With a regular deck of 52 cards, how would you find the probability of getting the Ace of Spaces if you took turns until someone got it? This is without reshuffling.
If it was a 1/52 chance each time, I would be able to do it like
(1/52) * (inf. sum ((51/52)^2)^n)
But this would have a...
Oh, then that makes sense.
Now, I can see how both are done, but there's one thing I can't do without a calculator at the moment and that's dealing with the variable in the exponent. I know that's what I'm missing to understand what he was telling me
Sorry, it took a few minutes but now I think I understand. Here, I'll attempt it, changing the game a little. Say, the first to roll a 1 or 6.
sum of ((1/3)^(2n+1))(1/3) = (2/9) * sum of (?)^n
I don't understand how you got 25/36, or where it came from.
If this is correct, I like it...
Just a little beyond me, so please try not to get annoyed.
I know there is a 5/36 chance of winning with one trial. So, what I think your equation is saying is ..
(win) * sum of (dont lose from trial 2 on ..)
Not that I understand how to calculate this, and wiki page is confusing me a bit...
No, I don't understand how to calculate anything infinite.
Also, the russian roulette idea, I was talking if you spun the revolver each turn. I forgot about the way of playing with a single spin and all taking a turn. Anyways ... off topic.
The easiest way I can explain this is russian roulette. You have a six-sided die, and I need to find out the probability of the other person rolling a 6 before me, or dying in russian roulette with a 6-chamber revolver.
I know it's different depending on who goes first, so for my example, I...