Does this algorithm guarantee equaprobable outcomes?

[Bipolarity]

Suppose you have a list of numbers, say ${1, 7, 9, 4, 5, 6}$.
You store the first number, and then iterate through this list.
For each number in the list, you flip a coin. If it is heads, you swap that element in the list with the number you stored. If tails, you do nothing. Either way, you move on to the next number in the list.

When you finish traversing the list, the resulting number is the number you stored. I am curious if all numbers in the list are equaprobably to be the final stored number, and if this is indeed the case, how might one prove this?

Thanks!

BiP

 

[Orodruin]

No, the numbers are not equiprobable. The last number has a 50 % probability.

 

[Bipolarity]

Ah I see, I suppose then that each number has half the probability of its successor?

Also, do you think there might be a way for me to tweak this so that the numbers are indeed equaprobable? This is for an application I'm trying to build.

BiP

 

[Orodruin]

Ah I see, I suppose then that each number has half the probability of its successor?

Yes, apart from the first number which has the same.

Also, do you think there might be a way for me to tweak this so that the numbers are indeed equaprobable? This is for an application I'm trying to build.

Yes, but the normal thing to do would be to simply generate a random number k between 1 and N, where N is the number of numbers you have and then select the kth number.

 

[Isaacsname]

Coin flips aren't equiprobable , are they ?

ETA: never mind, I see the thread already :P