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Math problem. Help plzzzzzzz!!!!!

A machine gives a uniformly random real number between 0 and 1 when you press a button. You press the button the first time, and you get the first number. Then you keep pressing the button until you get a number smaller than the first number. After the first number, what is the expected (average) number of times you press the button?

Reply 1

What level is this, what have you tried?

Reply 2

Yeah, but i’m not sure about the answer. I think that the answer is 2.

Reply 3

Original post
by u4ria
Yeah, but i’m not sure about the answer. I think that the answer is 2.

Again, can you say what level it is and what have you tried? The sticky at the top of the forum says you shouldnt just expect an answer/solution.

Reply 4

Original post
by mqb2766
What level is this, what have you tried?

I don’t know what level it is. But i got a little bit confused at first.

Reply 5

Original post
by u4ria
I don’t know what level it is. But i got a little bit confused at first.

What qualification are you studying?

Reply 6

Original post
by mqb2766
Again, can you say what level it is and what have you tried? The sticky at the top of the forum says you shouldnt just expect an answer/solution.

Yes. I tried so that i say i think the answer is 2. But I do not know how to show my work here!! The point I do not understand is that: i chose a certain value for example x. When the first time I press the button and get a number greater than x, then i press the button second time. Is the expected additional time i press the button in the second time is still same as the original state?

Reply 7

Original post
by sqastudenthi
What qualification are you studying?

IB, but this is definitely a question beyond our syllabus.

Reply 8

It’s probably infinity, given that x∈R and that 0<x<1 then the machine will continue repeating numbers. as x cannot be 0 then x will get infinitesimally close to 0, but will never be 0. This link here https://www.desmos.com/calculator/ngqtlmf7gf will help you understand the problem consider that x is the amount of times you click the button y will be the output value, but it won’t be 0.

Reply 9

Original post
by sqastudenthi
It’s probably infinity, given that x∈R and that 0<x<1 then the machine will continue repeating numbers. as x cannot be 0 then x will get infinitesimally close to 0, but will never be 0, then the number will be closer every time so it will be like 0.0… 1 but it will not be 0. This link here https://www.desmos.com/calculator/ngqtlmf7gf will help you understand the problem consider that x is the amount of times you click the button y will be the output value, but it won’t be 0.

Why will x gets closer to 0?

Reply 10

Does this mean that i can think of the whole case as geometric distribution?

Reply 11

Original post
by sqastudenthi
It’s probably infinity, given that x∈R and that 0<x<1 then the machine will continue repeating numbers. as x cannot be 0 then x will get infinitesimally close to 0, but will never be 0. This link here https://www.desmos.com/calculator/ngqtlmf7gf will help you understand the problem consider that x is the amount of times you click the button y will be the output value, but it won’t be 0.

Now I got the idea by thinking it as a geometric distribution. Thank you so much =)

Reply 12

Original post
by u4ria
Why will x gets closer to 0?

Because as 0<x<1 then x can’t be 0 but as the number gets smaller and smaller it will get closer to zero but based on the fact on the sentence above, it is impossible to be 0 as x can only be greater than 0

Reply 13

Original post
by u4ria
Now I got the idea by thinking it as a geometric distribution. Thank you so much =)

Your welcome.

Reply 14

Original post
by sqastudenthi
Because as 0<x<1 then x can’t be 0 but as the number gets smaller and smaller it will get closer to zero but based on the fact on the sentence above, it is impossible to be 0 as x can only be greater than 0

Oh! I got it! Thanks.

Reply 15

My teacher led us to think of a very simple way to solve this question. The possibility for a certain number P(X=n)=P(the last number we got is the smallest and the first number we got is the second smallest). It really makes sense and really easy to work out the possibility (since my classmate use complicated integration to work out this).

Reply 16

Original post
by u4ria
My teacher led us to think of a very simple way to solve this question. The possibility for a certain number P(X=n)=P(the last number we got is the smallest and the first number we got is the second smallest). It really makes sense and really easy to work out the possibility (since my classmate use complicated integration to work out this).

So what did they give as an ans? I agree with the pi/4 (quarter circle) for the integration equation.

Reply 17

Original post
by mqb2766
So what did they give as an ans? I agree with the pi/4 (quarter circle) for the integration equation.

You are right. And i give the methods they used under that post. I really learnt a lot. :biggrin:
(edited 8 months ago)

Reply 18

Original post
by u4ria
You are right. And i give the methods they used under that post. I really learnt a lot. :biggrin:

I was really asking about this question, what ans did your teacher give? For this problem, even doing it properly, I cant see how its a complex integration problem.

Reply 19

Original post
by mqb2766
I was really asking about this question, what ans did your teacher give? For this problem, even doing it properly, I cant see how its a complex integration problem.

Sorry i misunderstood you. For this problem, it is infinity. I mean that the way my classmate think the question is hard to understand. It is easy to integrate but he thought it in a complex way. I explained my teachers methods above and i write here again: P(X=n)=P(the last number we got is the smallest and the first number we got is the second smallest). n is the additional number we press the button. So the P(X=n) becomes 1/(n+1)n, where 1/(n+1) represents the possibility of “the last is the smallest” and 1/n represents the possibility of “the first is the second smallest”. And this possibility is easy to integrate to have the answer infinity.

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