Expected Value of Cup of Coffee in Flip a Lid Contest

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Discussion Overview

The discussion revolves around calculating the expected value of a cup of coffee in a promotional contest held by a coffee chain. Participants explore the probabilities of winning various prizes and how these affect the expected value from the customer's perspective.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the prizes available in the contest and suggests calculating the expected value based on the probabilities of winning each prize.
  • Another participant emphasizes the need to account for the cost of purchasing the coffee when calculating expected value.
  • A third participant presents their calculations for expected value, arriving at a figure of approximately 0.1274, but expresses uncertainty about its correctness.
  • A different approach is proposed, leading to a calculation of approximately -1.37, which includes the cost of the coffee and considers non-winning cups.
  • Participants acknowledge the oversight regarding the purchase price of the coffee in earlier calculations.

Areas of Agreement / Disagreement

There is no consensus on the expected value calculation, as participants present different methods and results. Some agree on the necessity of including the purchase price, while others have differing interpretations of the probabilities and outcomes.

Contextual Notes

Participants have not fully resolved the assumptions regarding the probabilities of winning and the impact of the purchase price on the expected value calculation.

Istar
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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)