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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?


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

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  • $\begingroup$ 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

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 $$ \frac6{21},\ \frac5{21},\ \frac4{21},\ \frac3{21},\ \frac2{21},\ \frac1{21}. $$


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