Inara ,'Objects In Space'
Off-topic discussion. Wanna talk about corsets, duct tape, or physics? This is the place. Detailed discussion of any current-season TV must be whitefonted.
Here's the question:
In a lottery in which the ratio of losing tickets to the number of winning tickets is 39:1, how many tickets should one buy to give oneself even odds of winning a prize?
At first I thought this was one of those "How many rolls of the dice before there's an even chance of getting a 6" questions, but it's not multiple trials, just one. Shouldn't the answer be... uh. 20?
20 sounds correct to me. Each ticket has one chance in forty to win, so to make the odds even, you'd need twenty chances at it.
A better analogy than dice would be a roulette wheel without the 0 and 00 holes.
At first I thought this was one of those "How many rolls of the dice before there's an even chance of getting a 6" questions, but it's not multiple trials, just one. Shouldn't the answer be... uh. 20?
If there are only 40 tickets and one winner. But what if there are 40,000 tickets and 1,000 winners? Since they didn't specify the number of tickets, you don't have the information to determine the impact on subsequent purchases of previous ones. Therefore, I think they're most likely expecting you to treat it as if each ticket is an independent trial, i.e. the "How many rolls of the dice" deal.
Of course, it's also true that saying the ratio is 39:1 could mean 390:10, in which case... well, in which case there's really no way for me to know how many tickets to buy. Okay, I think 20 is really the way to go here. Seems way too simple for the chapter, but there you are.
Therefore, I think they're most likely expecting you to treat it as if each ticket is an independent trial, i.e. the "How many rolls of the dice" deal.
Doesn't that still come out to twenty?
t props chin in hands and watches the pwetty bwains
Right, but it isn't independent trials. Can we really regard it that way?
Doesn't that still come out to twenty?
Oddly, no. I understand that bit. Er, I think.
So, watched the last few minutes of Alias tonight.
Where, as Paul and I both noticed, they invaded Fisher Plaza! (Where Paul worked until last year.) Paul was all, "Hey! That's my old rooftop! Hey! I've been in that stairwell!"
Those were the sat dishes he controlled!
Doesn't that still come out to twenty?
No, then it comes out to the value of n such that 0.975^n = 0.5. There are logs involved. It's a thing.
Right, but it isn't independent trials. Can we really regard it that way?
If there are, say, 40,000,000 tickets, then it's effectively independent. If it's 4,000 it's probably close enough. It'd have to be a very small lottery for it to be materially different from independent.