Help us settle this statistics problem and debate

Question

My friends and I were playing Mario Party 4 and one of us happened to win what's called the "Lucky Party Ticket" drawing at the end of the game. This is extremely rare, and led to a debate in the odds of this happening- which we can't seem to establish, because it's been far too long since any of us took a statistics class. So here's the scenario;

There are 4 characters in play during the game. At a point in the game, the character receives a ticket for the lottery. This ticket has a random assortment of 3 pictures on it, and the pictures can be of any of the 4 characters in play in the game. So for example, say you had A,B,C,D (characters in play) - tickets could be AAD, AAC, CCC, etc - the variables can repeat.

At the end of the game, a wheel is spun. The wheel has 4 spaces, and each space has a picture of one of the characters in play. The wheel is spun three times.

If all your pictures on your ticket match the spaces the wheel lands on, in the exact order , you win. The part that seems to be a challenge for us is that not only can the variables repeat themselves, but, the wheel at the end has to hit them in the order they're displayed on your ticket. We are not sure how to represent this mathematically. So if the wheel goes CCA, and your ticket is ACC, you lose even though the variables match- because they're not in the right order.

So, what are the odds of winning? If the explanation isn't clear enough here, you can view a video of this happening here- enter link description here

Answer 1

I will try to answer the question what are the chances that given a game of mario kart with four people, there is at least one winner of the "lucky party ticket". Since each person and each drawn card is distinguishable we can order all the pictures into 15 ordered slots (3 for the wheel and 3 for each players).

An outcome of a single game is thus a a sequence of 15 cards each of which can take 4 values. That means the underlying space has a size of $4^{15}$. We want to know the chance that at least one player wins which is the same as $1-P(\text{no player wins})$. So we will compute how many of the sequences have no player winning.

We take the 3 cards from the wheel to be fixed. That can be done in $4^3$ ways. We fix those. Than if no player is allowed to win that means no player has the same 3 cards. We can than for each player choose one of $4^3-1$ combinations for his cards. So there are in total $4^3*(4^3-1)^4$ many ways to choose the sequence in a way such that there is no winner.

All together the probability that there is at least one winner is $1-P(\text{there is no winner})=1-\frac{4^3*(4^3-1)^4}{4^{15}}$

This surprisingly doesn't seem to really simplify nicely (at least according to WA) and it's not quite $1/4^2$ though it is only about 0.0014 off. If anyone sees a mistake I made I would love to know.

Answer 2

Given four players, the probability that a unique ticket is the winning ticket is equal to $$ \frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \frac{1}{4^{3}} $$ where each quarter comes from the probability that the wheel spins the face on the ticket in each round. Because the order matters, it is exactly this value.

If it is impossible to get repeated tickets, then the probability that there is one winner amongst the four players is $$ \frac{1}{4^{3}} + \frac{1}{4^{3}} + \frac{1}{4^{3}} + \frac{1}{4^{3}} = \frac{1}{4^{2}}. $$ However, if repeated tickets are possible, then see DRF's answer.