# Homework Help: Probability question for succession?

1. Mar 29, 2015

### Arnoldjavs3

1. The problem statement, all variables and given/known data
This isn't a homework question really... it's just something i'm not sure how to tackle.

Say you have a 44% chance of being successful on one try, and then another 44% to succeed on your second try, and then a 36% to succeed on your final try.
If you fail once at any instance, you have to start over again.
E.g.,
Trial 1: Success
Trial 2: Success
Trial 3: Fail -> Repeat until all three trials succeed

I believe this comes to a 7% of succeeding.
How many times will it take to be guaranteed(99% of succeeding all 3 tries)?

2. Relevant equations
Geometric Distribution(?)
Bayes Theorem(?) I haven't learned about anything relevant to this in school so this is purely new to me.
If someone could guide me to a method that would be ideal for these kind of situations.
3. The attempt at a solution
I believe that it would take around 14 attempts to have one successful attempt at succeeding all three trials..
I think if i were to go through 4 attempts 3 would succeed on passing the first trial, and two would succeed on making the second trial, however, none will succeed on all three trials.

If i am wording this in a confusing manner please let me know.

Edit: After using bayes theorem(i will use 14 attempts here, i think that 6 attempts will succeed on the first trial, 3 will succeed on the second, and 1 will succesfully complete all three trials.)

For the first trial:
14 * .44 - (.56^14) = 6.3 -> round to 6
Second Trial:
6 * .44 - (.56^6) = 2.6 -> round to 3
Third Trial:
3 * .36 - (.56^3) = 0.8 -> round to 1

Last edited: Mar 29, 2015
2. Mar 29, 2015

### Staff: Mentor

As expectation value, yes.

To consider multiple attempts, you don't need the individual trials any more, the combined chance of about 7% is sufficient.
What is the probability to fail all 14 (or n) attempts in a row? Once this gets below 1%, you have a 99% chance that at least one attempt was sufficient.

3. Mar 29, 2015

### RUber

When you round up, you are likely to underestimate the number of trials needed to get at least 99%.

4. Mar 29, 2015

### Arnoldjavs3

Okay thank you guys so much!

Alot easier than i had thought.

5. Mar 29, 2015

### Ray Vickson

Just to clarify: do the 44%, 44%, 36% apply also on your second round (and all subsequent rounds)? I assume so.

I think the problem could be a lot HARDER than you suppose! Here is the reason: round 1 could be over unsuccessfully on tries 1, 2 or 3. Whenever it is over unsuccessfully, you start again, etc. So, you could stop on try 4 (try 1 fails, next three succeed) or try 5 (try 1 OK, try 2 fails, next three succeed), or try 6 (tries 1,2 succeed, try 3 fails, next three succeed), etc. If you want information about trial number (rather than round number) you need to keep track of where in the round you lose.

One way is to make a Markov chain model, with states describing where in a round you are currently. These are
state 0: about to start another round (or start the whole process)
state 1: have just won trial 1 in the round
state 2: have just won trial 2 in the round
state 3: have just won trial 3 in the round---> stop: the game is over.

Say that each trial takes 1 time step. You start in state 0 at time t = 0 and you want to know how long it takes you to reach state 3. This will be a random variable (call it $T_{03}$), and you may be interested in aspects of $T_{03}$ such as: (i) its mean and/or variance; (ii) its probability distribution; (iii) the probability that it exceeds a certain given number; or (iv) a smallest value of $n$ so that $P(T_{03} = n) \geq p$ for some given $p$. I think this latter is what you want, with $p = 0.99$.

I could tell you more about the model and how to set it up, then how to do calculations with it, but if you have not had much probability material, a lot of it might seem very mysterious to you.

6. Mar 29, 2015

### Arnoldjavs3

I'm all ears for learning this, however, I have no background in any form for probability. I'd be surprised if any of this was not alienated to me. Just a question though, using this more complex model, i imagine there would be a more accurate answer obtained in comparison to the methods used above?

7. Mar 29, 2015

### Staff: Mentor

The right model depends on what exactly you want to calculate. Do you want the number of trials, or the number of attempts (using the labels from post 1)?

