The odds of winning the lottery is one in a million
The lottery is held once a week.
Every week, a million people play the lottery.
So this week, your odds of winning are 1 in a million.
If you win this week, your odds of winning next week are still 1 in a million.
So it seems like the odds of someone winning the lottery twice must be 1 in a trillion ((10^-6 * 10^-6) = 10^-12). However, that’s just the odds of YOU winning the lottery twice in a row.
The odds that SOMEONE will win this lottery twice in a row are actually 1 in a million. Someone must win the lottery with a probability of 1. The odds that this person will win the lottery next week are the same as ever, 1 in a million. Therefore, the odds that the same person will win the lottery twice in a row is 1 * 10^-6, or 10^-6.
Now consider that the lottery runs every week, and we aren’t looking at the odds of back to back wins, we’re looking at the odds of someone winning the lottery more than once in their life.
Right–think about your hypothetical 50-year scenario with 1,000,000 people and 2500 drawings. On the very last week of the 50-year run, there have been 2499 winners already. Assuming no one has won twice yet, the chances of a two-time winner on the last week alone is 2499/1,000,000 = about 1/4%. That’s not all that small, and it’s only for the very last week. The week before, the chances of a previous winner winning again are 2498/1,000,000, again slightly under 1/4%.
Someone else can do the exact calculation, but extrapolating backwards over 2500 drawings, I think it’s easy to see why someone (or someones) is likely to win twice.
In fact, if I’m doing the math correctly, in your one-in-a-million example after 1000 draws you’d expect a double winner about 40% of the time. After 2000 draws the odds of no double winners is only 13.5%.
To make the equations a bit clearer, it works like this:
First draw - Nobody has won before so the odds of a “new” winner are 100%
Second draw - One previous winner, so odds of a “new” winner is 999,999 out of 1 million. The odds of only “new winners” to date is this number times the previous result.
Third draw - Two previous winners, so the odds of a “new” winner is 999,998 out of 1 million. The odds of only “new winners” to date is this number times the previous result.
Etc…
By the time you’ve repeatedly multiplied these percentages (which start out at 100% and end up at zut’s 99.75% after 2,500 draws) the overall likelihood of only new winners every week becomes small.
Actually, there would be only about a 63% chance that somebody would win, assuming people choose their lottery numbers randomly and independently, because some people would choose the same numbers, and therefore some combinations would go unchosen.
The scenario in the OP ultimately hinges on whether or not the probability of another bomb on the plane changes when you bring a bomb on board. It seems likely, in the real world, that it would not affect anyone else’s plans. But there are similar scenarios where the probability would change.
Consider the Monty Hall problem. We know that opening one door doesn’t affect the probability of your chosen door holding a prize. But if it weren’t guaranteed that there was a prize at all, like if you were a detective searching party-goers for the stolen necklace (where it’s possible that no one has it at all), then the probability of finding the necklace does indeed change as you check people.
Upthread, it was stated that a bomb on board your plane is less likely than on a random flight (because you’re taking up a seat), but we can tweak that so that it’s not true. Consider if we knew that there was going to be a bomb on some flight at that time but nobody knows which flight (except the planned bomber). In that case, it doesn’t matter how many innocent parties are on your flight- so long as there’s at least one seat open for the bomber, you’re just as likely as any other plane to get blown up.
Incidentally, I’ve won the birthday bet in a bar. I wagered that I’d find a pair within 30 people. It wasn’t immediately clear to the sucker why I was going to win since I just went down the bar and recorded everyone’s birthday one by one until I found a pair. But I bet he would’ve caught on halfway through if I instead asked the first guy to stand up and announce his birthday to the whole bar, and if it wasn’t a pair, if I asked the second guy to do the same. The sucker would watch the available dates dwindle and he’d sweat as he dodged a bullet with each next ‘contestant’.
I guess the point of my rambling post is that the intuitiveness (is that a word?) of probability can go from clear as mud to clear as crystal with just the sliiiightest tweak of the scenario, and the actual results of the scenario change just as easily.
Well, sure, but then you’ve got a chance of two one-time-winners being created in the same week. That’ll even out over time and you’ll average one winner per week over the long run. And if you’re still talking about a two-timer in a 50 year period, then the result is the same.
