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Questions about probability (successive attempts?)

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    ??

    3. 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...
     
  2. jcsd
  3. Apr 22, 2015 #2
    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"?
     
  4. Apr 22, 2015 #3
    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.
     
  5. Apr 22, 2015 #4
    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.
     
  6. Apr 23, 2015 #5
    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 eachother.
     
    Last edited: Apr 23, 2015
  7. Apr 23, 2015 #6
    Maybe they do and maybe they don't but that doesn't mean automatically the probabilities are the same.
     
  8. Apr 24, 2015 #7

    haruspex

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    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.
     
  9. Apr 24, 2015 #8
    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?
     
  10. Apr 24, 2015 #9

    haruspex

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    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)?
    What joint probability?
     
  11. Apr 24, 2015 #10
    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 whats 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?
     
    Last edited: Apr 24, 2015
  12. Apr 24, 2015 #11

    haruspex

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    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.
     
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