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Consider rolling a 6-sided die continuously and trying to tally as high a score as possible (the sum of all rolls). If you roll a 1, then your turn is over and your score is 0. So for each successful roll, the expectation value should be 4. According to Knizia (1999), an approximately optimal strategy would be to attempt to roll to a score of 20 and then stop. (S)he states:

"...we know the true odds of such a bet are 1 to 5. If you ask yourself how much you should risk, you need to know how much there is to gain.....If you put 20 points at stake, this brings the odds to 4 to 20, that is 1 to 5, and makes a fair game....Whenever your accumulated points are less than 20, you should continue throwing, because the odds are in your favor."

Don't understand this, is not rolling to 20 on average essentially rolling five times (without getting a 1)? The probability of this happening is only about 40%. So wouldn't the odds be about 60% that you would roll a 1 and score 0 if you always try to roll up to 20?

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If your current score is $x$, you get $x$ if you choose not to roll, else if you choose to roll, your expected score based on the results of the roll is $$\left({\small{\frac{1}{6}}}\right)(0) + \left({\small{\frac{1}{6}}}\right)(x+2) + \cdots + \left({\small{\frac{1}{6}}}\right)(x+6) = \frac{20 + 5x}{6}$$ So you should continue rolling as long as $$x < \frac{20 + 5x}{6}$$ and you should stop if $$x > \frac{20 + 5x}{6}$$ Thus, if your score is less than $20$, you should roll; if more than $20$, you should stop; and if equal to $20$, you're indifferent.

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