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Name: wildman jackson
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995

How could I figure out how many combinations are on a Master Lock, or a bike lock with 4 dials with 6 numbers on each dial? Is there a formula for figuring out the chances of getting the right combination?

Yes, this is a standard problem in combinatorics. Look under probability in many high school texts.


You could look this up in a book, but it is more satisfying to figure it out. Consider the simplest case, a lock with only one dial having, say, 6 numbers. How many combinations are there for such a lock? Clearly, 6. Now consider a lock with two dials, each dial having 6 numbers. For each choice of number on the first dial, we can have any of 6 different choices for the second number. Thus, this lock would have 6 * 6 = 36 lock combinations. Now consider a lock with 3 dials, each dial having 6 numbers. We just figured out that there are 36 ways to set the first two numbers; for each of these. I leave it to you to carry on this derivation to answer your question; and if you have understood this and see the pattern, you can immediately write down how many combinations there are for any such lock. By the way, if you did look for this in a book, you may have seen the term "combinations" in the context of "combinations and permutations"; that is something different from what you were looking for with this question.

Now the "odds" question. If there are N possible combinations, and you try one of them, the probability that it is the right one is 1/N, and the probability that it is a wrong one is (N - 1)/N. Now suppose the 1st try is a failure but the 2nd is a success; the odds of this are

[(N - 1)/N] * [1/(N - 1)] = 1/N again.

[The 1/(N - 1) factor comes from the fact that on the 2nd try there are N - 1 combinations to try, since you have tried one that does not work and, presumably, will not try it again.] The odds that you will succeed in one or two tries is the sum of the individual probabilities, because they are mutually exclusive events (that is, the first success cannot occur on BOTH the first and second tries). Generalizing to the case of a first success on the Kth try is straightforward.


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