Question

A gambler bets n times. Each time the gambler bets, 20% of the time he wins $10 and 80% of the time he loses $5. What is the expected gain (which can be negative) after n bets?

Answer 1

The expected value of an “experiment” is the long-run average—if the experiment could be repeated many times, the expected value is the average of all the results.

The expeced value is given by:

E[X] = x_i​.p(x_​i​)

​where, x_​i ​is the random variable & p(x_​i​) is the probability of random variable x_​i.

Here in this case,

x​i	$10 profit = +10	$5 loss = -5
p(x​i)	20% = 20/100 = 0.2	80% = 80/100 = 0.8

The expected value for this case can be defined as,

Expected value = ( probability of event 1) ( payoff for event 1) + (probability of event 2) (payoff for event 2)  

So, Expected gain = (+10)* 0.2 + (-5)*0.8 = 2 - 4 = - 2

​ The expected gain is negative for this case, hence the gambler had lose of $2 after n bets.