# Almost 2019

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:

1. 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?

Note by Jerome Te
2 years, 8 months ago

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The probability is $\frac{1}{2019}$

- 2 years, 8 months ago

I think it is incorrect :(

- 2 years, 8 months ago

But confirm 2018 out of 2019 will say correctly

- 2 years, 8 months ago

2018 out of 2019 will surely get it correctly if everyone will guess. But they can say pass

- 2 years, 8 months ago

- 2 years, 8 months ago

I still don’t think it is right :( because seeing the other people’s hat doesn’t give you any information about your own, you may calculate the probability of your hat being white, but the probability doesn’t quite matter, because everything with a small probability may happen.

- 2 years, 8 months ago

I mean everyone passes

- 2 years, 8 months ago

Here, optimal play means that they want to minimize the time they spend in prison. They are trying to minimize the estimated value.

- 2 years, 8 months ago

Did I phrase the question incorrectly?

- 2 years, 8 months ago

No. You did not

- 2 years, 8 months ago

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

- 2 years, 8 months ago

@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 : $\LaTeX$ Help

- 2 years, 8 months ago

Thanks for that!

- 2 years, 8 months ago

No probs

- 2 years, 8 months ago

- 2 years, 8 months ago

I realized something was wrong. I added a detail.

- 2 years, 8 months ago

What detail

- 2 years, 8 months ago

The standard coin detail, see it now?

- 2 years, 8 months ago

Just a little property of 2019:

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

- 2 years, 6 months ago

It is also a Lucky number and a Happy number.

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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.

- 2 years, 6 months ago

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

- 2 years, 6 months ago