it is 1 or 2, the other side is also gold and if it is 5 the other side is silver. The probability that the other side is gold is ^{2}⁄_{3}.

For the more mathematical solution, the above one is more intuitive: Let random variable B equal the a gold face of the coin (i.e. the probability of a success since the gold side is what we are looking for). Using Kolmogrov’s Axiom of all probabilities having to equal 1, we can conclude that the probability of drawing a silver side is 1 – P(B). Since P(B) = P(1) + P(2) therefore P(B) = ^{1}⁄_{3 }+ ^{1}⁄_{3 }= ** ^{2}⁄_{3}**. Likewise we can do this P(silver side) = 1 –

^{2}⁄

_{3 }=

^{1}⁄

_{3}.

*Thank you very much for your cooperation, it was a great experience and I enjoyed talking with all of them. Hope it will be educative for you all as it was for me.* Hope you are all having fun this summer, don’t forget to check my Facebook event: Mathematics and Summer ^_^ and also my magazine MathTravel if you think about spending the last month of summer like a math-traveler. Thank you for your help and support. Thank you for reading. You can find me on Facebook, Tumblr, Google+, Twitter, Instagram and Lettrs, I will try to post there as often as possible. *Don’t forget that maths is everywhere! Enjoy! *

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