I buy a lottery ticket with odds of 1 in 100 million. After the drawing, I look at the webpage and see that a winning ticket was sold somewhere. Given that, have my odds improved?
At first glance, you’d think no – there’s still 100 million possible numbers to choose from, just like before, so my odds are still 1 in 100 million.
On the other hand, suppose the website says a winning ticket was sold in my home city. Most people would say given that information, your odds are now much better. But if you think about it, that statement doesn’t add any additional quantitative info. So it’s not really clear why the two cases would be different.
So given the knowledge that someone won last night’s drawing, does that increase the probability of my ticket being the winning one? And if so, how can you mathematically show how much it increases the odds?
A lottery with odds of 1 in 100 million probably typically rolls over several times before anyone wins, so just hearing there’s some winner somewhere is good news. If you know that there’s at least one winner somewhere, then your odds of winning, given that, are at least 1 in the number of people who bought tickets (and could be better, if there was more than one winner). Likewise, if there is known to be a winner in your city, then the odds are 1 in the number of people who bought tickets in your city.
As for a mathematical formula to determine what your chances are, you can use Bayes Theorem if you want to prove similar questions, but the answer in this case is fairly intuitive. As Chronos said, the odds are 1 in the total number of tickets sold to the group that has the known winner.
Suppose only 5 people bought tickets. Once you know there’s a winner, no matter the original odds, your odds just went to 1 in 5 if you hold a ticket, since one of those 5 must be a winner, and they were all equally likely to win.
Bayes Theorem in generic terms (‘prob.’ means ‘probability’):
The conditional or posterior prob. is what you’re hoping to find. The prior prob. is the probability of the event occurring with no conditions. (In this case, winning the lottery). Likelihood is the probability of the event occurring under this condition, and the marginal prob. is the probability of the condition existing at all.
For this problem:
The prior prob. = 1/100 million, the probability that you won the lottery.
Our condition is “a winning ticket exists”.
The likelihood that if you won the lottery, a winning ticket exists is 1 since it must necessarily exist.
The marginal probability is the total number of tickets sold out of 100 million.
How does this relate to the Monty Hall problem though?
In that case, your original odds were 1/3, and after determining that the car *must *be behind one of the remaining two doors, most people think that the odds change to 1/2, but, in fact, remain at 1/3.
Why do the odds of the lottery change? I guess one difference between the two cases (the Monty Hall problem and the lottery problem mentioned in the OP) is in the process of determining the current viable set of choices, and that may be the determining factor in whether your odds change after the new info comes out.
I was recently telling some people how I “almost won lotto” once. My explanation was that I had bought a ticket and left it at home while I was away on holidays. When it was announced a few days after the draw that the winner hadn’t come forward I said to my wife, “it’s between us and the other people who haven’t checked their tickets yet.”
With each passing day I insisted we were getting more likely to be the winner as more people must have checked their tickets upon hearing the winner hadn’t come forward. As I told the story my plan was supposedly to keep waiting until everyone else had checked their ticket then I would be a “certainty”.
Unfortunately I had to give up my plan when they announced where the ticket had been sold. My listeners insisted my plan made me an idiot.
Yes it does. At first all you knew was that the winning ticket was somewhere in the set of 100 million tickets. Now you know it’s is in the subset constituted by ‘only the tickets sold in your home city’. So your odds of winning are proportionately greater. Simple example: if only two tickets were sold in your city, your odds of winning are 1 in 2.
Well, nothing wrong with your understanding of probability, so they had no cause to berate you for that. The problem is that common sense often trumps probability :).
No, you’re making an assumption that fewer than 100 million tickets were sold in your town. While that may be true, the original problem doesn’t state that. Maybe your town had a very large number of lottery players.
I was worried that someone would bring up Monty Hall.
In the Monty Hall problem, the puzzle is to figure out if you are better off switching doors or sticking with your original pick. The big difference here is that in the Monty Hall problem, Monty *knows *where the the car is, and *always *reveals a goat, leaving two doors. This doesn’t give you any additional information. Your odds of having picked the car are still at 1/3. I don’t want to go into this any deeper since it has been discussed here, in Marilyn’s column, and elsewhere ad nauseum.
This is quite different than the lottery problem, where everything is random. The Monty Hall problem would be completely different if Monty determined which door to reveal by a roll of a die.
Look at the problem another way. You buy a ticket with 1:100Mil odds on a winner. After the draw they announce that nobody won this week.
What are the odds your ticket is a winner? Obviously they’ve collapsed to zero. So the announcement about the outcome clearly affects the likelihood that you own the winning ticket.
It seems to me that a lot of people get hung up on the word “odds”, using it to mean either prior probability, conditional probability, purchase/payoff ratio, or damn near any other hand-waving idea. So the best way to think about, and communicate about, these kinds of problems is to immediately banish the word “odds” & use more precise terms.
Everything up to here is reasonable, although I think I think it was really between you, the others who hadn’t checked yet, and all the people who wouldn’t immediately come forward. Not sure how big that last group might be.
I assume this part was a joke.
Kind of depends on whether you were joking or not…
If we know that every week there will be entire cities with no winners, and if the lottery commission announces, before the results “This week, there are no winners in Cleveland”, and on another week announces “This week, there are no winners in Las Vegas”, this will be similar to the Monty Hall problem.
In that case, we know that at least one of the other two doors has a goat, and Monty picks one that has a goat, and of course picks a different one each time.
So, if the lottery commission announces a different city each time (and also, if there is a rule that they can never announce your city even if it has no winners [Monty cannot open your door, even if it has a goat]), since they know which cities have no winners in each week, then your probability does not change. It is still 1 in 100 million. It is not 1 in (100 million minus number of tickets sold in the city that was just announced)
