Help perfecting my statistical idea?

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Help perfecting my statistical idea?

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1drbubbles
Abr 4, 2008, 11:49 am

Hear me out first:

If you don't buy a lottery ticket, you have 0 chance of winning the jackpot.
If you buy 1 lottery ticket, you have a 1/gazillion chance of winning the jackpot, which is an infinite increase in probability over 0 tickets.
If you buy 2 lottery tickets, you have a 2/gazillion chance of winning the jackpot, which is merely a 100% increase in probability over 1 ticket.
And so on, each additional ticket providing a smaller marginal increase in your chance of winning the jackpot than the previous ticket.

So buying a single ticket, because it makes you infinitely more likely to win than buying no ticket at all, provides the largest possible marginal increase in chance of winning.

What's the best way to state, in such a way that it would sound reasonable enough to win an informal bet, the completely bogus conclusion that being "infinitely more likely to win" is equivalent to being "(virtually) guaranteed to win"?

2MMcM
Abr 4, 2008, 12:06 pm

Isn't there a very very small chance that you'll find a winning lottery ticket on the street on your way to work?

3readafew
Abr 4, 2008, 12:15 pm

Or your boss or loved one giving you one as a present?

4drbubbles
Abr 4, 2008, 12:35 pm

OK: substitute "have" wherever it says "buy"; likewise "having" for "buying."

5hashiru
Editado: Jun 1, 2008, 4:48 pm

While we're on lottery ploys: is there a winning strategy based on the Birthday paradox for winning the 3 or 4 digit lottery when "conditions are right"? The basic idea is that if the winning number for N consecutive days is different, then the probability that today's choice will be one of the previous N numbers is quite high (assuming the lottery isn't fixed and that the probability of any choice on any day is 1/1000). Similar to the idea that if more than about 40-50 people are in a room, the probability that two of them share a birthday approaches certainty.

The lottery strategy would then be to bet equally on the previous N numbers with payoff expected to be: 1000 - 2*(N - 1) for each N tickets played. (As I recall, the price for a 3-digit ticket is $2 and the payoff for a winner is $1000)

6MMcM
Editado: Jun 1, 2008, 7:59 pm

I'm afraid not.

The odds of one of your N different choices coming up are N/1000. It does not matter what happened before or how you arrived at your N. Just as after a coin-toss coming up heads M times in a row, the odds of it coming up tails are still 1/2. M heads in a row is no more likely than any other particular sequence of heads and tails, just easier to say in fewer English words.

It is not true that one of the N is highly likely after the N different ones. What is highly likely (better than 50-50 in 38 choices from 1000) is that two will be the same from all possible outcomes. But what is past is already fixed: you are not allowed to look at all possible outcomes, only those that start with the particular N already known. And that gets you back to 1/1000 for each independent event.

A slight edge can be gained in a lottery that shares payoffs (as most do) by picking unpopular numbers, and thereby not increasing your chance of winning, but rather decreasing your chance of sharing. But this edge is rarely enough to match the state's profit. (And everybody who lets the lottery machine pick randomly for them draws with you in the long-run, just like in RPS.)

In lotteries that accumulate the proceeds when there is no winner, the expectation can go up. But in most such cases, the number of people betting fresh also increases dramatically.