8. Mar 29, 2015

### Ray Vickson

The Markov model gives n = 100 as the smallest number of tries that give a win probability of >= 0.99. In other words, if each round except the last was 3 tries in length, you would need 33.33 rounds. Of course, some rounds take fewer than 3 tries, so the actual number of rounds would be > 33.33. If we wanted, we could work out the exact probability distribution of the number of rounds played in 100 tries. This number of rounds would be the number of visits to state 0 in 100 time steps.

The 100 try figure is what you need to have a win probability of 99% or better. Averages are much milder: the average number of tries until you win is 23.4389 ≈ 23.4. This would be about 7.81 full (3-try) rounds. At try number n = 24, your probability of having already won is about 0.65. Of course, this is far short of your desired 0.99, but it is at least a bit better than 50-50.

Last edited: Mar 29, 2015
9. Mar 29, 2015

### Arnoldjavs3

Right now what I want to figure out is the most optimal way of winning all three tries with the least amount of attempts. However, there are no outside factors influencing success and failure, they are merely chances predetermined. That is why I am trying to figure out how many attempts it would take to be guaranteed(99%) to win all three trials in one attempt.

So i re-read your post explaining markov's model, and I'm not sure if this is the model that would help me illustrate the entire situation(however I think it might, it seems like it could portray what may occur within each attempt)

So how about this, if i were to have X amount of players actually try to be successful on all three tries, how could i accurately portray this situation using markovs model?? (Say there are 25 players, or whatever suitable number to achieve that 99% of atleast one person succeeding.)

And yes, 44%, 44%, and 36% respectively on each subsequent rounds/attempts. These numbers do not change under any circumstance.

Edit: To be honest with you, I'm barely following along. I'd need to dedicate a bit of my own time on understanding this.

Last edited: Mar 29, 2015
10. Mar 29, 2015

### Ray Vickson

If the 44%, 44%, 36% win probabilities on tries 1,2,3 of each round are given to you and are unchangeable, there is no strategy you can use to decrease your number of tries. Your only possible strategy is "keep playing (and hope to win this time) or quit (and lose for sure)". The only way you could have a less simple strategy is if you could somehow get different win probabilities from the given 44%, 44%, 36%. One possible type of question might be that if you had to keep the 44,44 and 36, but could put them in a different order, what should you do? For example, is 44%,44%,36% better than 44%,36%,44% or 36%,44%,44%, etc?

There is nothing in the Markov model that you did not specify in your original description. In any round, you can lose (so start over again) at the end of trials 1, 2 or 3, but can win only at the end of trial 3. The Markov model takes care of that, under the assumption that the same 44,44,36 sequence applies in all rounds. It does no more and no less than what you asked for. The fact that you do not have the probability background to follow it is a different issue altogether.

11. Mar 29, 2015

### RUber

I agree that it could definitely be more complicated. However, the negative problem should be sufficient for this problem.
P(win) = .07, P(not win) = .93, .93^n < .01, find n.
I don't see how the Markov analysis adds additional information about number of games. Number of trials, sure.
Either way, the number of games required to ensure 99% chance of at least one success should end up to be the same.

12. Mar 29, 2015

### Arnoldjavs3

I'm doing a bit of research on Markov's model as of the moment, I'm trying to understand it through some youtube videos and I can see where it branches off. However the method mentioned above seems to be very straight to the point.

So... it would take about 63 attempts to actually succeed all 3 trials(For a 99% chance)?
That is ridiculously larger than my original estimate of 14. However my 14 wasn't theorized for a 99% chance, still it is rather up there.

I was honestly hoping to further my knowledge in this field(statistics, probability, whatever you may call it) on my spare time. Do you guys have any books or videos that you can guide me to? It would be helpful.

Last edited: Mar 29, 2015
13. Mar 29, 2015

### RUber

Make sure you are rounding up. If it is 63.1, you still would need to try 64 times to have at least 99%.
This is probability -- not statistics. If you get a good grasp on probability, you can move into statistics pretty easily.

14. Mar 29, 2015

### Ray Vickson

You can do a lot of what you want using the simple geometric random variable model; the Markov model comes into it only if you want reliable estimates of numbers of trials rather than numbers of rounds.

