P(correct at Nth step) = P(not correct in first N - 1 steps) * P(correct in last stage)

Case 1: If codes are repeated then

P(correct at Nth step) = [1 - (1/10)^4]^(N-1) * (1/10)^4.

Case 2: If codes are not repeated in any trail

P(correct in 1st trial) = (1/10)^4 = 1/10000.

P(correct in 2nd trial) = 9999/10000 * 1/9999 = 1/10000.

P(correct in 3rd trial) = 9999/10000 * 9998/9999 * 1/9998 = 1/10000.

and so on.

So in this case for any N, P(correct in Nth trial) = 1/10000.

P(correct at Nth step) = P(not correct in first N - 1 steps) * P(correct in last stage)

Case 1: If codes are repeated then

P(correct at Nth step) = [1 - (1/10)^4]^(N-1) * (1/10)^4.

Case 2: If codes are not repeated in any trail

P(correct in 1st trial) = (1/10)^4 = 1/10000.

P(correct in 2nd trial) = 9999/10000 * 1/9999 = 1/10000.

P(correct in 3rd trial) = 9999/10000 * 9998/9999 * 1/9998 = 1/10000.

and so on.

So in this case for any N, P(correct in Nth trial) = 1/10000.

Probability for noncorrect number in first attempt is 9999/10000; probability for noncorrect number in second attempt is (9999/10000)*(9998/9999);

probability for noncorrect number in k-th attempt is product of ratios (10000-j)/(10000-j+1) for j from 1 to k-1;

Finaly, probability for noncorrect number in N-1 attempts is

P(X=N-1)=Product((10000-k)/(10000-k+1), k=1,N-1)

Probability for correct number in N-th attempt is 1-P(X=N-1).

The statement

"Probability for correct number in N-th attempt is 1-P(X=N-1)."

is false. It is correctly to say:

"Probability for correct number in N-th attempt or in some later one is 1-P(X=N-1)."

Indeed, Zoran's random variable X means the number of first consecutive failed attempts. Let Y denote the number of the first succesful trial. Hence, {X=N-1} and {Y>N-1} are equivalent.

Dear all,

I am happy to share you my answer,

I assume that I have the following event E: the outcomes of all random trials F, F, F,....F (N-1) times, then S on the Nth - trial, Where F for fail and S for success. All trials are independent.

As all of you agree that:

The probability P(F) = 9999/10000 , and P(S) = 1/10000.

Therefore P(E)=(9999/10000)N-1 .1/10000 = (9999)N-1 /(10000)N .

Best wishes

@ Issam

this is a rediscovered Case 1 by Bharat:)

Regards, JoD

@Joachim

Yes , Thank you .

@Peter

You are right.

But we need to answer the question as it is, regardless of any improvement of the case under consideration.

Best regards

OK, Peter,

Your objections and/or suggestions are very challenging! So, let's start!

First: what is a locker? A very simple mechanical device, supposed to have 10^4 positions MORE OR LESS STABLE. However, they are not stable. If someone has a well developed sensitivity in fingers, he/she can feel some differences between different positions. Sometimes it helps in excluding some of the positions as appropriate with HIGH PROBABILITY. Thus, one can assume some suitable features of the probability.

Second: Thus, what should be put into the model to calculate the most probable positions opening the locker? I would strart with experiments with a goal to discover for each ring the best positions, when we feel that they are most stable. This would give us a set of 3-4 positions for each ring. I think it is attainable by circa 10-15 moves per ring = less than say 55.

Three. After this step we come to a choice one of at most 4^4 positions, which - if you would agree with the objected ways of calculations - have mean value equal to 128. The implied pd of the random number in question is uniform over the assumed set from 56 till 56+127=183.

Four. Taking into account that the people are not that stupid, some of them will be well described by a triangle pd.

Five. Currently, my final answer is: the probability that the random number N of trials for opening a 10-position 4-ring locker equals n is given by the formula:

C \cdot max{ 64.5 - |n - 119,5| ; 0 }, where C is a norming coefficient.

I guess, dear Peter, that again you will object this way, as well. And you will be right! Thus, perhaps we should stop to build any probabilistic models? My answer is NO, WE NEED PROBABILISTIC MODELS AT LEAST FOR PREDICTING RESULTS OF OUR HUMAN KIND ACTIVITY. This tendency is encrypted in our behaviour, thus even if try to stop modelingor call people to stop modelling with the use of probability, some of us will always do it!

Best regards, Joachim

There are many possible models of the same problem, which are giving diffrent results. Thus the right answer to this thread's quection is: it depends. The only mathematical requirement is, that the assigned probalities are non-negative numbers which sum up to one. That's everything what mathematics has right to say. All additional features sooner or later become just an opinion of the author of the accepted model.

