Can we use conditional expectation?

[Millie]

I found this question in a book:

Two palyers A and B alternatively roll a pair of unbiased die. A wins if on a throw he obtain exactly 6 points, before B gets 7 points, B wining in the opposing event. If A begins the game prove that the probability of A winning is 30/61 and that the expected number of trials need for A's win is approximately 6.

I was trying to solve the problem using conditional probability and expectation. To get 30/61 I conditioned on the sum on the dice on the 1st roll and then on the sum on the dice on the second roll.

However I am not sure whether we can calculate the required expectation through conditioning. I know that the exact answer is 371/61=6.0819. Through conditioning I only can 5.57 which however can also be approximated to 6.

Also if X denotes "the number of trials needed for A to win"
And Yi be the sum on the ith roll

Conditioning on the first roll,
E[X]= E[X/Y1=6)P(Y1=6) + E[x/Y1 is not equal to six]P[Y1 is not =6)
= 1x5/36 + E[X/Y1 not =6)x31/36

Then conditioninf on the 2nd roll.
E[X/Y1 not = 6]= E[X/Y1 not=6, Y2 not = 9)P(Y2 not equal 9) + E[X/Y1 not=6, Y2=9)P(Y2=9)

That's kind of where I am stuck for what is the expected number of trials for A to win given that B wins on the second trial??
Or should I try to condition on the 3rd or last event?
Or should I approach the problem without using condition.

Man I am so confused now! :rofl:

thanks!

 

[anathema]

Thank you for sharing your question and thought process with us. I would like to provide some insights and suggestions for solving this problem.

Firstly, your approach of using conditional probability and expectation is a good direction to take. However, I believe there may be some errors in your calculations. Let me try to explain my approach and hopefully it will help you to understand the problem better.

To calculate the probability of A winning, we can condition on the first roll and the second roll, as you have mentioned. However, we can also consider the third roll and so on. In fact, we can condition on the entire sequence of rolls until either A or B wins. This approach is known as the "geometric distribution" and it will help us to calculate the expected number of trials needed for A to win.

Let's start by considering the first roll. The probability of A winning on the first roll is 5/36, as you have correctly calculated. Now, let's consider the second roll. If A does not win on the first roll, then the probability of A winning on the second roll is the probability of A winning on the first roll multiplied by the probability of B not winning on the second roll. In other words, P(A wins on second roll) = (5/36)(29/36). Similarly, we can continue this approach for the third roll, fourth roll, and so on.

Now, to calculate the expected number of trials needed for A to win, we can use the formula E[X] = 1/p, where p is the probability of A winning on any given trial. Using the approach described above, we can calculate the probability of A winning on any given trial as the sum of all the individual probabilities for each roll. This will give us the expected number of trials needed for A to win, which is approximately 6.

I hope this helps to clarify the problem and guide you towards the correct solution. Good luck!

 

[quicknote]

Yes, we can use conditional expectation to solve this problem. In fact, it is a common approach used in probability and statistics to solve problems involving multiple events. In this case, conditioning on the first roll and then on the second roll is a valid approach.

To find the probability of A winning, we can use the law of total probability: P(A wins) = P(A wins on first roll) + P(A wins on second roll) + ...

We can calculate P(A wins on first roll) by conditioning on the sum of the dice on the first roll. Similarly, we can calculate P(A wins on second roll) by conditioning on the sum of the dice on the second roll. Continuing this process, we can calculate P(A wins) and show that it is equal to 30/61.

For the expected number of trials needed for A to win, we can use the definition of conditional expectation: E[X|Y] = ∑x P(X=x|Y). In this case, we can use the same approach as above, conditioning on the sum of the dice on each roll to calculate the expected number of trials needed for A to win. This will give us an approximation of 6, as you have found.

It is also possible to solve this problem without using conditional expectation, but using it can simplify the calculations and provide a more intuitive understanding of the problem. So, it is a valid and useful approach in this case.