Conditional probability question?

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Conditional probability question?

A game is played with one standard die. On a turn, if a player rolls a 1, he gets two more rolls. If he rolls a 2, he gets one more roll. If he rolls anything else, his turn ends. His score is the sum of all the rolls on his turn (up to 3 rolls). If a player scores 5 points on a turn, what is the probability that it took 2 rolls?

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Answer 1 · 2016-12-13T05:11:58

We use our good friend Bayes' rule to help us deduce:
$P r (took 2 rolls ∣ score is 5) = \frac{2}{15} .$ $Pr("took 2 rolls"|"score is 5")=2/15.$

Explanation:

Warning: Long answer ahead!

One form of Bayes' rule states the following:

$P r (B ∣ A) = \frac{P r (A B)}{P r (A)} = \frac{P r (A ∣ B) \cdot P r (B)}{P r (A)}$ $Pr(B|A)=(Pr(AB))/(Pr(A))=(Pr(A|B)*Pr(B))/(Pr(A))$

This allows us to write a conditional probability of "B given A" in terms of "A given B", which may be easier to calculate. In this question,

B = "die rolled twice", and
A = "score is 5".

Let's write this into the equation:

$P r (2 rolls ∣ score is 5) = \frac{P r (score is 5 ∣ 2 rolls) \cdot P r (2 rolls)}{P r (score is 5)}$ $Pr("2 rolls"|"score is 5")=(Pr("score is 5"|"2 rolls")*Pr("2 rolls"))/(Pr("score is 5"))$

The numerator on the RHS is easy to deduce; however, the denominator, in its current state, is not. But we can make it easier. We recall that, if $A$ $A$ can be partitioned into $k$ $k$ disjoint events $A \cap C_{1}, A \cap C_{2}, ..., A \cap C_{k}$ $AnnC_1, AnnC_2, ..., AnnC_k$ (where none of these intersected regions overlap, and together they all form $A$ $A$ ), then

$P r (A) = k \sum i = 1 P r (A ∣ C_{i}) \cdot P r (C_{i})$ $Pr(A)=sum_(i=1)^kPr(A|C_i)*Pr(C_i)$

We actually have that here; the $C_{i}$ $C_i$ 's will be the number of possible rolls on a turn. What we're saying is that

$P r (score is 5) = P r (score is 5 ∣ 1 roll) \cdot P r (1 roll)$ $Pr("score is 5")=Pr("score is 5"|"1 roll")*Pr("1 roll")$
$XXXXXXXXX + P r (score is 5 ∣ 2 rolls) \cdot P r (2 rolls)$ $color(white)"XXXXXXXXX"+Pr("score is 5"|"2 rolls")*Pr("2 rolls")$
$XXXXXXXXX + P r (score is 5 ∣ 3 rolls) \cdot P r (3 rolls)$ $color(white)"XXXXXXXXX"+Pr("score is 5"|"3 rolls")*Pr("3 rolls")$

Kind of a "the whole is equal to the sum of its parts" thing.

From hereon, let's use the shorthand $S_{5}$ $S_5$ for "the score was 5" and $R_{n}$ $R_n$ for " $n$ $n$ number of rolls". Putting this all together, we have

$P r (R_{2} ∣ S_{5}) = \frac{P r (S_{5} ∣ R_{2}) P r (R_{2})}{P r (S_{5} ∣ R_{1}) P r (R_{1}) + P r (S_{5} ∣ R_{2}) P r (R_{2}) + P r (S_{5} ∣ R_{3}) P r (R_{3})}$ $Pr(R_2|S_5)=(Pr(S_5|R_2)Pr(R_2))/(Pr(S_5|R_1)Pr(R_1)+Pr(S_5|R_2)Pr(R_2)+Pr(S_5|R_3)Pr(R_3))$
$= \frac{P r (S_{5} \cap R_{2})}{P r (S_{5} \cap R_{1}) + P r (S_{5} \cap R_{2}) + P r (S_{5} \cap R_{3})}$ $=[Pr(S_5nnR_2)]/[Pr(S_5nnR_1)+Pr(S_5nnR_2)+Pr(S_5nnR_3)]$

Now, we do the calculations!

