Dice rolling -Which of 2,6,1006 is more likely to be one of these totals?

[lfdahl]

A standard six–sided dice is rolled repeatedly and a running total is kept of
all the numbers rolled. Which of $2, 6, 1006$ is more likely to be one of these
totals? Prove your answer.

 

[kaliprasad]

A standard six–sided dice is rolled repeatedly and a running total is kept of
all the numbers rolled. Which of $2, 6, 1006$ is more likely to be one of these
totals? Prove your answer.

Spoiler

total number of rolls maximum to be considered shall be 1006.( Because 1007 counts shall give a minumum sum 1007

let us count number of ways partial sums can be

2 = 2 or 1 + 1 so 2 followed by any number or 1 ,1 followed by any number can give sum 2 so
$1 *6^{1005} + 1 * 1 * 6^{1004}$ ways

6 can come as (6), (5,1), (4,2) and other compinations so number of ways 6 can come

$> 6^{1005} + 2 * 6^{1004}$ taking care if above 3 cases and there are more cases also but not requires
for 1 to come we can have all 1 and for $1006^{th}$ element to be 1 the number of ways is $6^{1005}$ out of which some and lot all can have partial sum 1006.
comparing above 3 we see that number of ways partial sum > 7 is largest. then comes number of ways partial sum 2 then 1006.

so 6 is most likely to occur followed by 2 followed by 1006.

 

[Opalg]

[sp]Let $1\leqslant n\leqslant 6$. The number of ways to get $n$ as the sum of $k$ throws of the dice is ${n-1\choose k-1}$. The probability of each throw is $\frac16$. So the probability of reaching the total $n$ in $k$ throws is ${n-1\choose k-1}\frac1{6^k}$. The overall probability $P(n)$ of $n$ occurring as a total is thus $$\sum_{k=1}^n{n-1\choose k-1}\frac1{6^k} = \frac16\Bigl(1+\frac16\Bigr)^{n-1} = \frac{7^{n-1}}{6^n}.$$ This shows that $P(n)$ increases as $n$ goes from $1$ to $6$. In particular, $P(2) = \dfrac7{36}\approx0.1944$ and $P(6) = \dfrac{7^5}{6^6}\approx 0.3602.$

When $n>6$ the situation is different. If $n$ occurs in the running total then the previous total must have been $n-k$, where $1\leqslant k\leqslant6$, and the probability of getting from each of those to $n$ is $\frac16$. Therefore \(\displaystyle P(n) = \frac16\sum_{k=1}^6P(n-k).\) In other words, $P(n)$ is the average of the six previous probabilities, so it cannot be larger than all of them. In particular, $P(n)$ can never again get as large as $P(6).$

In fact, the average value of a throw of the dice is $\frac72$, which means that as $n$ gets large, $P(n)$ will converge to the limiting value $\frac27\approx 0.2857.$

In conclusion, $6$ is more likely than $2$ or $1006$ to occur as the running total.

[/sp]

 

[kaliprasad]

[sp]

When $n>6$ the situation is different. If $n$ occurs in the running total then the previous total must have been $n-k$, where $1\leqslant k\leqslant6$, and the probability of getting from each of those to $n$ is $\frac16$. Therefore \(\displaystyle P(n) = \frac16\sum_{k=1}^6P(n-k).\) In other words, $P(n)$ is the average of the six previous probabilities,
[/sp]

Hello Opalg,

I have a query about above statement

Spoiler

For example
If we have the previous number as n-1 then the number previous to that could be n-2 so n-2 has ome has a probability is different from n-2 has come(it could be previous to previous) for example 8 could come as

5,1,2 - sum 5,6,8
6,1,1; sum 6,7,8
5, 2, 1 sum 5.2.8
so getting 8 from 6 or 7 is not equal probable but from 6 or 7 as previous element equal probable

if my understanding is incorrect kindly let me know

 

[Opalg]

Hello Opalg,

I have a query about above statement

Spoiler

For example
If we have the previous number as n-1 then the number previous to that could be n-2 so n-2 has ome has a probability is different from n-2 has come(it could be previous to previous) for example 8 could come as

5,1,2 - sum 5,6,8
6,1,1; sum 6,7,8
5,2,1 sum 5.2.8
so getting 8 from 6 or 7 is not equal probable but from 6 or 7 as previous element equal probable

if my understanding is incorrect kindly let me know

[sp]There are 6 possibilities for the running total to be $n$:

1) the previous running total was $n-1$, followed by a $1$ on the dice,
2) the previous running total was $n-2$, followed by a $2$ on the dice,
.
.
.
6) the previous running total was $n-6$, followed by a $6$ on the dice.​

Those six possibilities are disjoint and exhaustive, so their probabilities add up to $P(n)$.

