I am trying to make sense out of this experiment :
There is a prize randomly put in one of 10 boxes. I am supposed to guess the box in which the prize lies.
Q1: What is the sample space here ?
Is it labels of the box being guessed by me at every attempt? Example: {1, 2, 3 .. 10}
Or is it the nth attempt that I am currently at? {Prize found in 1st attempt, Prize found in 2nd attempt ... , Prize found in 10th attempt}
Q2: Why I am not getting Probability $1$ in total ?
The probabilities of me able to find the prize in $i^{th}$ attempt comes out to be $\frac{1}{10}$ for any attempt.
Example: Prize to be found in 3rd attempt = $\frac{9}{10}*\frac{8}{9}*\frac{1}{8} = \frac{1}{10}$
However, If I am not able to find the prize till $9^{th}$ box, I will always be able to tell the prize as the $10^{th}$ box with certainty. I am not sure how will the probabilities add upto $1$ in this case.
What am I missing? Thank you.
Q1 : The Sample Space can be "modeled" in various equivalent ways.
It is covering the ways the Prize is put.
One way to "model" that is with tuples : the Sample Space will then be :
$(1,0,0,0,\cdots)$ , $(0,1,0,0,\cdots)$ , $(0,0,1,0,\cdots)$ , ETC
where there are $10$ elements (Boxes) in the tuple & $9$ elements are $0$ (no Prize) with a Single element having $1$ (Contains Prize).
Sample space will have $10$ such tuples. All are Equally likely.
Your attempts are not the Sample Space in this model. Your attempts will be the Experimental Process & the outcomes & the history of the Experiment.
Q2 : When you start the Experiment , you know the Prize is there , hence you have Probability $1=100\%$ to "find" it.
What you do not know is which Box has it.
Probability that the Box you select has it is $1/10=10\%$.
Conditionally , "if" that selected Box does not have it , the Probability that the other $9$ Boxes have it is ..... $1=100%$ !!
When you have eliminated $1$ Box , you can now make your second selection.
Probability that the new Box you select has it is $1/9$.
With the Experimental Process , getting to the Eliminated State & getting the Prize is $9/10 \times 1/9 = 1/10$
With the Experimental Process , getting to the Eliminated State & then Eliminating 1 more Box is $9/10 \times 8/9 = 8/10$
When you have Eliminated $2$ Boxes , you can now make your third selection.
Probability that the new Box you select has it is $1/8$.
With the Experimental Process , getting to that State & getting the Prize is $8/10 \times 1/8 = 1/10$
With the Experimental Process , getting to that State & then Eliminating $1$ more Box is $8/10 \times 7/8 = 7/10$
Continuing this way , we can see that Probability will always make up $1$ , till the END State.
Consider the Last Box :
You have Eliminated $9$ Boxes , which will have Probability $9/10 \times 8/9 \times 7/8 \times \cdots \times 1/2 = 1/10$
When you check the Last Box , you will "find" the Prize with Probability $1$.
Over-all Probability , to get to that END State & then "find" the Prize , is $1/10 \times 1 = 1/10$
Nothing to worry over !!