You have given us most of the basics already, but to keep the discussion focused I will summarize you approach here, and perhaps clean it up a bit. In the simple (geometric) model, the probability that a round ends in a win is p = 0.44*0.44*0.36 = 0.069696. The mean number of rounds until a win is 1/p = 14.348, and the number of rounds you must complete before being at least 99% sure of a win is 64. (For n = 63 the win probability is 0.98945, while for n = 64 rounds it is 0.99018.) You get this by requiring $P(N \leq n) \geq 0.99)$, where $N$ is the number of rounds and $P(N \leq n) = 1 - (0.930304)^n$. Here, 0.930394 = probability that a round does not end in a win = 1 - 0.069696.

Now for some things you did not give.

You can even finesse the simple model to get you into the right "ballpark" for estimating trials instead of rounds. Let the random variable L be the length of a round = number of tries in a round. We have $P(L = 1) = 0.44, P(L = 2) = (0.56)(0.44) = 0.2464, P(L = 3) = 1 - P(L=1) - P(L=2) = 0.3136$. The expected length of a round is
$$EL = 1 P(L=1) + 2 P(L=2) + 3 P(L=3) = 1.8736$$
Since on average we need 14.348 rounds to win and on average a round is 1.8376 tries, we would expect that the average number of tries needed to win is near (14.383)(1.8376) = 26.88246. This argument is not completely reliable, however, since the average of a produce (N*L) is usually not equal to the product of the averages. In this case the error is not too bad: the true mean number of tries is 23.4389, obtained from the Markov model. You could even try to estimate the number of tries to reach 0.99 win probability: since you need n = 64 rounds and each round is 1.8376 tries on average, you could expect to need somewhere near 64*1.8376 = 119.91 ≈ 120 tries. The actual number (from the Markov model) is 100 tries. The simple estimate is inaccurate because we should not really take a fixed number of tries per round; taking the average sweeps some probabilistic aspects under the rug. However, the simple estimate is certainly within the correct general area.

The Markov model adds bells and whistles to this, but at the moment you probably need not worry about it too much.

15. Mar 30, 2015

### Arnoldjavs3

This is extremely helpful. Knowing more information about the trials is actually interesting, because I was thinking a bit further on how the markov model may be needed more.
I was thinking about adding another element to this game that might make things a little more complicated. Let's say there are 8 players in this game, and so to say only 4 people make the second round. From there, out of those 4 people, only 1 succeeds and the rest have failed trying to win all three rounds. However, now out of those 3 people whom have failed, they all get to take one more attempt trying to win the third trial, but they are not resetting this time. They get to attempt winning the third trial yet again. Is this simply going to increase the success from 36% to 59% for those 3 players? Or would this actually affect the data on a deeper level?

Again I'd like to mention this is all out of my pure interest, as I have no experience in probability in any form. I would however like to indulge on it on my spare time (I may consider taking a few classes in my first year of university)

16. Mar 30, 2015

### RUber

It is difficult to say, since you are hypothesizing that they have already failed once. Since the second attempt would be an independent trial, you must assume that the probability of success is still 36%.
If you are speaking in terms of expected values (which is seems you are) you should instead start thinking in terms of probabilities. Since probabilities will determine your expected value and are more useful when desribing the model as a whole.

If you are changing the overall model to one where anyone who fails the first try at #3 gets a second try at #3, then you essentially have increased the odds of winning.
p(fail #1) = .56
p(pass #1, fail #2) = .44*.56 =.2464
p(pass #1, pass #2, fail #3, fail #3) = .44*.44*.64*.64=.0793
p(pass #1, pass #2, fail #3, pass #3)= .44*.44*.64*.36=.0446 (WIN)
p(pass #1, pass #2, pass #3)= .44*.44*.36=.0697 (WIN)
Note that this is for 1 game, and the sum of probabilities sum to 1 so you can be sure they account for the correct set of options.
Your net change is that odds of winning go from .07 to .114. Increasing the odds more that 50%.

17. Mar 31, 2015

### Ray Vickson

I found your explanation confusing, because you seem to be using "round" and "trial" differently in different places (but I am not sure). Sometimes the rules of the game were not spelled out clearly enough.

Just to clarify: let's say a "round" is 1,2 or 3 "tries" or "trials" (the tries having win probabilities of 0.44, 0.44, 0.36). So, four people survive to take their second round? How can that be? In your original description, nobody is ever kicked out of the game; if they lose a round they just start all over again. Are you are changing the rules now, so that only winners of the first round are allowed to stay active, and they are given the chance to play new rounds? Is that what you meant?

