Chance of scoring in dice game midnight (1-4-24)
Mia Lopez 
The game midnight ($1-4-24$) involves rolling $6$ dice, keeping at least one, then rolling the remaining dice etc, keeping at least one, etc. At the end if you have kept a $1$ and a $4$, you get to score the remaining $4$ dice. If not, you score zero. So to be successful you need to have rolled both a $1$ and a $4$ during your go, and kept them.
The best strategy is therefore to keep the first $1$ and $4$ that come up, and if you haven't kept a dice this throw, keep the highest.
Wikipedia entry says that the maximum probability of being able to score (have a $1$ and a $4$) is
$$P(\text{success}) = 1-\left(2\left(\frac{5}{6}\right)^{21}-\left(\frac{4}{6}\right)^{21}\right) = 0.956727$$
I get most of this formula except one piece, so I must be missing something obvious.
The $1-$$($ $)$ is just the $1 - P(\text{failing to get 1 and a 4})$
So to fail to score, you have failed to get both a $1$ and a $4$ in $6+5+4+3+2+1= 21$ rolls. The chance of failing to get a $1$ in $21$ throws is $(5/6)^{21}$, and the same for a $6$, hence the $2(5/6)^{21}$, I assume it's related to the probability of not getting a $4$ or a $1$ on a single dice, raised to the power of $21$ throws, but what exactly is this term and why there is a minus in front of it?
Any suggestions on what I've missed would be greatly appreciated!
$\endgroup$ 21 Answer
$\begingroup$That additional term is precisely the probability that neither of $1$ or $4$ appear. Here we have two events, so let's name them:
Event A: You roll no $1$'s in the $21$ rolls available.
Event B: You roll no $4$'s in the $21$ rolls available.
You are trying to compute the probability of $A\cup B$ - that is, the probability that either of these things happen. To do this, you can use a formula, which is a simple case of the inclusion-exclusion principle:$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$which states the the probability of at least one of the events happening equals the sum of the probabilities of either event happening, minus the probability that they both happen - since we, in a sense, counted the cases where both happen twice in the sum $P(A) + P(B)$.
As a simpler example, if you wanted to compute the probability that, if you flip two fair coins, you get at least one to land heads up, you could let event A correspond to the first coin being heads up and event B correspond to the second coin. Both events happen with probability $1/2$ - but we can't simply add them to get the probability of either happening, because that discounts the possibility that both comes up head. So we subtract this probability of $1/4$ to get the correct answer that you will get at least one heads with probability $1/2+1/2-1/4 = 3/4$.
In the example you give, $P(A)$ and $P(B)$ are both equal to $(5/6)^{21}$, being the probability of missing the particular number on every roll. Then $P(A\cap B)$ is the probability of missing both $1$ and $4$ on every roll, which is $(4/6)^{21}$. Thus, the probability of either not getting a $1$ or not getting a $4$ is $(5/6)^{21}+(5/6)^{21}-(4/6)^{21}$, which is the term given there.
This reasoning can be extended further - with some subtleties. If you had three events, you would write$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B+\cap C).$$This works because when we sum $P(A)+P(B)+P(C)$, we first need to subtract out the overlaps of the cases - as we have counted them twice. However, once we've done that, we counted cases where all three happen thrice in $P(A)+P(B)+P(C)$, but then subtracted them thrice in $-P(A\cap B)-P(A\cap C)-P(B\cap C)$, so then have to add those cases back in, as we cancelled them entirely. You can keep coming up with formulas - and they're very systematic - to deal with any number of events and this tends to be a powerful technique.
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