Welcome to our community

Be a part of something great, join today!

Discrete Probability: What is the expected value of cup of coffee to the customer

Istar

New member
May 11, 2020
8
In its flip a lid contest, a coffee chain offers prizes of 50,000 free coffees, each worth \$1.50; two new TVs, each worth \$1200; a snowmobile worth \$15 000; and sports car worth \$35 000. A total of 1 000 000 promotional coffee cups have been printed for contest. Coffee sells for \$1.50 per cup. What is the expected value of cup of coffee to the customer?
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
Hello, and welcome to MHB! :)

To begin, we need to determine the probabilities for the following events:
  • Customer who bought a cup of coffee, purchased a cup of coffee (this is certain, thus probability is 1)
  • Customer won a free coffee
  • Customer won TV
  • Customer won snowmobile
  • Customer won sports car
Once we've determined these probabilities, we can associate the net loss/gain for each event and compute the sum of the products of the probabilities and loss/gain for all events to determine the expected value. What do you get for the probabilities of the events?
 

Istar

New member
May 11, 2020
8
Thank you for your welcome,

This is how I did it :

X
P(X=x)
50000/1000000
1.5​
2/1000000
1200​
1/1000000
15000​
1/1000000
35000​


Expected mean value [x] = {x(Px)} =

(0.05x1.5) + (0.000002x1200) + (0.000001x15000) + (0.000001x35000) =

(.075) + (0.0024) + (0.015) + (0.035) = 0.1274

I am not sure if it right !
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
I would write:

\(\displaystyle E[X]=1(-1.5)+\frac{50000}{1000000}(1.5)+\frac{2}{1000000}(1200)+\frac{1}{1000000}(15000)+\frac{1}{1000000}(35000)=-\frac{6863}{5000}\approx-1.37\)

You have neglected the certainty that they will spend \$1.50 to purchase the cup of coffee.

Another way to approach this it to observe there are 949996 non-winning cups and write:

\(\displaystyle E[X]=\frac{949996}{1000000}(0-1.5)+\frac{50000}{1000000}(1.5-1.5)+\frac{2}{1000000}(1200-1.5)+\frac{1}{1000000}(15000-1.5)+\frac{1}{1000000}(35000-1.5)=-\frac{6863}{5000}\approx-1.37\)

As you can see this is mathematically equivalent to the way I initially approached the problem.
 

Istar

New member
May 11, 2020
8
Thank you very much .... I guess I was overthinking with the rest of the probability and neglected that. The table blindsided me. Thx again
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
Thank you very much .... I guess I was overthinking with the rest of the probability and neglected that. The table blindsided me. Thx again
Other than overlooking the purchase price of the cup of coffee, your table and resulting calculations were good. (Yes)