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A loaded dice has the property that when the dice is thrown the probability of showing a given number is proportional to the number. For example $2$ is twice as likely to show up compared to $1$ and $3$ is thrice as likely to show up compared to $1$, And so on. What is the probability that when the dice is thrown twice the sum is $4$ or less. $$ P = \frac 36 \cdot \frac 16 +\frac 16 \cdot \frac 36 + \frac 26 \cdot \frac 26 + \frac 16 \cdot \frac 26 + \frac 26 \cdot \frac 16 + \frac 16 \cdot \frac 16 = \frac{15}{36} $$ Where I am getting it wrong?

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You can why this is wrong directly, if you consider the odds of rolling a $6$. $6/6$ surely can't be right. –  Cruncher Aug 7 at 13:39

2 Answers 2

up vote 11 down vote accepted

Since $1+2+\cdots+6=21$, the probabilities of $1,2,3,4,5,6$ in $1$ toss are respectively $\frac{1}{21}$, $\frac{2}{21}$, and so on up to $\frac{6}{21}$.

The probability of a sum of $2$ in $2$ throws is $\frac{1}{21}\cdot\frac{1}{21}$.

The probability of a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$.

The probability of a sum of $4$ is $2\cdot \frac{1}{21}\cdot\frac{3}{21}+\frac{2}{21}\cdot\frac{2}{21}$.

Add up.

Remark: Your numerator, and therefore the basic analysis, was right. The denominators were not.

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The probability of a sum of 3 is not 2/21: You have a 2/21 chance of throwing a 1 followed by a 2, and another 2/21 chance of throwing a 2 followed by a 1. Therefore, the probability of a sum of 3 is 4/21. Similar logic applies to your sum of 4. –  pabo Aug 7 at 16:15
The answer does not say that the probability of a sum of $3$ is $\frac{2}{21}$. The answer above says that the probability of getting a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$. This is not further simplified, but if it were it would give $\frac{4}{21^2}$. Similarly, the probability of a sum of $4$ is given in unsimplified form in the answer. It simplifies to $\frac{10}{21^2}$. –  André Nicolas Aug 7 at 17:38
You're absolutely correct; I misread it as "The probability of a sum of 3 is 2 (1/21 * 2/21)." Sorry for the confusion! –  pabo Aug 7 at 20:00

This is the sum of combinations for 4 or less. When a sum to a number with elements that arent bounded (like in this case because 4 is a number into the values of the dice,i.e., $4\leq 6\ or\ X\leq D$. For others sums where is a bound on the elements of composition see this)

The sums on pairs will go from $(x-1)+1$ to $(\lfloor\frac{x}{2}\rfloor+\lceil\frac{x}{2}\rceil)$ and for any pair the probability will be


The exponent is a way to track the multiplicity of pairs that can be 2 or 1, i.e., you can sum 4 as 3+1 or 1+3 but you only have one 2+2 combination (the $\delta_{(j+1,x/2)}$ of exponent value 0 or 1, it is the Kronecker delta function that compare the vale of $j+1$ and $x/2$). D is the number of sides of the dice. The denominator is the sum of values from 1 to D.

This double sum over $p_j$ give the probabilities with your conditions

$$P(x\leq X)=\sum_{x=2}^{X}\sum_{j=0}^{\lfloor\frac{x}{2}\rfloor}\frac{p_j}{x-1}; only\ valid\ for\ X\leq D$$

The denominator $x{-}1$ is the total of different ordered pairs that sum x. This come from $\binom{n-1}{k-1}$ that is the amount of combinations to sum n with groups of k (where the k elements are naturals excluding zero and are not bounded), for pairs this is just n-1 because if k=2 then the denominator on the binomial coefficient is 1.

I know that this is a complicated (unnecessary complex) way to resolve the question... but I like to leave any answer on symbolic terms instead of just brute-force/numeric solutions.

A more interesting thing (and complex) can be the same case but for any sum on the dice (not just where $X\leq D$) and any number of dice (not just sum of pairs). I hope this maybe interesting/funny for you.

P.S.: of course an alternative to Kronecker delta is to make the full sum to x instead of $\lfloor\frac{x}{2}\rfloor$ and quit the factor 2.

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