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I've been studying probability for a few couple weeks. Today, I've given a homework, and I am really struggling with it. The type of question is unfamiliar to me and I don't have any idea how to solve it. I tried several ways, but they look not consistent.

question image: [question image][1][1]: https://i.stack.imgur.com/Hd9gT.png

Each of the dice in the figure above consists of 3 distinct numbers. All

dice are fair and the probabilities of the outcomes are as follows:

Die a: Pr(2)= 1 / 3, Pr(6)= 1 / 3, Pr(7)= 1 / 3

Die b: Pr(1)= 1 / 3, Pr(5)= 1 / 3, Pr(9)= 1 / 3

Die c: Pr(3)= 1 / 3, Pr(4)= 1 / 3, Pr(8)= 1 / 3

So as you can see, we have 3 dices and they are not a usual dice. Both of them only consist of 3 distinct numbers. Let's continue with the question.,

Two players will select and roll one die each, only one time. After the

first player’s selection, next player can only select and roll one of the

remaining two dice (namely: {a, b} or {a, c} or {b, c}).



a) Player 1 selects “Die a”, and player 2 selects “Die b”. Which one

is more likely to win? And what is the probability of winning? 

I don't want you to solve the question for me, I just want you to show me the way or give me the hint. Thanks in advance.

There are three outcomes for Player $1$, who has chosen die A: they can get $2,6$ or $7$. After they roll, Player $2$ rolls, so for each outcome of player $1$, there are three possible outcomes for Player $2$. So: start by writing out each of the possible outcomes and the probability of it occuring, and you should be able to get the answer from there. To give you a start, if Player $1$ rolls a two, then player $2$ rolls a $1$, we have that player $1$ wins. Since there are nine outcomes in total, so far we know that Player $1$ wins at least $1$ out of all $9$ outcomes.

Why ask such a question? Well, it generalises: when we have independent outcomes, like here, we can proceed by working out the probability of the first event, the the probability of the second event, and then the third... and so on. We can combine these probabilities (by multiplication), and so we obtain a process for handling this kind of question. Once you've solved your question, you should be able to see how to extend it to $10$ dice, with six numbers on each die, and even where the probability of each number occuring is different.

Ah, you say, but why do you say they have to be independent? Because if they're not then the second player's probablilities depend on the first player's roll, and we may not be able to multiply any more. In this case we need to develop new techniques, such as conditional probabilities, which you'll come to soon enough.