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4 votes
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Calculating the Probability of Two Pairs of Matching Faces When Rolling 4 Dice

Here are two ways to think of it. First choose the number on the first die (6 options). Next, choose which other die has the same number (3 options). Finally, choose the number on the other two dice (...
Especially Lime's user avatar
0 votes
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A question about dice combinations - cube digit pairs

You're not properly accounting for the fact that you have 12 spaces and 9 numbers, which means several numbers will appear on both dice. (Since we don't use 7, we can duplicate up to 4 numbers.) Your ...
Calvin Lin's user avatar
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12 votes
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How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

Let $X_1,...,X_k$ be i.i.d. Uniform($\{1,...,m\}).$ Then the distribution of $S_k=X_1+...+X_k$ is given exactly by a recursion proved in this paper$^\dagger$, which expresses this distribution in ...
r.e.s.'s user avatar
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6 votes

How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

While not a rigorous answer, I don't know how to get the number analytically, this should suffice. Let $X_i $ be the result of the ith roll and $X_{i}\in\left\{ 1,2,3,4,5,6\right\} $ with equal ...
Danny Blozrov's user avatar
4 votes

How many rolls are sufficient to ensure, with probability 99%, that the sum is greater than 100?

One way to do this is to invoke the central limit theorem which says that as you sum more and more i.i.d random variables, their sum get close to a Gaussian. Let's assume that by the time the sum has ...
Rohit Pandey's user avatar
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6 votes
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What is wrong with my logic in this dice and coin problem?

Outcomes such as $HHHTH$ and $HTHHTH$ are not permissible because they do not have four successive heads, yet they are included in your computation. To get $4$ successive heads, there are only $3$ ...
heropup's user avatar
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1 vote
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$4$ Die Sum equals $20$ - How to generalise it to $n$ die

Thanks to @underflow for their explanations in the comments: This generalisation can be done using generating functions. Let $x$ be some symbol. Consider the expression $\frac{𝑥+𝑥^2+𝑥^3+𝑥^4+𝑥^5+𝑥...
Devansh Agarwal's user avatar
2 votes
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Strange relationship between two expectations related to die rolls

Let $X_n^{[i]}$ with $i \in \{1, \dots, 6\}$ be a random variable denoting the number of rolls to see $n$ consecutive $i$'s and $Y_{n + 1}$ be the random variable representing the number of rolls ...
msantama's user avatar
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1 vote

Strange relationship between two expectations related to die rolls

Yes. Substituting in using Equation 1, we are trying to show that $$Y_{n+1} = 6Y_n + 1$$ The derivation for this is essentially the same argument: we keep rolling $n$ in a row and then roll one more; ...
HeartsOfFire's user avatar
5 votes

How to derive the expected number of rolls until a number appears $k$ consecutive times

Consider the case of getting some number $n-1$ consecutively. For eg. $5 5 5 ... 5 \_ $ The last $n$th roll, can either be a $5$ or it may not be $5$. In the first case, a number has appeared n ...
Devansh Agarwal's user avatar
0 votes

What is the logic behind the combinatorics used to find probability of getting a sum of 16 when 4 dice are rolled?

The basic idea is to treat "four dice sum to 16" as a "16 balls into 4 bins" problem so that we can use the familiar stars-and-bars machinery to do the counting for us. Sticks-and-...
DotCounter's user avatar
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5 votes
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What is the logic behind the combinatorics used to find probability of getting a sum of 16 when 4 dice are rolled?

The question is equivalent to the following: how many ways are there to write $16=d_1+d_2+d_3+d_4$ where $d_1,\ldots,d_4$ are integers between $1$ and $6$ (corresponding to the values on the four dice,...
Especially Lime's user avatar

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