Probability of Two Different and Specific Dice Rolls (2-12) in 3 chances. EX: a 7 and a 6

Question

Let's say I was rolling two dice (2-12 results) three times. What is the probability I get two specific results with different probabilities during those three trials?

3 trials where I roll two dice. I'm looking for how often I'll roll AT LEAST one 6 (5/36 each time) and one 7 (6 out of 36). I want to be able to any tell any number combination. I know the likelihood of each event and that this is similar to binomial calculation, but the different probabilities make sites like this inadequate: http://stattrek.com/online-calculator/binomial.aspx

Is there a formula for this? Something involving factorials? Crunched it out with a friend and we got 13.888888% which is much bigger than the previously mentioned site's 7.4% for getting two 7s.

Thank you so much!!!!

Answer 1

There are three ways to get at least a $6$ and a $7$: Either you get a third different result, or you get another $6$, or you get another $7$. There are $6$ different orders for three different numbers, and $3$ different orders for a pair and a singleton, so the total probability is

$$ 6\cdot\frac5{36}\cdot\frac6{36}\cdot\left(1-\frac5{36}-\frac6{36}\right)+3\cdot\left(\frac5{36}\right)^2\cdot\frac6{36}+3\cdot\left(\frac6{36}\right)^2\cdot\frac5{36}\\ =\frac{6\cdot5\cdot6\cdot25+3\cdot5^2\cdot6+3\cdot6^2\cdot5}{36^3}\\ =\frac{5490}{36^3}=\frac{305}{2592}\approx11.77\%\;. $$

Answer 2

(Disclaimer: First this is an answer to the question "What is the probability I get two specific results with different probabilities during those three trials?" as i understood it...)

If i correctly understand the setting, then we have at each step (one of the three steps) the following (sums, possibilities to get the sum,) probability:

• sum is $2$, $(1,1)$, probability $1/36$,
• sum is $3$, $(1,2)$, $(2,1)$, probability $2/36$,
• sum is $4$, $(1,3)$, $(2,2)$, $(3,1)$, probability $3/36$,
• sum is $5$, $(1,4)$, $(2,3)$, $(3,2)$, $(4,1)$, probability $4/36$,
• sum is $6$, $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, $(5,1)$, probability $5/36$,
• sum is $7$, $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, $(6,1)$, probability $6/36$,
• sum is $8$, $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(5,2)$, probability $5/36$,
• sum is $9$, $(3,6)$, $(4,5)$, $(4,4)$, $(6,3)$, probability $4/36$,
• sum is $10$, $(4,6)$, $(5,5)$, $(6,4)$, probability $3/36$,
• sum is $11$, $(5,6)$, $(6,5)$, probability $2/36$,
• sum is $12$, $(6,6)$, probability $1/36$.

Now we further group events with the same probability:

• sum $2,12$, probability $q_1=2/36$,
• sum $3,11$, probability $q_2=4/36$,
• sum $4,10$, probability $q_3=6/36$,
• sum $5,9$, probability $q_4=8/36$,
• sum $6,8$, probability $q_5=10/36$,
• sum $7$, probability $q_6=6/36$.

Now we can forget about the initial way to generate these probabilities, call these new possible results "boxes".

We start a tree steps experiment based on the "boxes" and ask for getting not three times the same box.

(At least i understood like this the question "What is the probability I get two specific results with different probabilities during those three trials?")

The probability is $$ 1-\sum_{1\le j\le 6}q_j^3\ . $$ This is (computed in sage):

sage: Q = [2/36, 4/36, 6/36, 8/36, 19/36, 6/36]
sage: 1 - sum([ q^3 for q in Q ])
4309/5184
sage: _.n()
0.831211419753086

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Now the second question.

We want at least once the sum $6$, and at least once the sum $7$ in the three trials. The new boxes are $6$, $7$, and anything else, $*$. The probabilities are $p_6=5/36$, $p_7=6/36$, 4p_*=1-p_6-p_7=25/36$. (In between, there is already an answer to this question, so i will go an other way...)

At the first step we get a $6$, a $7$, or a $*$, so we split into cases.

• $6$ first, then we need a $7$ in two steps, it comes either next, or finally, probability $$p_6(p_7+(1-p_7)p_7)=p_6p_7(2-p_7)\ .$$

• $7$ first, then we need a $6$ in two steps, it comes either next, or finally, probability $$p_7p_6(2-p_6)\ .$$

• $*$ first, then only $67$ or $76$ in the left trials are matching, probability $$(1-p_6-p_7)2p_6p_7\ .$$

Putting all together we get $$ p_6p_7\Big(\ (2-p_7)+(2-p_6)+2(1-p_6-p_7)\ \Big) = 3p_6p_7\Big(\ (2-p_6-p_7)\ \Big) \ . $$ This is explicitly $$ \frac{305}{2592} \approx \color{red}{0.117669}753086420\dots \ . $$

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Comment:

From the first comment to this answer, it seems important to know the following generalization: Let $n\ge 2$ be integer. Which is the probability $P(n)$ for the event, that in an $n$-step trial there is at least one occurence of the sum $6$ (probability for it in one step being $p_6=5/36$), and at least one of the sum $7$ (probability for it in one step being $p_7=6/36$).

(Since in the comment box there is not so much space, i will do it here.)

It is simplest to compute the complementary probability $1-P(n)$. Let $A_6(n)$, resp. $A_7(n)$ be the event that the sum $6$, resp. $7$ does not show in $n$ trials. Then $$ \begin{aligned} 1-P(n) &= \text{Probability that either $6$, or $7$ does not appear in $n$ steps} \\ &= \text{Probability}(A_6(n)\cup A_7(n)) \\ &= \text{Probability}(A_6(n)) + \text{Probability}(A_7(n)) - \text{Probability}(A_6(n)\cap A_7(n)) \\ &=(1-p_6)^n+(1-p_7)^n-(1-p_6-p_7)^n\ . \\[3mm] &\text{So:} \\ P(n) &= 1-(1-p_6)^n-(1-p_7)^n+(1-p_6-p_7)^n\ . \end{aligned} $$

Answer 3

Let's say you want at least one each of the sums $m$ and $n$ with $m\ne n$. Let's say the respective probabilities of sum $m$ and sum $n$ on a roll of two dice are $p_m$ and $p_n$.

Case 1: Get $m$ twice and $n$ once in some order. Probability is $3p_m^2p_n$.

Case 2: Get $m$ once and $n$ twice in some order. Probability is $3p_mp_n^2$.

Case 3: Get $m$ once and $n$ once and some different value once. Probability is $6p_mp_n(1-p_m-p_n)$.

Add these up to get your total probability.