Agent Smith
Full Member
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- Oct 18, 2023
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The ticket-collector's problem: Assume there are n different tickets you can collect, these tickets are found in boxes of a product (some boxes maybe empty i.e. have no tickets in them). There are therefore n + 1 different possibilities. Each box has the same probability for these n + 1 possibilities i.e. each box contains either no ticket or 1 ticket (one of the n). What is the average number of boxes that have to be purchased to ensure you collect all tickets. If collect them all you win a prize, that's how it usually works.
How do we solve this problem?
The easiest and dumbest way to find the answer is to use a random number generator. Assign each ticket + no ticket a number from 1 to n + 1 and carry out a simulation. Conduct a (large) number of simulations and find the average of the number of attempts (purchases) that have to be made to acquire all n tickets.
Wikipedia has an article on this but I couldn't understand it.
Here's how I attacked the problem:
Let the number of tickets = n = 4
We have n + 1 possibilities (counting empty boxes) = 4 + 1 = 5
For 4 boxes, we see total number of permutations possible = [imath]5^4 = 625[/imath]
The number of ways we can collect all 4 tickets = [imath]^4 C_4 \times^4 P _4 = 24[/imath]
Probability of getting all tickets in just 4 tries = P(four) = [imath]\frac{24}{625}[/imath]
For 5 boxes, we see that there are [imath]5^5 = 3125[/imath] possibilities (repetitions are possible)
In how many ways can we "win" (get all 4 tickets)? [imath]^5 C_4 \times ^4 P_4 \times ^4 P_1 = 480[/imath]
Probability of getting all tickets in just 5 tries = P(five) = [imath]\frac{120}{3125}[/imath]
For 6 boxes, total possibilities = [imath]5^6[/imath]
Win possibilities = [imath]^6 C_4 \times ^4 P_4 \times ^4 P_2[/imath]
Probability of getting all tickets in just 6 tries = P(six) = [imath]\frac{^6 C_4 \times ^4 P_4 \times ^4 P_2}{5^6}[/imath]
In general, [imath]P(x) = \frac{^x C_4 \times ^4 P_4 \times ^x P_{x - 4}}{5^x}[/imath]
Looks like I'm calculating the probability of getting all n tickets in exactly [imath]x[/imath] attempts/purchases.
What do I do with these probabilities?
If random variable = [imath]X[/imath] = the expected number of purchases to "win" then, ...
[imath]E(X) = P(four) \times 4 + P(five) \times 5 + P(six) \times 6 + ... + P(x) \times x[/imath]?
Help!
How do we solve this problem?
The easiest and dumbest way to find the answer is to use a random number generator. Assign each ticket + no ticket a number from 1 to n + 1 and carry out a simulation. Conduct a (large) number of simulations and find the average of the number of attempts (purchases) that have to be made to acquire all n tickets.
Wikipedia has an article on this but I couldn't understand it.
Here's how I attacked the problem:
Let the number of tickets = n = 4
We have n + 1 possibilities (counting empty boxes) = 4 + 1 = 5
For 4 boxes, we see total number of permutations possible = [imath]5^4 = 625[/imath]
The number of ways we can collect all 4 tickets = [imath]^4 C_4 \times^4 P _4 = 24[/imath]
Probability of getting all tickets in just 4 tries = P(four) = [imath]\frac{24}{625}[/imath]
For 5 boxes, we see that there are [imath]5^5 = 3125[/imath] possibilities (repetitions are possible)
In how many ways can we "win" (get all 4 tickets)? [imath]^5 C_4 \times ^4 P_4 \times ^4 P_1 = 480[/imath]
Probability of getting all tickets in just 5 tries = P(five) = [imath]\frac{120}{3125}[/imath]
For 6 boxes, total possibilities = [imath]5^6[/imath]
Win possibilities = [imath]^6 C_4 \times ^4 P_4 \times ^4 P_2[/imath]
Probability of getting all tickets in just 6 tries = P(six) = [imath]\frac{^6 C_4 \times ^4 P_4 \times ^4 P_2}{5^6}[/imath]
In general, [imath]P(x) = \frac{^x C_4 \times ^4 P_4 \times ^x P_{x - 4}}{5^x}[/imath]
Looks like I'm calculating the probability of getting all n tickets in exactly [imath]x[/imath] attempts/purchases.
What do I do with these probabilities?
If random variable = [imath]X[/imath] = the expected number of purchases to "win" then, ...
[imath]E(X) = P(four) \times 4 + P(five) \times 5 + P(six) \times 6 + ... + P(x) \times x[/imath]?
Help!
