AlexSkidanov's blog

The 12 chairs and diamonds.
By AlexSkidanov, 15 years ago, translation, In English
The problem is rather old, but very famous, that’s why I hope someone hasn’t come across it yet, and will get pleasure trying to solve it :o)

Let suppose that Klavdia Ivanovna (or what was her name) from a famous movie The Twelve Chairs hid her diamonds inside one of the chairs not for sure, but with 90% probability. What is the probability to find the diamonds in the 12 th chair after all the 11 chairs have been cut open and diamonds were not found inside one of them?

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15 years ago, # |
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i guess it is 1?
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  • 15 years ago, # ^ |
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    No. It would be one, if diamonds were hidden for sure. But they were hidden only with probability 90%, which means that initially with probability 10% diamonds were not in one of the chairs. As a result, if we opened 11 chairs and didn't find diamonds in any of them, they are not for sure in the 12th chair. The question is: what is the probability that they are in 12th chair, if they were not found in first 11 chairs.
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    • 15 years ago, # ^ |
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      these days i am reading the prblems wrong. i read, the diamond is hidden in one of the chairs with 90% probability.  .........
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      • 15 years ago, # ^ |
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        Ok, just to clarify:
        With the probability of 10% diamonds were not hidden in any of chairs
        With the probability of 90% diamonds were hidden in randomly chosen chair.
        After that 11 of them were cut off and none of these 11 contained diamonds.

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15 years ago, # |
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Is it 90%?
Because, if diamonds are hidden for sure ( 90% for that ), then
they must be in 12th chair ( 100% for that ), so
0.9 * 1 = 0.9
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15 years ago, # |
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Am I right that it is 3/7
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15 years ago, # |
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Isn't it 3/40?
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  • 15 years ago, # ^ |
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    3/40 is the answer for another problem: you open the chairs at random, what's the probability to find diamonds in the last chair? The difference is that in this problem you already know that you haven't found the diamonds in 11 chairs; that is, the question is about conditional probability.
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14 years ago, # |
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Should we consider order of the chairs?
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14 years ago, # |
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If someone is presented with the situation that ::
There is one chair (intact) and 11 are cut-open .(and it is told to him ," With 90% probability , the diamond is inside one of them. " )
He can surely say it is 90% chance of getting the diamond from intact chair.
++++++++++++++++++++++++++++++++
But if situation is ::
Initially all chairs are intact.And it told to him,"With 90% probability , the diamond is inside one of them."
Then 11 are cut-open and none contains the diamond.
Then it will not be 90 %.
is it 327/400?

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14 years ago, # |
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Maybe it is 0.42857142857142855 ?
I thought about it this way:
If we know there are three events with probabilities 0.2, 0.3, 0.5 and we make an experiment from which we find out that the first can't happen then the other two will have probabilities 0.3/0.8=0.375 and 0.5/0.8=0.625.
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