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Two fair six sided die are rolled fairly and the scores are noted. What is the probability that it takes 3 or more rolls to get a score of 7?

Here is what I have got so far:

The probability that it takes $3$ or more rolls to get a score of $7$ is equal to $$1-P(\text{you get a score of $7$ on less than $3$ rolls})$$

So $$P(\text{get a $7$ on zero rolls})=0\\P(\text{get a $7$ on $1$ roll})=\frac16\\P(\text{get a $7$ on the $2$ roll})=\frac16$$

So $$1-\left(0+\dfrac16+\dfrac16\right)=1-\dfrac26=\dfrac46$$

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  • $\begingroup$ the probability of getting the 7 on the second roll is the probability of failing the first one and getting it on the second one. $\endgroup$ – 3d0 Feb 24 '15 at 13:46
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#Pairs that sum 7 = 6

#total pairs = 36

$P($getiing a 7 after the third roll or later$)$ = $P($getting it at the third roll$)$ and $P($getting it at the forth roll$)$ and ... etc

getting that each probability is exclusive (that is you get the roll on the $n$th is because you could'nt get the roll between $1$ and $n-1$th) you get

$P($getiing a 7 after the third roll or later$)$ = $\sum_{i>2} P($ getting the roll on the $i$th$)$

$P($Getting the roll on the i-th$)$ = $P($not getting the roll$)^{i-1}P($getting the roll$)$ = ${(\dfrac{5}{6})}^{i-1}\dfrac{1}{6}$

therefore

$$P = \sum_{i=3}^{\infty}{(\dfrac{5}{6})}^{i-1}\dfrac{1}{6} = \sum_{i=0}^{\infty}{(\dfrac{5}{6})}^{i}\dfrac{1}{6} - \dfrac{1}{6} -\dfrac{5}{6}\dfrac{1}{6} = \dfrac{1}{6}\dfrac{1}{1-\dfrac{5}{6}}-\dfrac{1}{6}-\dfrac{5}{36}=1-\dfrac{11}{36}=\dfrac{25}{36}$$

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    $\begingroup$ This is correct but the problem can be solved easily without an infinite sum. $\endgroup$ – David K Feb 19 '15 at 18:33
  • $\begingroup$ How to solve without using infinite sum? $\endgroup$ – Zeus Feb 19 '15 at 18:37
  • $\begingroup$ See my answer already posted. You have only a small error in your calculation. $\endgroup$ – David K Feb 19 '15 at 18:41
  • $\begingroup$ Did you see my asumption that each probability is exclusive? that means the probability to get the sum on the third or later is the complement of getting the sum on the first two, therefore $P($getting the roll on the third or more$) = 1 - P($getting the roll on the first two$) = 1 -\dfrac{1}{6} - \dfrac{5}{36}=\dfrac{25}{36}$ $\endgroup$ – 3d0 Feb 19 '15 at 18:41
  • $\begingroup$ That is correct, $1-\frac16-\frac{5}{36}$ is one way to do it without an infinite sum. (3do: the "error" I mentioned was in the OP, not in your calculations.) $\endgroup$ – David K Feb 19 '15 at 19:10
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One of the two dices has red eyes, the other one blue eyes. On a single throw of the two dices, the probability that the number of blue eyes is $7$ minus the number of red eyes is ${1\over6}$. Therefore with probability ${5\over6}$ you don't make a total of $7$ in the first throw, and with probability ${25\over36}$ you don't make a total of $7$ in the first two throws. It follows that the answer to your problem is ${25\over36}$.

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1) If you mean 'to get an $overall$ score of 7, then the only way to get less than 7 (i.e. 6) in the first three rolls is to get 1 on each die every time you roll them, hence the probability is $1-\frac{1}{6^6}$.

2) If you mean 'to get a score of 7 on $each$ roll', then you need to find the probability that on the first three rolls the sum is strictly less than 7, i.e. 6 or less. You have 11 outcomes, 5 of them are less than 7. Can you handle from here?

EDIT: this is for the case when you need to $toss$ exactly 7: there's a total of 36 outcomes in each toss. 6 of can get you 7 (why?). Hence 30 of the don't. What you need is the probability to not toss 7 on all three first tosses.

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  • $\begingroup$ I meant score a 7 on each roll,I still don't understand how you got 11 outcomes im counting 15.can you explain a little more on how you got a 11 outcomes $\endgroup$ – Zeus Feb 19 '15 at 18:06
  • $\begingroup$ I'm sorry...you need a score of $exactly$ 7 or at least 7? $\endgroup$ – Alex Feb 19 '15 at 18:07
  • $\begingroup$ A pair of fair six-sided dice is rolled until a total score of 7 is obtained. Determine the probability that 3 or more rolls are required. sorry here is the actually question $\endgroup$ – Zeus Feb 19 '15 at 18:10
  • $\begingroup$ please see the edit $\endgroup$ – Alex Feb 19 '15 at 18:11
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The basic approach in the question is a good one, except for an error in combining the probability that the first roll is a $7$ and the probability that the second roll is a $7$.

In general, for two events $A$ and $B$, $$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$ If $A$ is the event of rolling $7$ on the first roll, and $B$ is the event of rolling $7$ on the second roll, then $P(A) = P(B) = \frac 16$ but $P(A \cup B) < \frac 13$ because $P(A \cap B) > 0$. You need to find the probability that both the first and second rolls will be $7$, and then you can apply the formula above.


A similar but slightly different approach is to let $A$ be the probability that the first roll is a $7$ and let $B$ be the probability that the first roll is not $7$ but the second roll is a $7$. You can confirm that any sequence of rolls that gets $7$ before the third roll has either event $A$ or event $B$. But now $A$ and $B$ are disjoint, that is, $P(A \cap B) = 0,$ so you can simply add the probabilities to get $P(A \cup B),$ and the solution is found by evaluating $1 - (P(A) + P(B))$. This is similar to the attempted solution, except that $P(B) < \frac16.$

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  • $\begingroup$ thanks, I see where my mistake happened now $\endgroup$ – Zeus Feb 19 '15 at 19:05

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