Question

You have 80 dollars and play the following game. An urn contains two white balls and two black balls. You draw the balls out one at a time without replacement until all the balls are gone. On each draw, you bet half of your present fortune that you will draw a white ball. What is your expected final fortune

Answer 1

$45

Explanation:

In order to find this we need to understand that every time we bet we are selecting white ball. Because of this we have the chance of getting two times right and two times wrong irrespective of the order. For example we can any of the following orders:

White - White - Black - Black

White - Black - White - Black

Black - White - Black - White

Black - Black - White - White

Irrespective of any of the orders above every time our bet is right (we get white urn) the amount of dollars that we have get increased by half of the current amount. For example on the first turn we are betting $40 (half of initial amount = $80) and if we are right we now have $120 with us. This can be written as

1.5*$80 = $120

Similarly if on first turn we get wrong (black ball) then our amount is half of initial which can be written as

0.5*$80 = $40

So now we know that exactly two times we will be right and exactly two times we will be wrong irrespective of the order because the balls are not replaced hence final solution is:

E(X) = 80*1.5*1.5*0.5*0.5

= $45