It is October 2018 and it is near to 2019. Included here is a problem involving 2019. The one who can post the correct answer first will post another problem involving 2019. Our target is to get 2019 problems here. So, here is the first problem:

- There are 2019 prisoners in a prison. The prison warden decides to play a game with them. He lines them up in a row all of them facing another prisoner. Prisoner N will face Prisoner N-1. He takes 2019 black hats and 2019 white hats. He randomly puts one on each prisoner’s head. (He flips a standard coin to determine what hat to put on a prisoner’s head.) Everyone can see the colour of the hat of the person with any number lower than them. He explains the rules of the game and tells the prisoners that each one of them has to guess the colour of their hat or say pass. If any of them will guess incorrectly, everyone spends 10 more years in prison. If someone guesses correctly, everyone goes free. If someone guesses right but another guesses wrong, they spend 10 more years in prison. If everyone says pass, nothing happens. What is the probability that they spend 10 more years in prison if they play optimally?

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## Comments

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TopNewestThe probability is \(\frac{1}{2019}\)

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I think it is incorrect :(

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But confirm 2018 out of 2019 will say correctly

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@Jerome Te, I think it is \(\frac{1}{2}\)

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@Jerome Te, I noticed that you are a new member. I suggest you learn \(\LaTeX\).

Note: If you do not know where to start you can refer to my set : \(\LaTeX\) HelpLog in to reply

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@Jerome Te, You may change profile pic in your about

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I realized something was wrong. I added a detail.

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What detail

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The standard coin detail, see it now?

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Just a little property of 2019:

2019 is the first 4-digit number that appears 6 times in the decimal expansion of \( \pi \).

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It is also a Lucky number and a Happy number.

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The divisors of 2019 (3 and 673) are also Lucky numbers.

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Imagine 23 free spots in a line and fill them with \( 2\times1 \) dominoes in all possible ways that only leave \( 1\times1 \) gaps (all larger gaps have to be filled). Now, count the number of empty spots in each arrangement and add them up for all arrangements. You will get 2019.

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2019 alsp happens to be the sum of the first 22 perfect powers.

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