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### In a game with 0.01 chance of survival, there are 100 participants, a specific player survives twice what are the odds?

The probability of a given player surviving twice (assuming survival is purely by chance and the two games are independent) is $.01^2 = .0001$, or 1:9999 odds. The probability of there being one ...
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The "paradox" here arises from sneaking in an implicit insinuation of independence that does not actually hold, which allows the argument to lead you to a wrong answer with a series of ...
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### Hazard function and survival analysis

Answer Given that your $\lambda(t)$ actually represents the instantaneous probability to die at time $t$ (the hazard), your function $F(t)$ computes the probability to die before a certain time $t$. ...

### MNIST with a TWIST, no labels given, only probabilities

By the way you’ve set up the problem, perfect accuracy is impossible, and we must accept that, even if we know a digit to be a $1$, there’s only a $90\%$ chance of being red, and that’s the best we ...
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The median minimizes expected absolute distance between observations sampled from the distribution and a single value. That is: $$\underset{m}{\text{min}}\text{ }\mathbb E\left[ \left\vert X - m\... • 31.3k 1 vote ### What is meaning of such notations in general?$$\mathbb{E}_{(\mathbf{u}, \mathbf{\sigma}, \mathbf{Y}_0)\sim N(0,1)\otimes \mu_D\otimes N(0,1)}\left[ e^{\mathbf{\sigma} t}\mathbf{Y}_0\mathbf{u}\right]=\int\int\int e^{\mathbf{\sigma} t}\mathbf{Y}_0\...
• 91.9k

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