Question

P_r_a_c_t_i_c_a_l_ _M_a_t_h

After repairs due to a minor car accident, power is lost to the radio system.
When power is restored, the radio requires a code to become operational once again.

The code required is a 5 digit code, where the digits are from the set 1,2,3,4.

If three successive guesses are incorrect the radio will not accept any further guesses until half an hour has elapsed.


Assume that guesses are made in a systematic fashion (no repeats), and that the time to enter each guesses is negligible.

How long can you expect it to take to guess the right code?

Answer 1

4^5 different 5 digit codes can be made by choosing digits from {1, 2, 3, 4}

Answer 2

Each iteration, we have \(3\) chances to test our luck and we're not repeating the already tested codes.
So the probability of success at \(n\)th iteration would be :
\[\dfrac{3}{4^5-3(n-1)}\]

Answer 3

I'm getting the expected time to be around 57 days
Is it any closer to the correct answer ?

Answer 4

i'm getting closer to 4 days

Answer 5

\[\frac{\left(\dfrac{4^5}3-3\right)}2\times0.5\ \text{hr}\div 24\ \text{hr}/\text d\]

Answer 6

Maybe i'll just take it to the mechanic.

Answer 7

Aren't you considering the outcomes are equally likely ?

Answer 8

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bn%3D1%7D%5E(342)+(3%2F(4%5E5-3(n-1)))*(30%2F(60*24)*(n-1))

Answer 9

Nvm, i see my mistake

Answer 10

why the sum?

Answer 11

does my equation seem reasonable?

Answer 12

I do assume that any of the possible guesses are equally likely.

Answer 13

I'm not sure how this can be approached:

Suppose you start at 12341, 12342....

ANY one combination is exactly the same probabilty as the other so there is the same chance you will get the right combination on the FIRST try as on the last.

I can't see that you can predict a time.

You might say that ON AVERAGE over a very large number of attempts to crack the code (I mean repeats of the WHOLE experiment) that teh time would be HALF the maximum time...

Maybe I've missed something..

Answer 14

I get 3.55... days on average.

That^ is working around the clock.

So just take it into the mechanic :-)

Answer 15

Yes @MrNood want the average, ( i.e. the expectation value )

Answer 16

\(\color{#0cbb34}{\text{Originally Posted by}}\) @UnkleRhaukus
Or is this it

\[\frac{\left(\dfrac{\text{total number of possibilities}-\text{free guesses}}{\text{guesses per go}}\right)}2\times\text{time per go}\]
\[\frac{\left(\dfrac{4^5-3}3\right)}2\times0.5\ \text{hr}\div 24\ \text{hr}/\text d \approx 7.09\ \text{day}\]
\(\color{#0cbb34}{\text{End of Quote}}\)
I got 7.1111

Answer 17

i don't know if my formula is right or not

Answer 18

This is just an assumption, but isn't after trying out one code and getting it wrong, the probability of others will increase? Like if I tried all combinations except one and got all wrong, the probability of the leftover one will increase to 1. Is this thing correct?

Answer 19

I guess maths isn't the practical way to fix my car after all.

Answer 20

The average is not the 'expected time' it is simply the average of many complete experiments (i.e. different codes each time'

In any ONE trial the expected time is equally likey to be 0 up to the max 1024/3 tries

Answer 21

The correct code is just as likely to occur at any position in the range: 1/4^5.
The best prediction, is that the correct code will be found after about half of the options are tried. This is the best prediction because it is just a likely to occur before this point, as is likely to occur after this point.

This is a question about the theoretical probability, and not the experimental probability.

Answer 22

I disagree

There is no basis for a prediction of the outcome of a single experiment.

The question was 'How long can you expect it to take to guess the right code? '

The answer to that is that you should expect it to take anything from 0 to the maximum time, with no way of estimating a more precise probable time.

Answer 23

If I cast two dice (and add the numbers that land face up).
I expect to get a sum of 7, with a standard deviation of about 2.4

Answer 24

If I crack the radio code, by guessing ever possible combination until success (brute force).

I expectation it will take 512 guesses. With a standard deviation of about 295.9

Answer 25

which means that the probability of solving it in 1 try is the same as solving in 512

(This is not a normal distribution - so the SD does not give you a predicition of increasing probability as you approach the mean)

Answer 26

However - if you choose to believe the mean is the 'expecte4d value' then I cannot argue further

Answer 27

This is a discrete uniform distribution.
The standard deviation predicts how far from the mean the right number is likely to be.

The mean is also know as the expectation value.

Answer 28

Experimentally, the correct radio code has been determined as:
`1` `4` `1` `2` `1`
Thanks for all the help y'all.

Answer 29

Thanks for asking :-)

Answer 30

@UnkleRhaukus

quote
"This is a discrete uniform distribution.
The standard deviation predicts how far from the mean the right number is likely to be. "

Since the right number is equally likely to be any distance (within the bounds) from the mean please would you point me at some documentation that illustrates how to calculate this using the SD?

( I note that the bounds are only approx +/- 1.6 SD from the mean - the number CANNOT be more than 1.6 SD from the mean and has a hard 'cutoff'.)

There is no reduction in probability of a success for numbers further from the mean.

(BTW - referring to the 2 dice point - the value of 7 IS more likely than other values since there are more combinations that add to 7 than for instance 2 or 12.) The code has equal probability of ALL answers WITHIN the bounds of 0 to approx 1024