Proof that some of you have way too much spare time

[JEK595]

In order to get into a certain bank vault, you have to enter the correct 720 digit combination into a 1001 digit key pad, the correct 27 digit combination into a 11 digit key pad, and the correct 1930 digit combination into a 5 digit key pad. Assuming that the digits in a combination can be repeated as many times as needed, what are the chances that he will get the first combination right on his first try, the second combination right on his third try, and the third combination right on his fifth try? (First one to get this right gets a lifetime supply of egg products).

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-grace the mind with puzzles, games, and tapioca.-
-If at first you don't succeed, try, try again. (Not suitable for skydivers.)

[Chuck]

He has a 0% chance since I'm doing the dialing.

[dave10000]

Not enough information. Perhaps I *know* the combination, in which case my odds are a lot greater than if I didn't.

[mith]

1 in 1001^720*11^27*5^1930, of course.

(i agree with dave, though. you didn't even say who this "he" character is)

(and chuck too, if this is just a silly trick thing)

[This message has been edited by mith (edited 02-17-2002 02:09 PM).]

[DJC]

Agree with Mith if 'he' just keeps trying random combinations without stopping when successful or learning not to repeat unsuccessful combinations.

The first dial is OK - 1/1001^720 as Mith indicates.

For the second, if you mean two unsuccessful attempts (and not repeating these attempts) it gets messier, I think as follows:

(1-1/11^27)*(1-1/(11^27-1))*(1/(11^27-3)
ie: P{wrong on first}*
P{wrong on 2nd if wrong on first}*
P{right on 3rd if wrong on first two}

The third dial is messier again, but the same pattern over 5 terms, and I won't attempt to set it out (not sure I would want to win the prize anyway).

The aggregate probability is then the product of the 3 terms.

Chuck's solution is certainly more elegant. Alternatively, if the vault locks if a wrong combination is entered on the first or second attempt, the second and third components, and the aggregate, are zero. Regardless, 'he' will be a very old or very lucky man when he gets the vault open if he has no additional information. In practice, I cannot see him living long enough without incredible luck - I am not even sure the universe will live long enough. Of course, if he knows the combination it all depends on his accuracy and patience.

[tigg]

I think if we know how to enjoy our spare time, then we don't have too much of it.

[Lepton]

I don't even know that this is spare time. Some of us are being paid to do this =P