In the case of your example, $P(8) = \frac16\bigl(P(2) + P(3) + \ldots + P(7)\bigr)$. So for example

5,1,2 - sum 5,6,8 The previous total is 6, so this is included in the probability $\frac16P(6)$.
6,1,1 - sum 6,7,8 The previous total is 7, so this is included in the probability $\frac16P(7)$.
5,2,1 - sum 5,7,8 Here, the previous total is again 7, so this case has already been covered.

[/sp]

 

[kaliprasad]

[sp]There are 6 possibilities for the running total to be $n$:

1) the previous running total was $n-1$, followed by a $1$ on the dice,
2) the previous running total was $n-2$, followed by a $2$ on the dice,
.
.
.
6) the previous running total was $n-6$, followed by a $6$ on the dice.​

Those six possibilities are disjoint and exhaustive, so their probabilities add up to $P(n)$.

In the case of your example, $P(8) = \frac16\bigl(P(2) + P(3) + \ldots + P(7)\bigr)$. So for example

5,1,2 - sum 5,6,8 The previous total is 6, so this is included in the probability $\frac16P(6)$.
6,1,1 - sum 6,7,8 The previous total is 7, so this is included in the probability $\frac16P(7)$.
5,2,1 - sum 5,7,8 Here, the previous total is again 7, so this case has already been covered.

[/sp]

Spoiler

but in the above (5,1,2) and (6,1,1) 6 appears but not last element in 2nd case and P(6) should take case.

 

[Opalg]

Spoiler

but in the above (5,1,2) and (6,1,1) 6 appears but not last element in 2nd case and P(6) should take case.

[sp]But 6 is not the previous running total, which is 7 in that example. The key point of my argument is to see how a given running total is obtained from its immediate predecessor, without any reference to its more distant past history.
[/sp]

 

[castor28]

My two cents...
[sp]
If I understand correctly, the point of kalipsarad's question is that, in the equation:

$$P(n) = \dfrac16\left(P(n-1)+\cdots+P(n-6)\right)\qquad[1]$$
​

$P(n)$ represents the current total, and the $P(i)$ on the right represent the previous total.
What we want is the probability that $n$ appears somewhere in the list. These three things are apparently different.

It may help to clarify things if we define:

• $f(k,n)$ as the probability that the total is $n$ after the $k$-th roll.
• $p(n)$ as the probability that $n$ appears as a total somewhere in the list.

We have $p(n)=\sum_k{f(k,n)}$, where only finitely many terms in the sum are different from 0.

Instead of equation [1], we write:

$$f(k,n) = \dfrac16\left(f(k-1,n-1)+\cdots+f(k-1,n-6)\right)$$
​

and summing over $k$ gives us equation [1] with $p(n)$ instead of $P(n)$; this is what we need.
[/sp]

 

[lfdahl]

I just want to thank all of you for correct solutions and comments.
It is always encouraging for me and I learn a lot, when we dig a little deeper into the matter.

Here´s the suggested solution, which (surprise, surprise! (Giggle)) is quite close to Opalg´s answer:

Spoiler

Let us denote for every positive integer $n$ by $p_n$ the probability, that $n$ is obtained as a running total.

Then, $p_1 = \frac{1}{6}$ and for every $k \in \left \{ 1,2,3,4,5,6 \right \}$ and any positive integer $n > k$, the probability that the first number rolled is $k$ and the running total is $n$ equals $ \frac{1}{6}p_{n-k}.$

Let us compute $p_m$, where $m \in \left \{ 2,3,4,5,6 \right \}.$ If the running total is $m$ the first number rolled is one of $m, m-1, m-2,…,1$. Consequently one obtains

\[p_m = \frac{1}{6}+\frac{1}{6}\cdot p_1+\frac{1}{6}\cdot p_2+...+\frac{1}{6}\cdot p_{m-1}.\]

It follows, that $p_1<p_2<…<p_6$.

Now for every running total $m \geq 7$ the first number rolled is one of $6,5,4,3,2,1$. Consequently one obtains
\[p_m = \frac{1}{6}\cdot p_{m-6}+\frac{1}{6}\cdot p_{m-5}+...+\frac{1}{6}\cdot p_{m-1}\]

which is the arithmetic mean value of $p_{m-6},p_{m-5},…,p_{m-1}.$

Since the mean value is smaller than or equal to the biggest of the involved numbers with equality only in the case that all the involved numbers are equal it follows that $p_m < p_6$ for $m = 7, 8, . . . , 12$ and, subsequently for every $m \geq 7$.

For example if $m = 7$, we immediately conclude: $p_7 < p_6$. If $m = 8$, we see that $p_8 < p_7$, and so on.