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A friend and I were disagreeing about how you calculate the probability for the following. Seeing as both of us aren't that strong in stats I figured I would come here. Sorry if this question is a bit pedestrian for you:

So lets say you are playing a game where you need to roll a certain sequence for a six sided die. Let's say you need to roll the die ten times and get all ones. The odds of that is easy enough $6^{10}$.

However, lets say in between each time you roll a die, you also have to flip a coin and get heads. So if you are rolling the die 10 times and in between each you flip a coin, that would be 9 coin flips. The odds of flipping a coin and getting heads every time would be $2^9$ right?

So then, what would the odds then be of rolling 1 on the die 10 times in a row, while also factoring in the fact you need to get heads on a coin flip in between each die roll? How do you calculate those odds?

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You have 19 sequential and independent trials, and need to be successful in all of them.

  • The first trial is a die roll, with a success probability of $\frac{1}{6}$.
  • The second trial is a coin toss, with a success probability of $\frac{1}{2}$.
  • The third trial is a die roll, with a success probability of $\frac{1}{6}$.

And so forth.

Now, since all trials are independent, the overall probability of success is just the product of all the separate probabilities of success, or

$$ \left(\frac{1}{6}\right)^{10}\times \left(\frac{1}{2}\right)^{9}\approx 3.2\times 10^{-11}. $$

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Instead of flipping a coin in between, you can roll a die with twelve sides. (for instance: construct the die so that there are light and dark sides numbered 1..6)

The coinflip is not needed after the last roll so you roll your new die nine times and finally one roll with the old die; this gives you the probability $$ \bigg(\frac{1}{12}\bigg)^9\cdot\frac{1}{6}. $$

$1.2^{-9} 10^{-9}\tfrac{1}{6} \sim \tfrac{1}{30}10^{-9}$

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    $\begingroup$ Although interesting, this solution seems to complicate the problem rather than simplify or explicate it. Isn't the crux of the matter to point out the independence of the coin flips and die rolls? $\endgroup$
    – whuber
    May 6 at 15:04
  • $\begingroup$ @whuber On the other hand bonus points for a fancy die! :D $\endgroup$
    – Alexis
    May 6 at 16:13
  • $\begingroup$ It is simpler in calculation but perhaps more costly in imagination. $\endgroup$
    – Hunaphu
    May 6 at 17:38

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