24 ms·

Did not read cuz its long, but I do believe in the gambler's fallacy, tosses are not independent. I can challenge any of you into gambling, using some JS code

Did not read cuz its long, but I do believe in the gambler's fallacy, tosses are not independent. I can challenge any of you into gambling, using some JS code we agree that is not buggy, using random.org's API, here is my rules: - we toss a coins in a row - we do not gamble until a string of tosses is the same side - the toss after I apply the gamble's fallacy (I'm stupid LOL), I choose the opposite side, we gamble who ever wins take a dollar - we keep doing it until one of us go bust Anyone up to the challenge?

aidenn0 4y ago
There's no upside for someone to take that challenge who does not believe in the gamblers fallacy. Assuming the gambler's fallacy is false, your proposed solution gives equal odds to you and your opponent, which advantages whomever has the larger stake to start with (in repeated 50-50 coin flips playing until bust, the larger stake is most likely to win). If you really believe in the gambler's fallacy, then there must be some greater-than 50-50 chance of the opposite side coming up. Offer me some odds (say 5/4) and define some table-stakes (thus removing the advantage of having a higher bankroll) and I'll definitely take you up on it. [edit] The random.org API seems unnecessarily complex, non-reproducible, and requires trusting whomever runs the JS. I propose an alternative of us each picking a value equal to the key+nonce length of ChaCha20 and pre-sharing the sha512 hash of that value. We then reveal our picked values, XOR them together and use ChaCha20 to generate bits for heads/tails.
- danamit 4y ago
  I can give you higher odds, but I will have to ask for a longer string of same side. Say like 20 in a row. In essence I will be gambling that there will be more 20 in a row same side than 21 in a row same side, which might be probably true, I have to test it. Edit: in reply to your edit, I will do this "virtual" bet and update this comment with the result. I will try to tweak it to be in my favor, if there is no way to do so, then I guess the gamble fallacy is a real fallacy. PS: not an actual bet, just tinkering stuff (betting is haram).
  - robinson7d 4y ago
    Not believing in Gambler’s Fallacy is already believing there will be more “20 in a row same side than 21 in a row same side” - its believing there are 50% more in a way. After the 20 have landed the next flip will still be 50/50. Just as 20 perfectly alternating heads/tails/heads/tails will lead to another 50/50 chance, or any other potential pattern the past may have held.
  - Dylan16807 4y ago
    > In essence I will be gambling that there will be more 20 in a row same side than 21 in a row same side, which might be probably true, I have to test it. I think you're missing a subtlety here. It's true that 20 in a row is twice as common as 21 in a row. But 20 in a row is exactly as common as 21 or more in a row. When you bet on that coin toss, you're not betting on 20 vs. 21. You're betting on 20 vs. 21+. So it's 50:50. For every thousand times you reach 20, the bet that it ends at 20 wins about 500 times, and the bet that it goes above 20 wins about 250 + 125 + 62 + 31 + 16 + 8 + ... = 500 times.
    - danamit 4y ago
      You are right actually, no easy money for me :(

aidenn0 4y ago

There's no upside for someone to take that challenge who does not believe in the gamblers fallacy. Assuming the gambler's fallacy is false, your proposed solution gives equal odds to you and your opponent, which advantages whomever has the larger stake to start with (in repeated 50-50 coin flips playing until bust, the larger stake is most likely to win). If you really believe in the gambler's fallacy, then there must be some greater-than 50-50 chance of the opposite side coming up. Offer me some odds (say 5/4) and define some table-stakes (thus removing the advantage of having a higher bankroll) and I'll definitely take you up on it. [edit] The random.org API seems unnecessarily complex, non-reproducible, and requires trusting whomever runs the JS. I propose an alternative of us each picking a value equal to the key+nonce length of ChaCha20 and pre-sharing the sha512 hash of that value. We then reveal our picked values, XOR them together and use ChaCha20 to generate bits for heads/tails.

danamit 4y ago
I can give you higher odds, but I will have to ask for a longer string of same side. Say like 20 in a row. In essence I will be gambling that there will be more 20 in a row same side than 21 in a row same side, which might be probably true, I have to test it. Edit: in reply to your edit, I will do this "virtual" bet and update this comment with the result. I will try to tweak it to be in my favor, if there is no way to do so, then I guess the gamble fallacy is a real fallacy. PS: not an actual bet, just tinkering stuff (betting is haram).
- robinson7d 4y ago
  Not believing in Gambler’s Fallacy is already believing there will be more “20 in a row same side than 21 in a row same side” - its believing there are 50% more in a way. After the 20 have landed the next flip will still be 50/50. Just as 20 perfectly alternating heads/tails/heads/tails will lead to another 50/50 chance, or any other potential pattern the past may have held.
- Dylan16807 4y ago
  > In essence I will be gambling that there will be more 20 in a row same side than 21 in a row same side, which might be probably true, I have to test it. I think you're missing a subtlety here. It's true that 20 in a row is twice as common as 21 in a row. But 20 in a row is exactly as common as 21 or more in a row. When you bet on that coin toss, you're not betting on 20 vs. 21. You're betting on 20 vs. 21+. So it's 50:50. For every thousand times you reach 20, the bet that it ends at 20 wins about 500 times, and the bet that it goes above 20 wins about 250 + 125 + 62 + 31 + 16 + 8 + ... = 500 times.
  - danamit 4y ago
    You are right actually, no easy money for me :(