|January 11th, 2017, 06:22 AM||#1|
Joined: Jan 2017
Hard probability question. Advanced answers required.
To start off:
I have a number of trials let's say 10000.
I have an abstract die with a very high number of sides.
Each roll of a die consumes 1 trial.
I roll the die every second.
My objective: After selecting a number, roll the die less than 10000 times (number of trials) until you get that number.
Let the die have 5000 sides.
That means I have a probability of 1/5000=0.0002 of getting the selected number.
Now. With the same number of trials, is it more efficient to use more dice, supposing you only need one of them to show the selected number in order to win?
Each die has 5000 sides.
I roll ALL 5 dice in a second and that consumes 5 TRIALS out of all 10000.
I only need one out of the 5 dice to show the selected number in order to win, otherwise, if I run out of trials, I lose.
Efficient = Use of the least amount of time that's possible (1 second per roll of a die/more dice at a time) AND the least amount of trials (keep in mind that a second can consume 1 or more trials).
1. Is it more efficient to use more dice in those conditions? (10000 trials and 5000 sides)
2. Are the variables "trials" and "sides" able to change the answer of question number 1?
Thank you for your attention!
|January 11th, 2017, 05:00 PM||#2|
Joined: May 2007
Obviously if you have more dice, the number of trials will be less. The only measure is the product of the number of trials and the number of dice.
|January 11th, 2017, 06:50 PM||#3|
Joined: Jan 2017
More efficient because there are more dice which will mean less rolls therefore more time saving. The probability of each trial is not affected and the trials still remain independent therefore answer to question 1 does not change
Last edited by dthiaw; January 11th, 2017 at 07:02 PM.
|January 11th, 2017, 07:09 PM||#4|
Joined: Sep 2015
by grouping the rolls on average you'll end up using up a couple that you don't need. This becomes particularly evident if the number of rolls per group becomes large.
|January 13th, 2017, 05:33 AM||#5|
Joined: Jan 2017
I am not studying math nor I know so much about it, but I think I have a solution for your problem, but I warn you, it may be wrong!
The probability of obtaining that desired number in 10000 trials is the same using 10000 dice (so you would have 1 trial for each one) that using just 1 and 10000 trials for this one.
The only thing that varies is the time, using 10000 dice you are using just a second.
So if the probability is the same, but the time spent is less, Using 10000 dice seems better.
However, what if using 10000 dice ("in the same attempt/ at the same time") we obtain that number more than once? It wouldn't be so efficient as it would be using more trials than required for obtaining just 1 time that number.
To obtain that number of trials by which we obtain that desired number just once, we have to look at probability.
The initial probability of obtaining that number, with the amount of sides that you specified is 0,02%.
That means that in 100 trials I would get that number 0,02 times.
In 1000 trials I would get it 0,2 times, and in 10000 trials I would get it 2 times.
So for a single die, that number is expected to show up 2 times for each 10000 trials.
thus, in 5000 trials it would show up once.
so the best in terms of efficency for a single die would be to perform 5000 trials.
But, as the probability of obtaining that number with a single die using X trials, is the same of the probability of X dice in 1 attempt (here by X I mean X dice = X trials)
The most efficient in terms of time and trials would be to use
As you can see therefore, the amount of sides equals to the amount of trials (or to be more efficient as we spend less time; dice) required. That's because, if there are for instance 10 sides, in probability terms each side is expected to show up at least once in 10 trials.
So yes, the number of trials and sides change the answer of question 1.
you will need the same amount of trials that sides you have.
I hope to have helped!!
|advanced, answers, hard, probability, question, required|
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