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This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the tota

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

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leephillips 5y ago
If you are talking about a simple lottery with N tickets and a fixed probability of any one ticket being the winner = 1/N, then the screamers were correct. To see this, imagine that there are a total of two tickets (N = 2). You buy one ticket. The probability that it is the winner = 1/2. After buying the second ticket, is your probability of winning somehow > 1? Or has it exactly doubled?
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qot 5y ago
But you are wrong: it's exactly double the odds. If you bought all the tickets, do you think you'd have over 100% chance of winning? Edit: looks like leephillips beat me to it
- failwhaleshark 5y ago
  Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence. P(A) = first ticket wins P(B) = second ticket wins P(A|B) + P(A|^B) + P(^A|B) = 1 - P(^A|^B) = 1 - (1 - 1/N) * (1 - 1/(N-1)) = 2 / N It is slightly > 2 / N if the ticket is revealed before the next one is chosen because then it does throw away that possibility.
aaplok 5y ago
I may misunderstand you, but replace the lottery with a coin toss. Buying a ticket gets you a 50% chance of winning. This reduces the pool of possible outcomes by 1 for the next choice, so there is precisely one choice left. However buying a second (different) ticket does not give you a 150% chance of winning.
- failwhaleshark 5y ago
  If you buy heads and tails, then you have a 100% chance of winning. :) Unfortunately, each ticket cost 5 cents. (See above.)
feoren 5y ago
There are still N possible outcomes, and you have 2 of them instead of 1. Your chance of winning is 2/N instead of 1/N and has exactly doubled. And most lottery games allow numbers to be re-used, so you haven't "eliminated" anything. If you exhaustively bought all N number combinations, you have a 100% chance of winning, but you also have a decent chance to split the pot with someone else who also bought the winning numbers. Next time your family screams and shouts at you that you're wrong, maybe you should actually listen to what they're saying?
- failwhaleshark 5y ago
  Piling-on and screaming isn't a rational argument for anything. No listening can occur. None of them had above a high-school education. Your piling on isn't helpful either, so can it.
jonasenordin 5y ago
I think if you buy a first ticket, and somehow found out (but lotteries usually don't allow this) that it wasn't a winner, and only _then_ buying a second one, your reasoning would be correct.
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leephillips 5y ago

If you are talking about a simple lottery with N tickets and a fixed probability of any one ticket being the winner = 1/N, then the screamers were correct. To see this, imagine that there are a total of two tickets (N = 2). You buy one ticket. The probability that it is the winner = 1/2. After buying the second ticket, is your probability of winning somehow > 1? Or has it exactly doubled?

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qot 5y ago

But you are wrong: it's exactly double the odds. If you bought all the tickets, do you think you'd have over 100% chance of winning? Edit: looks like leephillips beat me to it

failwhaleshark 5y ago
Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence. P(A) = first ticket wins P(B) = second ticket wins P(A|B) + P(A|^B) + P(^A|B) = 1 - P(^A|^B) = 1 - (1 - 1/N) * (1 - 1/(N-1)) = 2 / N It is slightly > 2 / N if the ticket is revealed before the next one is chosen because then it does throw away that possibility.

aaplok 5y ago

I may misunderstand you, but replace the lottery with a coin toss. Buying a ticket gets you a 50% chance of winning. This reduces the pool of possible outcomes by 1 for the next choice, so there is precisely one choice left. However buying a second (different) ticket does not give you a 150% chance of winning.

failwhaleshark 5y ago
If you buy heads and tails, then you have a 100% chance of winning. :) Unfortunately, each ticket cost 5 cents. (See above.)

feoren 5y ago

There are still N possible outcomes, and you have 2 of them instead of 1. Your chance of winning is 2/N instead of 1/N and has exactly doubled. And most lottery games allow numbers to be re-used, so you haven't "eliminated" anything. If you exhaustively bought all N number combinations, you have a 100% chance of winning, but you also have a decent chance to split the pot with someone else who also bought the winning numbers. Next time your family screams and shouts at you that you're wrong, maybe you should actually listen to what they're saying?

failwhaleshark 5y ago
Piling-on and screaming isn't a rational argument for anything. No listening can occur. None of them had above a high-school education. Your piling on isn't helpful either, so can it.

jonasenordin 5y ago

I think if you buy a first ticket, and somehow found out (but lotteries usually don't allow this) that it wasn't a winner, and only _then_ buying a second one, your reasoning would be correct.

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