**The two envelopes paradox**

*“You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.”*

Two envelopes problem - Wikipedia, the free encyclopedia

**-This is a problem of expected value:**

Suppose random variable *X* can take value *x*_{1} with probability *p*_{1}, value *x*_{2} with probability *p*_{2}, and so on, up to value *x*_{k} with probability *p*_{k}. Then the **expectation** of this random variable *X* is defined as

Since all probabilities *p*_{i} add up to one (*p*_{1} + *p*_{2} + ... + *p*_{k} = 1), the expected value can be viewed as the weighted average, with *p*_{i}’s being the weights:

If all outcomes *x*_{i} are equally likely (that is, *p*_{1} = *p*_{2} = ... = *p*_{k}), then the weighted average turns into the simple average. This is intuitive: the expected value of a random variable is the average of all values it can take; thus the expected value is what one expects to happen *on average*. If the outcomes *x*_{i} are not equally probable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than the others. The intuition however remains the same: the expected value of *X* is what one expects to happen *on average*.

Expected value - Wikipedia, the free encyclopedia

**My contribution**

On expected vale:

E(x)= Expected Value

P= probability

V= Value

EP= P1+P2+……….+Pn

EV=V1+V2+………..+Vn

E(x)= P1xV1+P2xV2+…………..+ PnxVn

V1= currency value inside of Envelope 1

V2= currency value inside of Envelope 2

Vx= Expected currency value inside V1 or V2 (meaning the expected value of any of the two envelopes)

**EV= V1+V2**

The problem can be solved in two ways either V1=2V2 or V1=0.5V2. **Either way you assume, the total of the values of both envelopes must and will remain the same. **

(1) EV= V1+V2

(2) If V1>V2 then ** EV = 2X + 1 X =3X**. If EV = 30 then **X=10**, V1=20, V2=10

(3) If V1<V2 then ** EV =X+0.5 X=1.5X**. If EV = 30 then **X=10**, V1=10. V2= 20

In other words, regardless of which equation of EV you use. You will find both values of 10 and 20, which again add to 30 and therefore the expected value should be 15.

Determining expected value of Vx

Vx=P1xV1+P2xV2

Using equation (2). Vx= ½ (2X) + ½ (X)= 1.5X

EV=2X+X= **3X**. If EV= 30 then X= 10 then Vx= 1.5X= 1.5 (10)=** 15**

Using equation (3) Vx= ½ ( X ) + ½ (0.5 X)= 3/4X

EV = X+0.5X= **1.5X**. If EV= 30 then X= 20 then Vx= 1.5X= 1.5 (20)= **15**

Regardless, which equation envelope you pick, the sum of the values of both envelopes have to remain constant in this sample: EV=30 and Vx= 15

**The paradox**

Vx=P1xV1+P2xV2

Vx=½ (2x) + ½ (0.5X)= 5/4 X

However………Vx≠ 5/4 X

**Cause ****EV≠2X+0.5X≠2.5X.**

**EV=3X or 1.5X depending on which equation is used (either equation 2 or 3).**

The reason for the paradox is due to rather than using either equation (2) or (3). The paradox is using an equation that does equal the total value of both envelopes cause is combining a portion of equation 2 and a portion of equation 3 into a new one.

In other words, assuming that **E**V =2.5X.

Which we know is not true.

If rather than using money values we have two pills: a red pill and a blue pill. And we distribute them in two envelopes.

The expected value of which pill is on each envelope is

(a) Vx= ½ Blue+ ½ Red

(b) Total of colors= Blue + Red

The total of the colors of options (B) have to present in the formula of expected value (a).

Let me know your opinions guys.