Best regards, Joachim

Superb conversation. very useful.

words, words, words. The meaning of probability in applications is not the subject of this thread. The mathematical meaning is already answered. Giving example with the tringle pd I have just turning the attension at many other possibilities than just the celebrated uniform one. The acceptance for application in this case was conditioned by the possibility of sensing individual tumbler position. Thus Peter's final claim is not opposing my suggestion. Thus: why is it presented as an oppsite opinion?

I have to stress, that acceptance for applications is not a subject of philosoby, by a subject of statistical analysis, which - what is funny - is also a subject of our acceptation. Any additional notions like Peter's universes makes for me no sense, since they just increase the number of objects of the theoretical model which should help us in predicting the real events, especially in mass service.

PS. An interesting discussion about the different understanding of probability has been led few months ago at

https://www.researchgate.net/post/is_possibility_distribution_related_to_probability_distribution_expressly

I have explained there my believes, which are not exactly those assigned to me by Peter:

I have got a good place for presenting my almost unacceptable point of view, which is:  THERE ARE NO PROBABILITIES IN THE REAL WORLD. However, I like working with a machinery called Calculus of Probabilities. Even more, I claim it is usefull. To answer the current question is really a hard job, since it relates notions from different stories. The mathematical meaning of probability is nicely explained by Jochen. Any meaning of probability outside mathematics requires special definitions, dependent on the culture in the wide sense of particular social group (say, nation) [ or Peter's universe:-) ]

Regards, Joachim

Peter, Your modeling of the probability via the entropy has nothing to do with the reality. Since the probabilities DO NOT EXIST. Thus, everything what count is the accepted assumption. Who told you that we HAVE to apply maximum entropy rule to the real experiments? However, I agree that if we model the "no primary knowledge" = "maximum of the entropy, then you are getting uniform pd. My assumption about some cases is o take suitable triangle pd, and you cannot oppose it unless you want to compare it to results of some real experiments. We can start a game do decide which assumption is better fitting the results. BUT even then, if we want to esteem each other, we come to a moment that we agree on common criteria of the better fitting(s). IN EVERY CASE THIS IS OUR CONVENTION which mathematical probability will be assigned to the real events. There is no mathematical proof that for the throwing "ideal cube" the probability of each result is 1/6. This is simply our commonly accepted model of what some call the chance. Without references to repeatable experiments nobody can prove this convention. The last opinion is justified also with respect to nonrepeatable experiments like the existence of life on Mars, provided we have some indications about creation of life or about the ways it (the life) can be spread over the Universe. A simple explantionof the possibility gives counting probabilities on the result of throwing two dyes, IF we agree e.g. with product measure. We do not have to model for the pair (x1,x2) of the results since then it sufficies to know p(x1) and p(x2) and simply multiply them. BUT THE CONVENTION OF THE PRODUCT FORM IS AGAIN AN (ADDITIONAL) ASSUMPTION, possibly obtainable from the maximum etropy rule, whenever accepted.

Regards, Joachim

PS. I don't like to be just a label :), and even any icon :) , nor just J :). My name is Joachim (etc.), my pseudo is JoD.

Trying to apply Shannon's theorem to the real world (I don't know if this is the same as Peter's universe) one has to assume that the probabilities assigned to the real events follow the maximum entropy postulate. Shannon theorem is not proving this postulate, because it is from a different world build in our minds, only.

With respect to Peter's question about position of the peak: I am not saying that my triangle probability distribution possesses a peak; it possesses equal maximal values at two different points. Is there any obligation to possess exactly one point of maximal probability.

@Peter, don't be kidding our community. The current state of our positions is clear. You are voting for application of the mximal entropy rule in choosing the probabilistic model; I am opposing this, since this leads to a determined pd, which could be controversial. The second issue is that you believe in existence of real probabilities. I do not. But I see usefulness of the calculus of probabilities in organizing our activity. My negation is motivated by the following simple examples:

1. What gives me the probability of a win in a game with n possible results with one winning, if I play it once, only? It does not matter whether the probability is 1/n or 3/4. After I played, the only iterest is if I am winning or I am not. How about if I don't know the number of cases? Would you say therefore, that the probability served by the universe (whatever it is) equals 1/2?

2. How to use Shannon theorem to obtain the well accepted geometric distribution for the Case 2 introduced by Bharat. To the best of my knowledge, if the pd of maximal entropy is sought among those with given expectation - you will get the geometric pd. But if the pd is sought among those with given second moment, then it is not a geometric sequence. Thus, for obtaining the well accepted geometric pd with ratio 9999/10000 you need the expectation 10000. You cannot get it from the Shannon theorem. How then would you guess it? And why should you use the first moment as the (unique) given data, and reject that you know e.g. the second moment?

Regards, Joachim

@Peter ,@Joachim : Fascinating debate.

Thank you, Adriana, for creating this opportunity.

Peter > The revised answer is that your present and future in the universe is fixed

I see that I am really fixed by the universe that I didn't notice the smooth change of the discussion from science to new science. Thanks, I am quiting.