reinout-g

No, the calculator claims that it's infinite because it's an absurdly large number.

$$e^{9185} \approx 9.88 \times 10^{3988}$$

 Reinout 

Could you elaborate on what you find unsettling in the terms of agreement?

Try the easier one then,

It doesn't really require any calculation, just some thinking. 

AAAAAAAAAAAAAAARRRRRGGGHHH!

BY THE BEARD OF ZEUS!

THOU SHALT NOT TORMENT THY HOLY LAWS!

THOU SHALL FEAR THE WRATH OF MATH!

I CURSE THEE WITH UNBREAKABLE BONDS!

$$P=NP / P \neq NP$$

But what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive 

Unless the skiers died in a skiing accident off course 

Here is an update on the probability puzzles!
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Easier one (hasn't been answered yet): 

The fabulous game host Steve offer you a bag with 5001 pearls of which 2501 are white and 2500 are black. He asks you to draw as many pearls from the bag as you desire without looking. Steve then tells you how many black and white pearls you drew. If there are equally as much white as black pearls in the draw you gain €10 for each pearl you drew. How many pearls should you draw?

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Hard one: 

All right, so here's a classic one.

Jason and Tyrell are in a classroom of 23 students.

At some point Tyrell finds out that some of his classmates have the same birthday.

In surprise he exclaims; 'what a surprise!'.

Jason however, says it is more likely for two or more people to have the same birthday in their class than it is for noone to have the same birthday.

Assume any birthday is equally likely to another and pretend leap years don't exist. (it makes it a lot easier)

What are the odds of two or more people sharing their birthday?

Okay, so I think Creepercosmo was so close to the answer that I'm gonna give him 5 points...

Here's how it can be done.

Let's say there are indeed 4 options.

1. I confess, my friend does too and we both spend 10 years in jail

2. I confess, my friend does not and he spends 20 years on his own

3. I do not confess, my friend does and I spend 20 years on my own

4. I do not confess, my friend doesn't either and we both spend 3 years

Let's see how many suffering points each situation gives me;

1. s = 2*10 + 10 = 30

2. s = 2*0 + 20 = 20

3. s = 2*20 + 0 = 40

4. s = 2*3 + 3 = 9

Let p be the estimated probability that your partner confesses.

Then the expected amount of suffering is

If I confess;

E(S|C) = 30p+20(1-p) = 10p+20

If I do not confess;

E(S|NC) = 40p+9(1-p) = 31p+9

We want to find a value for p where you're indifferent between confessing or not confessing.

Therefore

$$10p+20 = 31p+9 \Rightarrow 21p = 11 => p = \frac{11}{21} = 0.5238$$

Now the slope of 31p+9 is obviously higher than that of 10p+20 (and we want to minimize the amount of suffering)

Therefore if $$p>0.5238$$ You'd choose to confess and

if $$p<0.5238$$ You'd choose not to confess.

Now obviously, I highly doubt someone would say his friend has a 52.39% chance of confessing, so well done Creepercosmo! 

I'm going to think of a new one 

I heard the  brought them 

Nope that's wrong it's 42