Keep rolling a fair, 6-faced die, and keep adding the outcomes until the running total exceeds 1000.

OfficialAnu

New member
Joined
Mar 3, 2019
Messages
6
Keep rolling a fair, 6-faced die, and keep adding the outcomes until the running total exceeds 1000. When you stop, the running total must be 1001, 1002, ..., or 1006; but with what probabilities? (Approximate solutions are welcome.)



I don't even know where to start
 

Attachments

Last edited by a moderator:

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,816
Keep rolling a fair, 6-faced die, and keep adding the outcomes until the running total exceeds 1000. When you stop, the running total must be 1001, 1002, ..., or 1006; but with what probabilities? (Approximate solutions are welcome.)



I don't even know where to start
So let's get you started. The least number of rolls is 167.
Now you explain why that is true. Can you tell us the probability of that?
The most number of rolls is 1001. Why? Probability?
 
Last edited by a moderator:

OfficialAnu

New member
Joined
Mar 3, 2019
Messages
6
So let's get you started. The least number of rolls is 167.
Now you explain why that is true. Can you tell us the probability of that?
The most number of rolls is 1001. Why? Probability?
Probability of getting any number in one role is 1/6
P(getting a six 167 times)/P(1002) = (1/6)^167?
P(getting a one 1001 times)/P(1001) = (1/6)^1001?
Now what? Getting 1001, 1002, 1003, 1004, 1005, and 1006 can occur through different rolls, not just single numbers like 6s
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,009
Since P(getting a six 167 times) = (1/6)^167 I doubt that P(getting a six 167 times)/P(1002) = (1/6)^167 . Why do you think it is?
Same with P(getting a one 1001 times)/P(1001) = (1/6)^1001? Why do you think that is true. After all, P(getting a one 1001 times) = (1/6)^1001?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,816
Probability of getting any number in one role is 1/6
P(getting a six 167 times)/P(1002) = (1/6)^167?
P(getting a one 1001 times)/P(1001) = (1/6)^1001?
Now what? Getting 1001, 1002, 1003, 1004, 1005, and 1006 can occur through different rolls, not just single numbers like 6s
Since P(getting a six 167 times) = (1/6)^167 I doubt that P(getting a six 167 times)/P(1002) = (1/6)^167 . Why do you think it is?
Same with P(getting a one 1001 times)/P(1001) = (1/6)^1001? Why do you think that is true. After all, P(getting a one 1001 times) = (1/6)^1001?
If one rolls a die 167 times in order to get a sum of 1002 the die must come up six every time.
The probability of that happening is \(\displaystyle \frac{1}{6^{167}}\).
Let's look a much smaller question: roll a die five times and sum the results. LOOK HERE
That expansion models that event. The term \(\displaystyle 126x^{25}\) tells us that the sum of \(\displaystyle 25\) can be gotten in \(\displaystyle 126\) ways.
Therefore, \(\displaystyle \mathcal{P}(S=25)=\frac{126}{6^5}\). Realize that \(\displaystyle 6^5=7776\) and if we add up all the coefficients in the expansion we get \(\displaystyle 7776\).
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
This question has also been posted to Reddit (here).
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,170
This question has also been posted to Reddit (here).
Yes, and I note that OfficalAnu replied to the worked solution there with, "Thanks for the input man!"

Based on what kdshart2 posted here, it seems like OfficialAnu is officially fishing for worked solutions, not guidance.

:(
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,816
Yes, and I note that OfficalAnu replied to the worked solution there with, "Thanks for the input man!"
Based on what kdshart2 posted here, it seems like OfficialAnu is officially fishing for worked solutions, not guidance.
If OfficialAnu is looking for an answer he ain't going to get one without a very high capacity computer he is sadly mistaken.
I think that OfficialAnu has no idea how very complicated the answer s/he seeks is.

Example look at this expansion. That tells us about tossing ten dice( or a die ten times) The coefficients that correspond to the exponents tell us the numbers of ways to get the sum. The term \(\displaystyle 2002x^{55}\) says that there are \(\displaystyle 2002\) ways to get a sum of \(\displaystyle 55\) if a die is tossed ten times. In that array there is only one way to get a sum of \(\displaystyle 60\text{ or }10\). If you are interested to explore this at the wolframalpha site, just change the 10 to say 15. The result tell us about tossing fifteen dice. After several examples, anyone can see what a monumental task OfficialAnu has ask for. Only a high capacity computer could to that.
 
Top