$P r (S_{5} \cap R_{1}) = P r (roll a 5)$ $Pr(S_5nnR_1)=Pr("roll a 5")$
$= \frac{1}{6}$ $=1/6$
$P r (S_{5} \cap R_{2}) = P r (roll a 2, then a 3) = \frac{1}{6} \cdot \frac{1}{6}$ $Pr(S_5nnR_2)=Pr("roll a 2, then a 3")=1/6*1/6$
$= \frac{1}{36}$ $=1/36$
$P r (S_{5} \cap R_{3}) = P r [roll a 1; then (1,3), (2,2), or (3,1)] = \frac{1}{6} \cdot \frac{3}{36}$ $Pr(S_5nnR_3)=Pr["roll a 1; then (1,3), (2,2), or (3,1)"]=1/6*3/36$
$= \frac{1}{72}$ $=1/72$

Finally, we place these values back into the equation for $P r (R_{2} ∣ S_{5}) :$ $Pr(R_2|S_5):$

$P r (R_{2} ∣ S_{5}) = \frac{\frac{1}{36}}{\frac{1}{6} + \frac{1}{36} + \frac{1}{72}} \cdot \frac{72}{72}$ $Pr(R_2|S_5)=(1/36)/(1/6+1/36+1/72)color(blue)(*72/72)$

$P r (R_{2} ∣ S_{5}) = \frac{2}{12 + 2 + 1}$ $color(white)(Pr(R_2|S_5))=(2)/(12+2+1)$

$P r (R_{2} ∣ S_{5}) = \frac{2}{15}$ $color(white)(Pr(R_2|S_5))=2/15$

For completeness, $P r (R_{1} ∣ S_{5}) = \frac{12}{15} = \frac{4}{5},$ $Pr(R_1|S_5)=12/15=4/5,$ and $P r (R_{3} ∣ S_{5}) = \frac{1}{15} .$ $Pr(R_3|S_5)=1/15.$ These three probabilities sum to 1, which is what we'd expect, since if the player's score was 5, it had to take either 1, 2, or 3 rolls.

Bonus:

There's a great Numberphile video on YouTube discussing Bayes' rule here:

Answer 2 · 2016-12-13T22:10:07

Alternate explanation to the long version I previously posted.
We still get $P (2 rolls ∣ scored 5) = \frac{2}{15} .$ $P("2 rolls"|"scored 5")=2/15.$

Explanation:

What are the different ways to score 5 points, and what are their probabilities? Well, we could roll:

a 5 right away;

a 2, then a 3;

a 1, then a 1, then a 3;

a 1, then a 2, then a 2; or

a 1, then a 3, then a 1.

$ul("One roll:")$
$P(5)=1/6$

$ul("Two rolls:")$
$P(2, 3)=1/6*1/6=1/36$

$ul("Three rolls:")$
$P(1, 1, 3)=1/6*1/6*1/6=1/216$
$P(1, 2, 2)=1/6*1/6*1/6=1/216$
$P(1, 3, 1)=1/6*1/6*1/6=1/216$

So the probability of scoring 5 points is simply the sum of the probabilities above. The final answer we seek is the fraction of this sum belonging to the two-roll method—in other words, the relative probability of "scoring 5 in two rolls" to "scoring 5 in any way".

Using conditional probability, we get:

$P("2 rolls"|"scored 5")=[P("2 rolls"nn"scored 5")]/[P("scored 5")]$

$=[P(2,3)]/[P(5)+P(2,3)+P(1,1,3)+P(1,2,2)+P(1,3,1)]$

$=(1/36)/(1/6+1/36+1/216+1/216+1/216)color(blue)(*216/216)$

$=6/(36+6+1+1+1)$

$=6/45=2/15$

which is the same answer as before.

This method is easier in this case; however, the Bayesian method is more general and has much broader applicability.

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Conditional probability question?

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