- #1

Arnoldjavs3

- 191

- 3

## Homework Statement

I have another thread where there's a game similar to this one(however simpler to some degree) and I learned quite a bit from it. However, my neighbor had designed a game when he was in university for a project, and I would like to understand a bit more about it. .

Basically I just have a few questions that I want to clarify with myself(the actual game is not this, this is just for the sake of understanding):

Say there is a game, where there are 3 trials one has to undergo in order to win. Each trial has their respective success rate - and of which leads to either a success or failure. You have 15 tries to win the game - if you fail after 15 attempts, you lose the game.

Trial 1: You have a 40% chance of winning this trial, if you win, you proceed to trial 2.

Trial 2: You have a 40% chance of winning this trial, if you win, you proceed to trial 3.

Trial 3: You have a 33% chance of winning this trial, if you win, you win the game.

E.g(for the sake of understanding):

Player 1 fails trial 1 8 times.

Player 1 then wins when he is on his 9th attempt, however fails every single other attempt. He loses the game and can not attempt anymore.

Now... a couple of people are throwing some arguments in my direction with their claims that the success of winning all 3 trials is the same as winning the first one and that they have no relevance to each other. They used the gamblers fallacy to support this claim. This is leading to my questions that I have.

Question 1: Are trials 2 and 3 are dependent on the rounds before them?

Question 2: Is the gamblers fallacy applicable in any instance of this situation?

Question 3: How does the results differ with the probabilities calculated? Are 15 attempts even remotely close to having an accurate representation here? Can 15 attempts even be accurately represented assuming there is only 1 player? Is it just simply luck?

## Homework Equations

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## The Attempt at a Solution

1) I think that trial 2 is dependent on trial 1, and trial 3 is dependent of trial 2. This is due to the simple fact that you can not attempt either attempt without reaching that stage first. Of which, calls for having multiple successive wins which is not as likely to occur.

There is a 5.28% of winning the game on your first try, and 55.6% of winning on your 15th attempt(using geometric distribution?).

However, how does probability differ from the actual percentage of success here? Are you more likely to see a win on your 15th attempt than you are on your attempts before so? Is this just due to the fact that you aren't as likely to see the same result with each attempt you perform?

2) I do not think it is, due to the things I have said above. (I do not like the idea of the fallacy to begin with) How could one explain why you are less likely to see the same result over and over again - given the initial circumstances?

3) I do not think 15 attempts are enough at all to formulate an accurate prediction. I think maybe somewhere along the lines of 100,000 attempts, even more, would be better. However, could it be done?

Those are the sum of my questions in this situation...