Probability problem

An ATM personal identification number consists of four digits, each a 0, 1, 2, ..... , 8, or 9.

The following choices are prohibited:
(I) All four digits identical.
(II) Sequences of consecutive ascending or descending digits.
(III) Any sequence starting with 19.

Someone has stolen an ATM card and knows that the first and last digits are both 1. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try. What is the probability that he gains access to the account?

My answer:

We are looking for the probability that the number of tries, before he gains access to the account, is less or equal to three. Thus,

P(access gained) = P(access gained on 1st try) + P(access gained on 2nd try) + P(access gained on 3rd try)

On the first try we have (9)(10) - 1 = 89 possible choices because he knows he cannot pick 9 for the 2nd digit, and he cannot pick 1 for the both the 2nd and 3rd digit.
On the second try we have 88 possible choices because he will not pick the same choice twice.
On the third try we then have 87 possible choices.

Thus,

P(access gained) = 1/89 + 1/88 + 1/87 = approx. 0.0341

The answer given in my text is 0.0337. Am I thinking about this wrong?

KyleM42090 said:

An ATM personal identification number consists of four digits, each a 0, 1, 2, ..... , 8, or 9.
The following choices are prohibited:
(I) All four digits identical.
(II) Sequences of consecutive ascending or descending digits.
(III) Any sequence starting with 19.
Someone has stolen an ATM card and knows that the first and last digits are both 1. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try. What is the probability that he gains access to the account?

Click to expand...

You did not consider condition II. Did you?
By that condition, 2 cannot be the second nor third digit. WHY?
What other pairs cannot be used as the second & third digits?
Can 1131? be a valid pin? Why or why not?

Play with some numbers and get
1/89 + 1/89 + 1/88 ~ 0.0338355464759959
so one of those numbers has to be smaller (the eighty something has to be larger). So
1/89 + 1/89 + 1/89 ~ 0.0337078651685393


Would seem to me that the text answer does not consider all of the conditions unless the interpretation of "then randomly selects a different pair of digits for the second try" is "does a different random selection of a pair of digits" but that seems a little far fetched to me.