How can pick odds or even be fair?

177 views

[ad_1]

A even number plus a even number goes even and a odd number plus a odd number goes even too.

The only chance to get a odd number is a even and an odd.

How it cannot be 2/3 and 1/3 ?

In: Mathematics
[ad_2]

Player 1 can pick Even or Odd. Player 2 can pick Even or Odd.

The options then are:

1E, 2E – > result is Even

1E, 2O -> Result is Odd

1O, 2E -> Result is Odd

1O, 2O -> Result is Even

Your error is in combining the two options, 1E, 2O and 1O, 2E into one concept of “one picks odd, the other picks even”.

There are two scenarios where the game can end with an even and an odd.

Player 1: even, player 2: odd

Player 1: odd, player 2: even

Looking at it another way, let’s say you pick odd. The game will end even if the other player picks odd, and the game will end odd if the other player picks even. So it’s 50:50 if you choose odd.

If you choose even, the game will end even if the other player picks even and will end odd if the other player picks odd, so picking even is also 50:50.

Odd plus odd is even.
Even plus even is even.

Odd plus even is odd.
Even plus odd is odd.

It’s 2/4 both ways. Not 1/3 and 2/3.
There are four options, not three.

Because there are not three possibilities but four.

1st | 2nd | result
—|—|—-
even | odd | odd
even | even | even
odd | odd | even
odd | even | odd

You can’t just lump two possible results together. odd+even and even+odd are both different results.

You have to imagine both parts separately.

If you flip two coins what are the chance that thy both show the same result.

Either they are both head or both tail or one head and one tail. But the last part could be either the first being head and the second being tail or the first being tail and the second being head.

Not sure if I understand what the question is.

Regardless, I hope the following will give some insight:

If you pick 2 numbers, X and Y, at random, X and Y can either be even, or odd. So, we have the following 4 possibilities. (Not 3)

– X, Y are both even -> sum is even

– X is even, Y is odd -> sum is odd

– X is odd, Y is even -> sum is odd

– X, Y are both odd -> sum is even

I think now it’s easy so see that the odds are 1/2 and 1/2.

Edit: formatting