# Questions about probability (successive attempts?)

• Arnoldjavs3
In summary, the conversation is discussing a game with three trials and 15 total attempts allowed. Each trial has its own success rate, and if the player fails any of the trials, they must restart from the beginning. The conversation includes a discussion about the joint probability of reaching and succeeding at each trial, as well as the relevance of the gamblers fallacy in this situation. The conversation also raises questions about the accuracy of predicting success with 15 attempts and the possibility of having a second attempt at a failed trial.
Arnoldjavs3

## 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?

??

## 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...

IS this a homework problem?

Anyway, what do you mean by "the success of winning all 3 trials is the same as winning the first one"?

Did you mean "the probability of winning all 3 trials is the same as winning the first one"?

I guess not? It's simply for my learning experience though.

Yes, one was attempting to argue that the probability of success of winning all 3 trials was the same as winning 1 of them.

I guess you were supposed to post this in the Mathematics/Probability section.

Anyway, I don't see how you could possibly equate winning 3 trials the same as only winning 1 of them.

Well I will post it there as it seems I won't be receiving much interest
And his argument was that the events in sequence have no relevance to each other.

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Maybe they do and maybe they don't but that doesn't mean automatically the probabilities are the same.

First, let me see if I understand the game description. The player is allowed a total of 15 plays. She takes some number of plays to pass the first trial. If successful before running out of plays she proceeds to the second trial, with some number of plays less than 15 left, ok?
The probability of success at a given play depends only on the nature of the trial - first second or third. But you seem to be confusing that with the joint probability of reaching that trial and succeeding at it. These are different things.

Yes I believe that you have the right idea here. So if you fail at any instance, you have to restart. Once you have used up your 15 attempts, you lose.

So in essence, if allowed a second attempt at the same trial(just for the sake of understanding) then essentially the joint probability is increased?

Arnoldjavs3 said:
Yes I believe that you have the right idea here. So if you fail at any instance, you have to restart.
Restart? So if I pass the first trial at the third attempt, then fail the second trial at the first attempt, I have to go back to trial 1? Or do I continue with trial 2, 9 'lives' left (i.e. 9 more failures permitted)?
Arnoldjavs3 said:
So in essence, if allowed a second attempt at the same trial(just for the sake of understanding) then essentially the joint probability is increased?
What joint probability?

Yes you would have to restart if you fail on your first attempt on trial 2. Your 9 lives continue from trial 1.

And let's say I'm on the third trial - however, the 'gamemaster' decides to let me have 2 shots at trying to win trial 3 rather than having me restart. This would increase the joint probability of succeeding correct?

Edit: I feel like I'm just simply repeating what I already know, however referring back to the original post, the gamblers fallacy is what's frustrating me here on accuracy of sampling data. Are there any similar games that you guys are aware of where I can make predictions for?

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Arnoldjavs3 said:
Yes you would have to restart if you fail on your first attempt on trial 2. Your 9 lives continue from trial 1.

And let's say I'm on the third trial - however, the 'gamemaster' decides to let me have 2 shots at trying to win trial 3 rather than having me restart. This would increase the joint probability of succeeding correct?

Edit: I feel like I'm just simply repeating what I already know, however referring back to the original post, the gamblers fallacy is what's frustrating me here on accuracy of sampling data. Are there any similar games that you guys are aware of where I can make predictions for?
A joint probability is the probability of two or more given events all happening. Which combination of events are you referring to?
If there's a game master who can arbitrarily choose to let you have extra tries then I don't see how you can calculate any probability of ultimate success.
You keep referring to the gambler's fallacy, but it's not clear to me how this is confusing you. Do you understand exactly what the gambler's fallacy is?
In regard to the game structure as I now understand it, it will be a non-trivial calculation to find the overall probability of success.

Question 1:

## What is probability and how is it relevant in successive attempts?

Probability is a measure of the likelihood of an event occurring. It is relevant in successive attempts because it can help predict the chances of a certain outcome happening in a series of attempts.

Question 2:

## How is probability calculated for successive attempts?

The probability for successive attempts is calculated by multiplying the individual probabilities of each attempt. For example, if the probability of success for each attempt is 0.5, the probability of success in two successive attempts would be 0.5 x 0.5 = 0.25.

Question 3:

## What is the difference between independent and dependent events in terms of probability for successive attempts?

Independent events are those where the outcome of one attempt does not affect the outcome of the next attempt. In this case, the probability for successive attempts can be calculated by multiplying the individual probabilities. Dependent events, on the other hand, are those where the outcome of one attempt does affect the outcome of the next attempt, and the probability for successive attempts must be calculated using conditional probability.

Question 4:

## How can the concept of conditional probability be applied to successive attempts?

Conditional probability is used when the outcome of one event affects the outcome of the next event. In the context of successive attempts, this can be seen when the probability of success for the second attempt is dependent on the outcome of the first attempt. In this case, the conditional probability can be calculated by dividing the joint probability (probability of both events happening) by the probability of the first event.

Question 5:

## What are some common misconceptions about probability in successive attempts?

One common misconception is the belief in the "gambler's fallacy," which suggests that if a certain event has not occurred in previous attempts, it is more likely to occur in the next attempt. In reality, the probability remains the same regardless of previous attempts. Another misconception is that the probability of success increases with each successive attempt, but in reality, the probability remains constant for each attempt.

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