Mixing
Giving out my pin 2 digits at a time…
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If I have a 4 digit pin and, every time I phone an organisation, they ask for 2 of those digits, what is the probability they have my complete PIN after 3 or 4 calls?
For "my security" I would also like to assume that they can't ask for the same two digits two calls in a row (in case I was overheard on call 1).
Over two calls I think this is 6(no of ways to give out complete pin) / 8 (no of 2 digit combinations) x 7 (no of 2 digit combinations with no repeat).
So the bit I am struggling with is working out the no of ways to give out a complete PIN over 3 or 4 calls.
probability
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
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add a comment |
$begingroup$
If I have a 4 digit pin and, every time I phone an organisation, they ask for 2 of those digits, what is the probability they have my complete PIN after 3 or 4 calls?
For "my security" I would also like to assume that they can't ask for the same two digits two calls in a row (in case I was overheard on call 1).
Over two calls I think this is 6(no of ways to give out complete pin) / 8 (no of 2 digit combinations) x 7 (no of 2 digit combinations with no repeat).
So the bit I am struggling with is working out the no of ways to give out a complete PIN over 3 or 4 calls.
probability
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
$endgroup$
add a comment |
$begingroup$
If I have a 4 digit pin and, every time I phone an organisation, they ask for 2 of those digits, what is the probability they have my complete PIN after 3 or 4 calls?
For "my security" I would also like to assume that they can't ask for the same two digits two calls in a row (in case I was overheard on call 1).
Over two calls I think this is 6(no of ways to give out complete pin) / 8 (no of 2 digit combinations) x 7 (no of 2 digit combinations with no repeat).
So the bit I am struggling with is working out the no of ways to give out a complete PIN over 3 or 4 calls.
probability
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
$endgroup$
If I have a 4 digit pin and, every time I phone an organisation, they ask for 2 of those digits, what is the probability they have my complete PIN after 3 or 4 calls?
For "my security" I would also like to assume that they can't ask for the same two digits two calls in a row (in case I was overheard on call 1).
Over two calls I think this is 6(no of ways to give out complete pin) / 8 (no of 2 digit combinations) x 7 (no of 2 digit combinations with no repeat).
So the bit I am struggling with is working out the no of ways to give out a complete PIN over 3 or 4 calls.
probability
probability
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
asked Jan 4 at 14:57
James StubbsJames Stubbs
31
31
add a comment |
add a comment |
1 Answer
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$begingroup$
You might as well assume they ask for the first two digits on the first call. They get the whole PIN in two calls if they ask for the last two on the second call, which is one choice. There are ${4 choose 2}-1=5$ pairs of digits they can ask for on the second call because they cannot ask for the first pair again, so the chance on two calls is $frac 15$.
To get the PIN specifically on the third call, the first gets digits $1$ and $2$, the second must repeat one of $1$ and $2$ and give one more, which might as well be $3$. That is a chance of $frac 45$. Then they must get the fourth digit on the next request, which is chance $frac 35$ as one combination missing the fourth digit is not allowed. The chance is then $frac 45 cdot frac 35=frac {12}{25}.$ The chance of getting the whole PIN by the third request is then $frac 15 + frac {12}{25}=frac {17}{25}. $
I leave the four request case to you.
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
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1 Answer
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$begingroup$
You might as well assume they ask for the first two digits on the first call. They get the whole PIN in two calls if they ask for the last two on the second call, which is one choice. There are ${4 choose 2}-1=5$ pairs of digits they can ask for on the second call because they cannot ask for the first pair again, so the chance on two calls is $frac 15$.
To get the PIN specifically on the third call, the first gets digits $1$ and $2$, the second must repeat one of $1$ and $2$ and give one more, which might as well be $3$. That is a chance of $frac 45$. Then they must get the fourth digit on the next request, which is chance $frac 35$ as one combination missing the fourth digit is not allowed. The chance is then $frac 45 cdot frac 35=frac {12}{25}.$ The chance of getting the whole PIN by the third request is then $frac 15 + frac {12}{25}=frac {17}{25}. $
I leave the four request case to you.
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
You might as well assume they ask for the first two digits on the first call. They get the whole PIN in two calls if they ask for the last two on the second call, which is one choice. There are ${4 choose 2}-1=5$ pairs of digits they can ask for on the second call because they cannot ask for the first pair again, so the chance on two calls is $frac 15$.
To get the PIN specifically on the third call, the first gets digits $1$ and $2$, the second must repeat one of $1$ and $2$ and give one more, which might as well be $3$. That is a chance of $frac 45$. Then they must get the fourth digit on the next request, which is chance $frac 35$ as one combination missing the fourth digit is not allowed. The chance is then $frac 45 cdot frac 35=frac {12}{25}.$ The chance of getting the whole PIN by the third request is then $frac 15 + frac {12}{25}=frac {17}{25}. $
I leave the four request case to you.
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
You might as well assume they ask for the first two digits on the first call. They get the whole PIN in two calls if they ask for the last two on the second call, which is one choice. There are ${4 choose 2}-1=5$ pairs of digits they can ask for on the second call because they cannot ask for the first pair again, so the chance on two calls is $frac 15$.
To get the PIN specifically on the third call, the first gets digits $1$ and $2$, the second must repeat one of $1$ and $2$ and give one more, which might as well be $3$. That is a chance of $frac 45$. Then they must get the fourth digit on the next request, which is chance $frac 35$ as one combination missing the fourth digit is not allowed. The chance is then $frac 45 cdot frac 35=frac {12}{25}.$ The chance of getting the whole PIN by the third request is then $frac 15 + frac {12}{25}=frac {17}{25}. $
I leave the four request case to you.
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
$endgroup$
You might as well assume they ask for the first two digits on the first call. They get the whole PIN in two calls if they ask for the last two on the second call, which is one choice. There are ${4 choose 2}-1=5$ pairs of digits they can ask for on the second call because they cannot ask for the first pair again, so the chance on two calls is $frac 15$.
To get the PIN specifically on the third call, the first gets digits $1$ and $2$, the second must repeat one of $1$ and $2$ and give one more, which might as well be $3$. That is a chance of $frac 45$. Then they must get the fourth digit on the next request, which is chance $frac 35$ as one combination missing the fourth digit is not allowed. The chance is then $frac 45 cdot frac 35=frac {12}{25}.$ The chance of getting the whole PIN by the third request is then $frac 15 + frac {12}{25}=frac {17}{25}. $
I leave the four request case to you.
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
answered Jan 4 at 15:08
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
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