In: Statistics and Probability
Your goal is to collect all 80 player cards in a game. The Player cards are numbered 1 through 80. High numbered cards are rarer/more valuable than lower numbered cards.
Albert has a lot of money to spend and loves the game. So every day he buys a pack for $100. Inside each pack, there is a random card. The probability of getting the n-th card is c(1.05)-n, For some constant c. Albert buys his first pack on June 1st. What is the expected number of Player cards Albert will collect in June?(30 days)
a.) Find an exact, closed-form expression for c. (Answer should not include a summation symbol or integral sign).
b.)Find the expected number of unique Player cards Albert will collect in June. (Answer may include summation symbol or integral sign.
(A) Everyday when Albert buys the pack, there will be a card in the pack with a random number between 1 and 80 written on it. Hence everyday following relation will hold ( sum of probabilities of obtaining n th card will be 1):
On solving this using the summation of GP formula we get:
Hence,
(B) Now, the expected number of unique cards with Albert at the end of 30 days will be given by the relation below:
E(Number of unique cards) = 1*Pr(same card every day) + 2*Pr(2 numbers across 30 days) + 3*Pr(3 numbers across 30 days) +.........+ 30*Pr(30 numbers across 30 days)
Now, the probabilities on the RHS of above equation are given by:
Pr(same card every day) =
Explanation: Pr(same card everyday) = Pr(number 1 everyday) + Pr(number 2 everyday) +......+ Pr(number 30 everyday
Pr(2 numbers across 30 days)=
Explanation: Pr(2 numbers across 30 days) = First number occurring for R days and second occurring for 30-R days. And, the two numbers can take values from 1 to 80 but cannot be equal to each other.
Pr(3 numbers across 30 days)=
Pr(30 numbers across 30 days)=
By substituting these probabilities in the above defined equation we can get expected number of unique cards at the end of 30 days with Albert