This curious fact actually ocurred to me yesterday. It never happened before.

To grant access to the operations provided by the ATM (Automatic Teller Machine) of my bank, I have to type (touching the screen) the password of my magnetic card.

The screen which appears to me shows five "buttons", each one of them labeled with two digits, from 0 to 9. For example, in the first there are the numbers 0-2, in the second, 3-7, in the third, 4-5, in the fourth, 1-9 and in the fifth, 6-8. So, if my password, that consists of 6 digits, not necessarily different, is 123456, I touch, in order, the fourth button (1), the first (2), the second (3), the third (4), again the third (5) and the fifth (6).

The numbers that appear in each button change daily, and yesterday I noticed that to enter my password, I touched only two buttons. To clarify, with the configuration above, if my password were "357457", I would have touched the second button, the third, the second, the third, again the third, and the second.

What is the probability that this occurs, that is, that I have to touch only (and exactly) two buttons to enter my password, if it is made of a) 2 different digits; b) 3 different digits; c) 4 different digits?

If there are 2 different digits, then we are looking for the probability that they are on two different buttons.  Select the lower digit.  The probability that the other digit is not on the same button is 8/9.

If there are four different digits, then we are looking for the probability that they are paired on two different buttons.  Select the lowest digit.  The probability that one of the other three is on the same button is 3/9.  Now select the lowest digit that is not on that button.  The probability that the 4th digit is on the same button as the "third" is 1/7.  The probability of both of these things happening is (3/9)*(1/7) = 1/21.

If there are 3 digits, let's figure out the probability that they are on 3 different buttons.  Select the lowest.  The probability that the next number is on a different button is 8/9, and the probability that the third digit is on yet another button is 6/8.  The probability that they are on three different buttons = (6/8)*(8/9) = 2/3.  So the probability that they are on exactly two buttons is 1 - (2/3) = 1/3.

Posted by Steve Herman on 2008-08-11 10:58:59