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#### TheFallen018

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- Nov 9, 2017

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- Thread starter TheFallen018
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- Nov 9, 2017

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- Feb 7, 2012

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The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.

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- Nov 9, 2017

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Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks

The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.

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- Feb 7, 2012

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In the expression ${\color {red} p(x)}{\color {green}\dfrac{p(x)}{p(x)+p(7)}}$, the red $\color {red} p(x)$ gives the probability that the first roll of the dice gives the value $x$. The green fraction represents the probability of rolling $x$ again before rolling a $7$. My argument for that is that after rolling the first $x$, you can completely disregard any subsequent rolls until either an $x$ or a $7$ turns up. The only question is, which one of those will appear first. The relative probabilities of $x$ and $7$ are in the proportion $p(x)$ to $p(7)$. So out of a combined probability of $p(x) + p(7)$, the probability of an $x$ is $\dfrac{p(x)}{p(x)+p(7)}$, and the probability of a $7$ is $\dfrac{p(7)}{p(x)+p(7)}$.Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks

I hope that makes sense.