I asked this question before and got clearly wrong answers. I’ll try again.
There’s a lottery with odds of 1 in 200,000,000. I buy a ticket, and the next day, I check the lottery webpage and see that there was a winner. ****I DO NOT KNOW HOW MANY TICKETS WERE SOLD ***. Maybe there were fewer than 200 million tickets sold, but maybe there were more than 200 million tickets sold. I don’t know that – I just know there was a winner.
Before I learned there was a winner, my odds were 1 in 200 million. Do my odds go up after learning that someone won? Of course, if I had learned there was NOT a winner, then my odds would have dropped to 0. So does the extra information that there was a winner increase my chances?
Keep in mind I don’t know how many tickets were sold – if I knew only X tickets were sold, then yes, my odds would improve to 1/X. But I don’t know that. So does the information that someone won increase the odds that I won?
Without knowing how many tickets were sold… no, you cannot recalculate your odds, beyond saying you still have one in 200 million (if someone won) and zero in 200 million (if nobody did).
And depending on the type of lottery, you might not even be able to calculate a 1/(tickets sold) value. If it’s a powerball-type lottery in which participants can choose their numbers, it’s entirely likely that a large amount of duplication will occur.
Your actual chances did not change just because your information changed. You either had exactly 1/200,000,000 if no-one had yet won, or exactly 0 chance if someone had already won.
I’m not sure understand this question. Yours odds never change; you either hold the winning ticket or you do not; the chance you do is 1/200,000,000 regards of what info you have or do not have.
The above posts are wrong or misguided, so far as the technical apparatus of probability is concerned.
Two things to keep in mind: A) You can pick any probability distribution you want and calculate the probabilities according to such a distribution. Anytime someone reports probabilities, it is with respect to some implicit distribution which is of interest for some reason (perhaps because it has suitable symmetry properties, perhaps because it corresponds suitably with some other data of interest, or so on). B) If one is originally considering probabilities with respect to the implicit distribution D, and then learns new information, one conventionally switches over to using the implicit distribution of D conditioned upon the new information.
Some mathematics: if conditioning upon the negation of X would make the probability of Y drop, then conditioning upon X must make the probability of Y rise. Specifically, P(Y) = P(X) * P(Y | X) + P(not X) * P(Y | not X), where P(a) is the probability of a before conditioning upon any new information, and P(a | b) is the probability of a after conditioning on b. Thus, if you had a particular model in mind for the probability (before learning it) of there being a winner, you could deduce from this the amount by which the probability of you being the winner increases upon learning that there actually is a winner.
If I may, imagine a powerball lottery where everyone can pick their own numbers. In a simplified example, there are ten numbers in the pool and ticket-buyers can choose any one of them
Conceivably, everyone (including the OP) chooses the same number. Thus his odds of winning (assuming someone has won) is 100%.
If instead there are ten participants and each has chosen a different number, the OP’s odds remain one in ten.
If five people chose 6 and five people chose 3 (and the OP is one of these people), and someone wins, the OP’s chances are 50%.
I’m assuming in a powerball-style lottery with, say, a large pool of contestants individually picking six numbers from a pool or 50 or so, many combinations will not be chosen, while many combinations will be chosen more than once, making odds difficult to compute, to say the least.
A more conventional raffle-type lottery in which each ticket bought has a unique number is a far simpler problem, but the OP doesn’t say.
The case of concern is even easier, actually: P(a | b) = P(a & b)/P(b). Thus, the probability that you are a winner, given that someone is a winner = (the prior probability that you are a winner)/(the prior probability that someone is a winner). The value of the denominator of course depends on the particular probabilistic model you adopt, but whatever it is, it will be at most 1 and presumably less, causing the probability that you are a winner to increase upon conditioning on someone being a winner.
For example, if, before you learn whether there is a winner yet or not, you would assign a probability 1/3 to there being a winner yet, then after conditioning upon the knowledge that there is a winner, your probability of being the winner will multiply by 3.
Of course, you could choose to investigate a probability distribution in which all probabilities are already either “0, because it’s definitely false” or “1, because it’s definitely true”, and thus never shift upon conditioning (the “There aren’t any probabilities; just the way things are, whether you know them or not” line). But that’s hardly the only probability distribution one can look at, and indeed, clearly not the kind you were interested in looking at.
