Afternoon all; I'm following along with Harvard's Extension school class: "Sets, Counting, and Probability" and am stuck on a particular question:

Elementary Probability 2nd ed. - Stirzaker Elementary Probability 2nd ed. - Stirzaker

I've been able to (after much work), get the correct answer for the question:

"Show that the probability that Achilles is last to throw a six is 305/1001" by way of

P(r) = 1/6 * [(5/6)^r * (1 - (5/6)^r)^2]

and then calculating the geometric sum between the broken down components; multiplying their total aggregate by 1/6.

However, for the question, "For each player, find the probability that he or she is the first to roll a six." I've tried pages and pages of what I think may be correct, but just cannot figure out how the answers provided in the index were attained.

I've tried attacking the problem the same way as above, by letting

P(r) = 1/6 * [ (5/6)^3r ]

Which to me seems the correct route for P(A) - the intersection of each A, B, and C having rolled all {1,2...5} to this point, where P(r) is the probability of the event where A rolls a six first.

The related lecture discussion can be found at 1h39m at the Lecture recording

Answers from Index
  • P(A): 36/91
  • P(B): 30/91
  • P(C): 25/91

I would really appreciate any ideas!

Thank you.


A way to find the individual probabilities is to think of it in terms of adding disjoint events whose union is that probability in question. So for the person in the $i$th of ABC (e.g. Brises is in the 2nd position), we can add up all of the ways that person in $i$th position can roll a six first. This essentially means he could roll a six first on first set of rolls, or second set of rolls, and so on. Now we notice that the probability that the person in the $i$th position wins on the $n$th set of rolls is equal to $$\left(\frac{5}{6}\right)^{(i-1)+3(n-1)}\left(\frac{1}{6}\right)$$ Why is this? This is because if you have the $i$th turn and you want to roll to roll six first on $n$th turn you need $(i-1)+3(n-1)$ (*) positions before you to fail and than you to win. Thus we have $$P\left(\textrm{Person in position i rolls 6 first}\right)=P\left(\bigcup_{n=1}^{\infty}\textrm{Person in in positon i rolls 6 first on turn n}\right)=\sum_{n=1}^{\infty}P\left(\textrm{Person in in positon i rolls 6 first on turn n}\right)=\sum_{i=1}^{\infty}\left(\frac{5}{6}\right)^{(i-1)+3(n-1)}\left(\frac{1}{6}\right)=\sum_{i=0}^{\infty}\left(\frac{5}{6}\right)^{(i-1)+3n}\left(\frac{1}{6}\right)=\sum_{i=0}^{\infty}\left(\left(\frac{5}{6}\right)^{i-1}\left(\frac{1}{6}\right)\right)\left(\frac{5^{3}}{6^{3}}\right)^{n}$$ where this is a geometric series thus we have $$=\frac{\left(\frac{5}{6}\right)^{i-1}\left(\frac{1}{6}\right)}{1-\frac{5^{3}}{6^{3}}}=\frac{(6^{3-i})(5^{i-1})}{91}$$

And for here we have A($i=1$), B ($i=2$), and C ($i=3$)

(*) to derive this $(i-1)+3(n-1)$ you have to think of how many failures do you have before person in $i$th position rolls six first in $n$th set of rolls. Well first you will need $n-1$ set of rolls all to be failures in order to have success on $n$th set, so that is $3(n-1)$ failures. Then you need the people in the positions before person in $i$th position, i.e. $i-1$ positions, to fail also. So in total you need $(i-1)+3(n-1)$ positions to not roll a six before person in $i$th position does so on $n$th set of rolls

  • $\begingroup$ Awesome, I finally had a chance to take a look at this in depth and solve it for myself. This explanation works well, but it also highlighted a gap in my understanding of the Geometric Series. I thought the Professor had explained it thoroughly, but it seems that as |5/6^3n| < 1 then the formula is that of a / r, where a is the 'fixed' parts of (5/6)^i-1 * (1/6) and r is the series part of (5/6)^3n. Which explains the last step. I ended up with ( (6^2) * (5/6)^i-1 ) / 91 which is correct. Can you explain the (6^(3-i)*5^(i-1))/91 ? $\endgroup$ – Momer Aug 26 '14 at 13:21
  • $\begingroup$ those two expressions are equivalent just use properties of exponents $\endgroup$ – Kamster Aug 26 '14 at 18:23

We give a series-free solution to your last problem by conditioning on the results of the first three throws.

Let $a$, $b$, and $c$ respectively be the probability that A, B, C is the first to roll a $6$.

Maybe A gets a $6$ immediately. Or maybe A fails on the first roll. Then the probability A is gets the first $6$ is $c$, because now in effect A is third.

Thus $$a=\frac{1}{6}+\frac{5}{6}c\tag{1}$$

Maybe A fails on her first trial. Then either B rolls a $6$, or she is in effect third. Thus $$b=\frac{5}{6}\cdot\frac{1}{6}+\frac{5}{6}\cdot\frac{5}{6}c.\tag{2}$$

A similar analysis shows that $$c=\frac{5}{6}\cdot \frac{5}{6}\cdot\frac{1}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}c.\tag{3}$$

Equation (3) immediately yields $c=\frac{25}{91}$. Equations (1) and (2) now tell us what $a$ and $b$ are.

  • $\begingroup$ never cease to amaze me by the simple proofs you give, so much easier than what I had done $\endgroup$ – Kamster Aug 25 '14 at 7:12
  • $\begingroup$ Both were great answers, but I had to choose @user159813 's because of the equations laid out in a way I can understand and learn from. $\endgroup$ – Momer Aug 25 '14 at 21:54
  • $\begingroup$ That is exactly what you are asked to do, choose the answer that is most helpful. $\endgroup$ – André Nicolas Aug 25 '14 at 21:57

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