6
$\begingroup$

This question already has an answer here:

If a dice is thrown till the sum of the numbers appearing on the top face of dice exceeds or equal to 100, what is the most likely sum?

$\endgroup$

marked as duplicate by azimut, Davide Giraudo, Daniel Fischer, Dan Rust, Lord_Farin Nov 11 '13 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If this is homework, please say so. It would be best if you show your working here too :) $\endgroup$ – Shaun Nov 11 '13 at 11:37
  • $\begingroup$ @Shaun, Nope, It was there in placement test in our college. $\endgroup$ – A P Nov 11 '13 at 11:41
  • $\begingroup$ Fair enough. Now we know :) $\endgroup$ – Shaun Nov 11 '13 at 11:42
  • 1
    $\begingroup$ Possible duplicate: math.stackexchange.com/q/246723/56801, which has a possible duplicate itself: math.stackexchange.com/q/12433/56801 :) $\endgroup$ – Keep these mind Nov 11 '13 at 12:06
  • $\begingroup$ For ending up at 105, the last die must show a 6 (otherwise, the sum already was $\geq 100$). Similarly, for ending up at 104, the last die must have shown 5 or 6 etc. Only for the result 100, any outcome of the last die is possible. This suggests that the answer is 100. Of course, this is not a rigorous proof. $\endgroup$ – azimut Nov 11 '13 at 12:24
1
$\begingroup$

This has been computed several times on the site already so let us present some of the underlying theory.

Renewal theory deals with sums $(S_n)$ of i.i.d. nonnegative increments $(X_n)$ with common integrable distribution $\mu$ and asks for the occurrences of these sums just before and just after some given time.

Thus, one sets $S_0=0$, $S_n=X_1+\cdots+X_n$ for every $n\geqslant1$, $\mathcal S=\{S_n\mid n\geqslant0\}$ and, for every nonnegative $t$, $L_t=\max\mathcal S\cap[0,t]$ and $U_t=\min\mathcal S\cap(t,\infty)$.

A standard result of the theory is that, in continuous time, that is, when the distribution $\mu$ is continuous, $U_t-L_t$, $U_t-t$ and $t-L_t$ all converge in distribution. More precisely, $U_t-L_t$ converge in distribution to a size-biased version $\hat X$ of $X_1$, whose distribution has density $x/E[X_1]$ with respect to $\mu$. Furthermore, $$ (U_t-L_t,U_t-t,t-L_t)\to(\hat X,Z\hat X,(1-Z)\hat X)\quad\text{in distribution}, $$ where $Z$ is uniform on $(0,1)$ and independent on $\hat X$. In particular, the so-called residual waiting time $U_t-t$ has density $g$ with respect to the Lebesgue measure, with $$ g(x)=\frac{\mu([x,\infty))}{E[X_1]}. $$ In particular, $g$ is maximum at $x=0$ and $g(0)=1/E[X_1]$.

This suggests that, in the present (discrete) case, the most probable overshoot when $t\to\infty$ is $0$, which happens with probability $6/21=2/7$. True, there are some subtle differences with the continuous setting since one asks about $V_n-n$, where $V_n=\min\mathcal S\cap[n,\infty)$ ($n$ included). But the result above probably carries through and time $t=100$ is probably already large enough for this asymptotics to apply. Anyway, the asymptotic distribution of the overshoot on $\{0,1,2,3,4,5\}$ is $$ \frac6{21},\ \frac5{21},\ \frac4{21},\ \frac3{21},\ \frac2{21},\ \frac1{21}. $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.