What about the fact that you are given added information. For example: you find out there was only 1 ticket sold (yours). Now what is the probability that you won?
Clarifying my earlier post, assuming everyone gets to choose their own numbers and assuming somebody won, the odds of you personally winning are (the number of tickets with your combination) / (number of tickets sold).
Of course, that’s another piece of information you have to get.
I think he meant that 200,000,000 tickets are offered for sale but only 1 is sold. The odds are still 1/200,000,000 - again, no amount of information that comes to your hand changes the odds, they are what they are when the tickets where printed, and information you receive cannot go back in time and alter it in any way.
Well this is nothing like the OP’s lottery, but what you elide here is that you are assuming someone must win. In your first case, the true odds are 1/10, as for 9 picks no-one wins - this is how lotto-style games are run. If you now posit a game where the operator continues to pick until someone wins that’s a different matter, but note that his prize money is divided between the co-winners so his expected returns are identical.
That doesn’t match any definition of “someone won” that I’m aware of. Besides, that example only works in raffle-lotteries, not powerball-lotteries, where tickets are printed only on purchase.
And I’m not assuming someone must win, I’m just going with the OP’s assumption, quote, “I check the lottery webpage and see that there was a winner.”, i.e. someone did win.
To reiterate what Indistinguishable said: The information that there was a winner definitely improves your odds by some amount, but we can’t say how much it improves your odds without some sense of how likely it was for there to be a winner.
Like Indistinguishable, I was surprised by the wrong answers. I’m afraid that if doubled chooses not to believe Indistinguishable and I, he’ll need to ask the question a third time.
I’ll rephrase the argument. Suppose the a priori chance that anyone will win is p. The new chance that you won is q, and appears in this expression:
1/200000000 = p*q + (1-p)*0
It’s a simple alegebraic fact that if p is less than 1, then q is more than 1/200000000. Period.
Now what others may be pointing out is that you don’t know what p is, so can’t compute the q. You might take your “best guess” at p to get a best guess at q. (It may actually be more complicated than this: it might be better to integrate over a pdf.)
Since the coin I just tossed has come to rest, its chance of heads is either 0 or 1, but if I don’t tell you which, you’ll have to guess 0.5. This is a confusing fact about probabilities: they often describe one’s ignorance. In your case you’re ignorant about the a priori value of p. But don’t fall for the sucker’s game of implicitly or explicitly using a wrong value for p, just because it’s hard to estimate.
Yeah, that happened the last time I asked this question, too.
Here’s the thing I don’t get – most of the time, there is not a winner. So right before I check to see whether there was a winner, my odds are 1 in 200million and I also know that empircally, most drawings go without a winner. I don’t know for certain that the number of tickets sold was less than 200 million; but it’s certainly possible (and indeed likely based on past results). It seems like simply learning that SOMEONE won, must give me more hope now than before I knew that.
Yes, it’s possible that every possible ticket was bought, in which case my odds haven’t changed at all – there HAD to be a winner and my odds are still 1 in 200 million. But it’s also possible less than 200 million tickets were sold, in which case my odds have increased to 1/X.
Once you find out someone won, you feel happier than you did before the drawing. You think, “Well, now at least I have a chance!” After all, most of the time no won wins the lottery, so you just survived a big hurdle. Is this a rational feeling that you think you’re better of now? Is this because we’re implictly making the assumption (without acknowleding it) that fewer than 200 million tickets were sold? Or is there some other reason why people assume the knowledge that somebody won increases their personal odds?
If people weren’t allowed to select the same combinations of lottery numbers this would be an easy Bayesian problem in which the initial probability of winning would be modified by the OP’s share in the total number of tickets bought. Since that’s not the case, you’d have to know the number of *unique *combinations of numbers purchased before you could modify your prior expectation.
In most lotteries, since people can choose whatever combinations they want, the number of tickets sold doesn’t really relate to all the combinations being covered. So there’s no “number of tickets sold” that will assure full coverage. You would want to know the percentage of combinations covered to know for sure that a winning ticket would be sold.
Your odds of picking the winning ticket combination is say, 200 million to one. Once you know that a winning ticket has been sold, your odds are 1/number of tickets sold. That can be further complicated by multiple winners with the same combination, but you said that a single winning ticket was sold, right?
I’ll try again.