Quadratic Equation
34
100Ṁ2589
May 30
68%
2 real roots
3%
1 real root
29%
imaginary roots

I will roll 4 numbers a b c d in that order each from 1 to 19. If the equation ax^2+bx+c=d has 2 real roots then this question will resolve in option 1. If it has one real root then this question will resolve in option 2. If it has imaginary roots then this question will resolve in option 3.

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There are 130321 equally likely possibilities for abcd (19^4 combinations w/ replacement).
41246 combinations have imaginary roots. p = 41246/130321 ~= 0.31650

320 combinations have 1 real root. p = 320/130321 ~= 0.0024555

88755 combinations have 2 real roots. p = 88755/130321 ~= 0.68105

@ikirpichev Roll time

sold Ṁ1 NO

@ikirpichev Are you using FairlyRandom to generate the numbers?

@evan yes

bought Ṁ50 NO

what is the point of a market like this where there is an obvious "right answer" to the probabilities? pretty much anyone who trades will just calculate the right answer.

bought Ṁ100 NO

@CrypticQccZ Not everyone lol

Hmmm:

```

>>> results = [random.randint(1,19) ** 2 - 4 * random.randint(1,19) * random.randint(1,19) for in range(1_000_000)]

>>> sum(1 for x in results if x < 0)

768280

>>> sum(1 for x in results if x == 0)

3666

>>> sum(1 for x in results if x > 0)

228054

```


If this is right, odds should be close to 76.8% imaginary, 0.4% 1 real, and 22.8% 2 real. I'm attempting to calculate the portion under the square root in the quadratic formula, which is the only piece that determines this. I'm open to being told I made some mistake here, but betting up the odds on my analysis anyways.

bought Ṁ10 YES

@MarcusM What about the fourth random variable, d?

bought Ṁ2 YES

@4fa Reading comprehension fail:

```
>>> results = [random.randint(1,19) ** 2 - 4 * random.randint(1,19) * (random.randint(1,19)-random.randint(1,19)) for in range(1_000_000)]

>>> sum(1 for x in results if x < 0)

316873

>>> sum(1 for x in results if x == 0)

2364

>>> sum(1 for x in results if x > 0)

680763
```
🤦‍♂️

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