Maths Probability Conundrum

[Efbe]

Ok, I'll start off by saying I know the answer, but I don't know how to work it out!

Here is the conundrum:
If one number of a 10 digit Pin-Pad is broken, what percentage of people will be affected entering in their 4 digit pin, given they have been assigned completely random numbers?


I know the answer because I can easily calculate it in excel, fill a series with the numbers 0000-9999, and perform a calculation on how many times a number is shown. the answer is 3439/10000 = 34.39%
But when I try to calculate the answer using probability, I get 40%, using the following equation:
1/10 + 1/10 + 1/10 + 1/10 = 4/10
so this isn't right. Where am I going wrong, and how would you express the answer as an equation?

[wormburner]

I think you're miscounting those PIN which contain more than one 4?

[trashbat]

The tests are not independent. Think of it as a hurdles race where knocking over a hurdle disqualifies you.

1/10 people have their first number as the broken digit. They are disqualified. Add 1/10.

9/10 people remain. Of those, 1/10 have their second number as the broken digit. Add 9/10 * 1/10.

9/10 of *those* people remain. Of those, 1/10 have their third number as the broken digit. Add 9/10 * 9/10 * 1/10.

etc

So:

0.1 + (0.9*0.1) + (0.9*0.9*0.1) + (0.9*0.9*0.9*0.1) = 0.3439 = 34.39%

Edited by trashbat on Wednesday 15th August 11:04

[ewenm]

Easier way to do it:

You can easily work out how many people have a pin NOT containing the faulty key:
9/10 * 9/10 * 9/10 * 9/10 = 0.6561

So the chance of a person having a pin containing the faulty key:
1 - 0.6561 = 0.3439 = 34.39%

Edit to correct a typo

Edited by ewenm on Wednesday 15th August 11:08

[Hugo a Gogo]

you might be double counting people who have the 'broken number' twice?

[Vipers]

My head hurts ............

[Efbe]

thanks a lot guys, all makes sense now

probability is probably covered pre-GCSE, so think I need to go back to school.

[wormburner]

ewenm said:

Easier way to do it:

You can easily work out how many people have a pin NOT containing the faulty key:
9/10 * 9/10 * 9/10 * 9/10 = 0.6561

So the chance of a person having a pin containing the faulty key:
1 - 0.6561 = 0.3439 = 34.39%

Edit to correct a typo

Edited by ewenm on Wednesday 15th August 11:08

This is an elegant way to do it.

[IainT]

Did anyone account for invalid PINs? (e.g all digits the same, sequences, etc.)

[ewenm]

IainT said:

Did anyone account for invalid PINs? (e.g all digits the same, sequences, etc.)

No.

[Simpo Two]

You're also assuming (I think) that they have four *different* numbers. That's not random - because otherwise there's a probability that people will have the broken number twice, three times and eventually all four. IIRC the phrase is 'with replacement' - in which case P(4 duds) is 0.1 x 0.1 x 0.1 0.1 and that (and the others) must be allowed for.

I think...

[LordGrover]

ewenm's solution takes that into account AFAIK. Each press is a 1:10 chance of the faulty button.

[Alfanatic]

Simpo Two said:

You're also assuming (I think) that they have four *different* numbers. That's not random - because otherwise there's a probability that people will have the broken number twice, three times and eventually all four. IIRC the phrase is 'with replacement' - in which case P(4 duds) is 0.1 x 0.1 x 0.1 0.1 and that (and the others) must be allowed for.

I think...

I don't think that matters. If they have the broken number twice or more, they're equally as disqualified as someone having it only once. The result ties up with the Excel test, which sounded like it included every possible combination, not just single appearances, and the sequential calculation posted earlier by Trashbat removes people from further consideration as soon as the first broken digit appears in their number, so they don't get double counted if they have the broken digit twice. This is how it should be.

EDIT: Or maybe I'm reading it wrong and you're giving a reason why the OP's calculation didn't match the Excel answer, in which case I agree with you.

Edited by Alfanatic on Friday 24th August 12:23