5 ms·

Is that true? You'd still have evidence that the distribution of balls tilts one way, wouldn't you?

danielmarkbruce 3y ago
I think with a binomial the tilt is perfectly offset by the tilt you get from taking the ball.My math isn't fresh enough to write a nice dense statement showing why. If you change their simulation formula to pull from a binomial instead of uniform... it looks 50/50.
- jncfhnb 3y ago
  The probability of flipping a coin and getting heads is the same as flipping ten coins, choosing one at random, and getting heads. Since the probability is the same, you have learned nothing about the distribution of unchecked coins. Since the coins are independently flipped, you can assume that it’s still just a binomial distribution of size n-1 and the same p.
jncfhnb 3y ago
Flip three coins. First one lands heads. Does that mean the rest are more likely to be heads? No.
- j7ake 3y ago
  You have two coins, one is biased heads and the other biased tails. Someone picks a coin without you knowing which one, then starts flipping. The first toss shows heads. Does that mean subsequent flips from the same coin will be more likely to be heads? Yes. Answer is only no if all coins are the same with no bias, which they are not.
  - jncfhnb 3y ago
    I’m not clear why you posted this. This is a different problem from the one being discussed.
    - j7ake 3y ago
      The problem being discussed is the same as my toy example, just extended to more coins. You now have 101 coins, with biases ranging from 100 percent heads (or urn with only red balls) to 100 percent tails (or urn with only green). Someone chooses one of the 101 coins randomly (each coin has equal probability of being chosen) and starts flipping it. First flip shows heads, what is the probability the next flip is also heads? Answer is 2/3. The point is that knowing what was the first flip gives information about what biases a coin may have. The original problem has a minor technical twist that balls are drawn without replacement, but it’s an irrelevant detail for N>10 total balls.
      - jncfhnb 3y ago
        I mean, sure, what you’re describing is relevant to the original discussion, but you’ve posted it in a sub thread about how it changes if it is a 50/50 binomial distribution. You’ve replaced it with a binomial of a uniformly selected p, which ends up being, amusingly, the same as a uniform.
        j7ake 3y ago
        As long as you think the flipper chose their coin from a bag of coins with varied biases, then first flip always gives actionable information. It’s mostly irrelevant what kind of distribution this bag of coins comes from (except for some degenerate cases). What matters is that selecting coins from this bag gives coins of varying biases.
        jncfhnb 3y ago
        Yes but the question being discussed assumes they are fair. That’s why it’s a different question.
- eximius 3y ago
  If I flip three coins and use the results to decide which color balls to put in a bag, I'll have one of these distributions: RRR,RRG,RGG,GGG. If I draw a red ball, I am more likely to have drawn from RRR or RRG than RGG (or GGG), so I have learned something about the distribution more than just how it was generated.
  - jncfhnb 3y ago
    No you haven’t :) Let’s ignore RRG because it becomes a 50-50 anyway. RRR has a 100% chance of yielding Red. But is 3X less likely than RGG. RGG has a 33% chance of yielding Red but was 3X more likely. Edit: so still 50/50 Given a draw of red, you have exactly the same probability of having been in RRR and RGG. So you were equally likely to have 100% chance or a 0% chance.

I think with a binomial the tilt is perfectly offset by the tilt you get from taking the ball.My math isn't fresh enough to write a nice dense statement showing why. If you change their simulation formula to pull from a binomial instead of uniform... it looks 50/50.

jncfhnb 3y ago
The probability of flipping a coin and getting heads is the same as flipping ten coins, choosing one at random, and getting heads. Since the probability is the same, you have learned nothing about the distribution of unchecked coins. Since the coins are independently flipped, you can assume that it’s still just a binomial distribution of size n-1 and the same p.

jncfhnb 3y ago

Flip three coins. First one lands heads. Does that mean the rest are more likely to be heads? No.

j7ake 3y ago
You have two coins, one is biased heads and the other biased tails. Someone picks a coin without you knowing which one, then starts flipping. The first toss shows heads. Does that mean subsequent flips from the same coin will be more likely to be heads? Yes. Answer is only no if all coins are the same with no bias, which they are not.
- jncfhnb 3y ago
  I’m not clear why you posted this. This is a different problem from the one being discussed.
  - j7ake 3y ago
    The problem being discussed is the same as my toy example, just extended to more coins. You now have 101 coins, with biases ranging from 100 percent heads (or urn with only red balls) to 100 percent tails (or urn with only green). Someone chooses one of the 101 coins randomly (each coin has equal probability of being chosen) and starts flipping it. First flip shows heads, what is the probability the next flip is also heads? Answer is 2/3. The point is that knowing what was the first flip gives information about what biases a coin may have. The original problem has a minor technical twist that balls are drawn without replacement, but it’s an irrelevant detail for N>10 total balls.
    - jncfhnb 3y ago
      I mean, sure, what you’re describing is relevant to the original discussion, but you’ve posted it in a sub thread about how it changes if it is a 50/50 binomial distribution. You’ve replaced it with a binomial of a uniformly selected p, which ends up being, amusingly, the same as a uniform.
      - j7ake 3y ago
        As long as you think the flipper chose their coin from a bag of coins with varied biases, then first flip always gives actionable information. It’s mostly irrelevant what kind of distribution this bag of coins comes from (except for some degenerate cases). What matters is that selecting coins from this bag gives coins of varying biases.
        jncfhnb 3y ago
        Yes but the question being discussed assumes they are fair. That’s why it’s a different question.
eximius 3y ago
If I flip three coins and use the results to decide which color balls to put in a bag, I'll have one of these distributions: RRR,RRG,RGG,GGG. If I draw a red ball, I am more likely to have drawn from RRR or RRG than RGG (or GGG), so I have learned something about the distribution more than just how it was generated.
- jncfhnb 3y ago
  No you haven’t :) Let’s ignore RRG because it becomes a 50-50 anyway. RRR has a 100% chance of yielding Red. But is 3X less likely than RGG. RGG has a 33% chance of yielding Red but was 3X more likely. Edit: so still 50/50 Given a draw of red, you have exactly the same probability of having been in RRR and RGG. So you were equally likely to have 100% chance or a 0% chance.