OK, so now we have 4 people taking round 2 (because all 4 won round 1?). Only one of the four wins all three tries in round 2; the others lose one of the three trials. What you did not say was whether all of those three won trials 1 and 2 and just lost trial 3 (which they then get two more chances to repeat?), or whether somebody who lost at trial 1 or 2 will also be given (2?) chances to do trial 3.

18. Apr 16, 2015

### Arnoldjavs3

I apologize for the lack of responses here on my part, I waas quite busy with school work. If you don't mind bringing this topic back up, I would like to clarify some things.

You're right about me not explaining well here. Originally I had the game set up where you could try endlessly until you succeed. One person could keep going until he failed. Now there are a specific amount of players, and I changed it up where if you lose at any instance of the game then you can not attempt again(with the exception of the 2nd round, if you fail the 2nd trial, you can take another attempt to reach the 3rd trial. But if you fail this trial again, then you lose the game completely.) So yes, only the winners of the first round are allowed to stay active, but they can only attempt the 2nd round for another time assuming they fail once. If they fail again trying to reach the 3rd, the game is over for them. Only one of the players wins all rounds succesfully, whilst the other 3 have failed. Those 3 are allowed to try again in attempts to winning the game. So if they lose, they are done.

So i was just wondering in this situation, accounting for the new rules in the game, how could I keep the number of players at the very minimum with one of them successfully winning the game(This is ignoring the previous situation where I had made up random numbers for who won what round)?
If you aren't clear with what I'm trying to say please say so. I can clearly see that I did not label the aspects of the game very well.

Edit: Just for an example to better illustrate what I'm trying to say here(no accuracy, all numbers are made up):
10 people play the game.

1) 5 of these 10 people fail the first trial. They lose the game completely and can not attempt again. The other 5 had passed, and they are the only ones playing the game now.

2) Out of these 5 people, only 1 of them had won the second trial, and has won the game. However, the other 4 are permitted to take another attempt. If they fail again, they lose the game completely.

3) Out of those 4 who failed once already, another 1 wins the game whilst the other 3 fails. The 3 can not try again.

4) Two people have successfully won the game where 1 person had won all trials straight, whilst the other took a second shot after failing the second trial.

Last edited: Apr 16, 2015
19. Apr 16, 2015

### Staff: Mentor

What happens in (2) or (3) if none, or multiple players win the second/third round?

To summarize: you have exactly 1 attempt for the first and third stage, but two attempts for the second stage? And you want to maximize the probability that exactly one player out of how many passes?
You cannot guaranteee that someone will win this way (unless you make all winners).

20. Apr 16, 2015

### Arnoldjavs3

Right, I'd like to find the minimum amount of players required to essentially have a 99% of one of them winning the game. But just for the sake of rules here, if no one wins then those two trials then that's that.

I re-read my original post and I can see why it's a little confusing here. I'll reconfigure such as so:
Say you have a 44% chance of being successful on one try, and then another 44% to succeed on your second try, and then a 36% to succeed on your final try.
If you fail once at any instance with the exception of trial 2, you lose the game.
E.g.,
Trial 1: Success -> proceed to trial 2
Trial 2: Fail
Attempting to win Trial 2 again: Success
Trial 3: Success -> You win the game.

So essentially players who win the first trial, but fail the next, have a second chance to attempt winning trial 2. If they fail they lose the game. I want to find the minimum amount of players I need to ensure that I have a 99% of one player winning every trial(it is okay for him to fail at trial 2)

I'm not sure if that's clarifying :|

21. Apr 16, 2015

### Staff: Mentor

That gives 0.44*(0.44+(1-0.44)*0.44)*0.36 probability to succeed for a player. The number of players can be calculated in the same way we calculated how often you have to try before.

22. Apr 16, 2015

### Arnoldjavs3

So that would be around 40 correct?

I had gotten a .108 as the answer to the expression you had given, and that yields a 89% of failing one attempt.

So of which around 40 tries would have been sufficient?

I have a much more complicated game that I'd like to ask how to accurately formulate probability for, but I feel as if it were inappropriate as the work required is quite hefty(Although I'm not asking for someone to do the work, rather just some guidance?)

Again this is all purely for my entertainment/curiosity.

23. Apr 16, 2015

Yes.