# Understanding more about probability? (beginner here)

1. Apr 23, 2015

### Arnoldjavs3

Hi, I am currently graduated from highschool and will be entering university in the upcoming fall, however out of pure interest I have gained a desire to understand more about probability. I have absolutely no background in any way for probability so understanding more complex theories may be quite troublesome, however I would still like to endeavor in this.

Basically I just have a few questions that I want to clarify with myself.

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 trial 1 when he is on his 9th attempt, however fails in attempts to beating the second trial. Due to this, he is forced start all over again.
Player 1 then fails with all his remaining attempts to beat the first trial, which leads to him losing the game.

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?

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.

I think its a bit stupid to label the chances of winning all three trials the same as winning only one. You aren't as likely to see the same result each time in a row.

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... Please excuse my ignorance where applicable

Last edited: Apr 23, 2015
2. Apr 23, 2015

### micromass

Staff Emeritus
Of course trial 2 is dependent on the outcome of trial 1. If you don't win trial 1, then you don't go to trial 2. So the trials are clearly dependent on eachother.

If you want to know the overal probability of winning the game, then you should model this as a Markov chain. That makes it very easy to determine the total outcome. So let me do that:
There are 4 states:
1) You did not yet win trial 1
2) You did not yet win trial 2, but you did win trial 1
3) You did not yet win trial 3, but you did win trial 2
4) You won the game.

The probability of going from state 1 to 2 (= winning the first trial) is 4/10. The probability of going from state 1 to state 1 (= not winning the first trial) is then 6/10.
The probability of going from state 2 to state 3 (= winning the second trial) is 4/10. The probability of remaining in state 2 is 6/10.
The probability of going from state 3 to state 4 (= winning the game) is 33/100. The probability of remaining in state 3 is then 67/100.
The probability of going from state 4 to state 4 is 1.

So we put all the information in a Markov transition matrix:
$$P=\left( \begin{array}{cccc} \frac{6}{10} & \frac{4}{10} & 0 & 0\\ 0 & \frac{6}{10} & \frac{4}{10} & 0\\ 0 & 0 & \frac{67}{100} & \frac{33}{100}\\ 0 & 0 & 0 & 1 \end{array} \right)$$

The probability after doing this 15 times is then

$$P^{15} = \left( \begin{array}{cccc} 0.00047 & 0.00047 & 0.038 & 0.956\\ 0 & 0.00047 & 0.011 & 0.988\\ 0 & 0 & 0.0024 & 0.997\\ 0 & 0 & 0 & 1 \end{array} \right)$$

So we see that the probability of going from state 1 to state 4 (= overall winning the game) in 15 tries is approximately 95.6%.

3. Apr 23, 2015

### micromass

Staff Emeritus
I did assume that if you won trial 1 then you never have to do it again, so you can just try trial 2 until the 15 trials run out.

4. Apr 23, 2015

### Arnoldjavs3

I rushed making this post and just now realized that there is a HUGE mistake in the example I used... I apologize for that.

Once player 1 had failed the second trial he had to start all over and failed winning the first trial all those successive attempts. I didn't read carefully.

However I want to thank you for replying here, someone else on this board had recommended the Markov's chain for this type of situation as well. My question(having no experience in this field) is how different is the reliability between the method you used vs using simpler methods(i.e., geometric distribution)? I am going to further look at some videos so I could actually understand how to develop the chain first of all, but from i can understand it explains the transitions between states more effectively?

5. Apr 23, 2015

### micromass

Staff Emeritus
Question: Is the following example accurate:

Round 1: Wins trial 1
Round 2: Wins trial 2
Round 3: Loses trial 3
Round 4 - Round 15: Loses trial 1

Is this something that you had in mind. Or is it the following:

Round 1: Wins trials 1 and 2, but loses 3
Round 2: Wins trial 1, but loses 2
Round 3 - Round 15: Loses trial 1

6. Apr 23, 2015

### Arnoldjavs3

The second example I guess is more or less what I'm trying to portray here.

So if you fail at any instance of the game, you have to restart. Once you use up all 15 attempts, you lose.

7. Apr 23, 2015

### micromass

Staff Emeritus
OK, then you're right and it is just a geometric distribution. So the probability of winning on one attempt is indeed about 55.6%.

8. Apr 23, 2015

### insightful

Note that there is a 55.5% chance of winning the game overall. Winning on your 15th attempt means first losing 14 times in a row.

9. Apr 23, 2015

### insightful

The answer is NO. The gambler's fallacy applies.

10. Apr 23, 2015

### insightful

This is explained by human emotion and expectation. It is easily proven that in an honest game, the initial circumstances have no effect on the current outcome.

11. Apr 23, 2015

### insightful

Now you know that this can be solved for one or more attempts. You also know the importance of being very precise in stating probability problems.