Geometric distribution with locks and keys

Asked Sep 18 '20 at 20:42

Active Sep 18 '20 at 20:42

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Assuming you have 8 keys - only one of them opens the lock. You try one by one without knowing which is the right key. If the key does not open the lock - you remove it. You do this until you find the right key that opens the lock. What is the distribution of the number of your tries until (and including) the opening of the lock.

Geometric
Uniform
No answer is correct
Binomial
HyperGeometric

I think the answer is "No answer is correct" because all these distributions (not including HG) is when you don't change the number of keys each "cycle" - and here you do! (You remove the key if it does not open) so your $p$ changes... and HG - I don't see how it can help here...

But my answer is wrong! It is not "No answer is correct"

Why is that? Which is answer is correct? this goes against my intuition. Thanks!

asked Sep 18 '20 at 20:42

CSch of x

2

Think about uniform. The proper key can be any one of the eight, right? So is it more likely to be the first or the eighth? The second or the fifth? And so... – David G. Stork Sep 18 '20 at 20:53
2

Each key has equal chance to be the right one – Peter Franek Sep 18 '20 at 20:54
Hey thanks! But what about the fact we remove some each iteration? why that does not matter? thanks! – CSch of x Sep 18 '20 at 20:54
1

Exactly... and so what does this imply about the distribution? – David G. Stork Sep 18 '20 at 20:54
1

So imagine you don't remove it. You just enumerate them and go 1,2,3,...,8. It's the same thing. – Peter Franek Sep 18 '20 at 20:55
1

Related Probability Problem with $n$ keys – Sil Sep 18 '20 at 20:55
1

Thank you so much all ! :) – CSch of x Sep 18 '20 at 20:56

Geometric distribution with locks and keys

0 Answers0