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No, it means that in the next 10 tosses, you still expect to get a 5/5 split, so over these 20 tosses, you'll probably have 15 heads, which is closer to the mea

by beder 16y ago

No, it means that in the next 10 tosses, you still expect to get a 5/5 split, so over these 20 tosses, you'll probably have 15 heads, which is closer to the mean. I've seen this analysis in sports. In baseball, for example, you generally expect teams to win one-run games as often as they win games in general (that is, there's no "great closer" effect). So if you see a team winning 80% of their one-run games in the first half of the season, you can expect their record to be worse in the second half.

StavrosK 16y ago
Ah, I see, thanks.
- ssebro 16y ago
  No dude- you were right and he was wrong.
  - StavrosK 16y ago
    No, he's saying that if you have a few outliers to begin with you expect to get more normal-looking data later on, drowning your outliers in them. If you get 5 heads initially and then another 5 heads and 5 tails, it's still closer to the mean than when you began.
    - ssebro 16y ago
      That can't be right, since the each point in the data is independent. In a 50/50 coin-flip game where you get 9 coins come up heads, the probability that the next coin flip results in tails is still 50%. What you need to understand is that probability deals with uncertain events. So instead of thinking about the final count needing to be around 50% heads and 50% tails, you should expect the count of the yet undecided part to converge around 50% heads, 50% tails.
      - beder 16y ago
        I just noticed this thread continued, but I'm not sure what you're worried about. The two things you said: * the final count will be about 50/50 * the remaining part will be about 50/50 are actually the same thing (since we're talking about a limit). That is, we could flip a fair coin a billion times, get all heads, and still, in the long run, expect the total ratio to converge to 1/2 (imagine flipping the coin forever - that initial billion flips won't even be a blip on the graph).
ssebro 16y ago
This is still wrong, since coin flips are independent events, and events that are totally independent can't affect each other's probability. I understand your reasoning, but it is flawed.

by beder 16y ago

StavrosK 16y ago

Ah, I see, thanks.

ssebro 16y ago
No dude- you were right and he was wrong.
- StavrosK 16y ago
  No, he's saying that if you have a few outliers to begin with you expect to get more normal-looking data later on, drowning your outliers in them. If you get 5 heads initially and then another 5 heads and 5 tails, it's still closer to the mean than when you began.
  - ssebro 16y ago
    That can't be right, since the each point in the data is independent. In a 50/50 coin-flip game where you get 9 coins come up heads, the probability that the next coin flip results in tails is still 50%. What you need to understand is that probability deals with uncertain events. So instead of thinking about the final count needing to be around 50% heads and 50% tails, you should expect the count of the yet undecided part to converge around 50% heads, 50% tails.
    - beder 16y ago
      I just noticed this thread continued, but I'm not sure what you're worried about. The two things you said: * the final count will be about 50/50 * the remaining part will be about 50/50 are actually the same thing (since we're talking about a limit). That is, we could flip a fair coin a billion times, get all heads, and still, in the long run, expect the total ratio to converge to 1/2 (imagine flipping the coin forever - that initial billion flips won't even be a blip on the graph).

ssebro 16y ago

This is still wrong, since coin flips are independent events, and events that are totally independent can't affect each other's probability. I understand your reasoning, but it is flawed.