Expected Value of Cup of Coffee in Flip a Lid Contest

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SUMMARY

The expected value of a cup of coffee in the Flip a Lid contest is calculated to be approximately -$1.37. This value accounts for the purchase price of $1.50 and the probabilities of winning various prizes, including 50,000 free coffees, two TVs, a snowmobile, and a sports car. The calculations involve determining the probabilities of winning each prize and the associated net gains or losses. The final expected value reflects the overall loss to the customer when participating in the contest.

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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?
 
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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?
 
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 !
 
I would write:

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

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:

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

As you can see this is mathematically equivalent to the way I initially approached the problem.
 
Thank you very much ... I guess I was overthinking with the rest of the probability and neglected that. The table blindsided me. Thx again
 
Istar said:
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)