Hi /sci/, /tg/ here
Math question; mostly because I'm too dumb to remember how binomials work. Not that I was ever particularly good at those in high school to begin with..
How do I find out how many six-sided dice you roll on average before you've rolled one number twice? (looking for 1, but which number it is shouldn't matter for the math afaik)
The answer to the problem itself would also be greatly appreciated.
Hint: If you rolled seven six-sided dice, what is the likelihood they will all show different numbers?
>>7717636
You're not using a binomial distribution to solve this.. You're using a geometric distribution..
Let's suppose you roll:
[math] X=i [/math] for some i = 1,2,3,4,5,6
Then the question you want to know is how many times should I roll on average to get another i?
Well, assuming the die is fair:
[math] P(X=i) = \frac{1}{6} [/math] so
[math] P(X \neq i) = 1 - \frac{1}{6} = \frac{5}{6} [/math].
The geometric distribution models the number of times an experiment must be performed until a condition called success occurs.. Naturally, our success condition will be [math] X=i [/math]. That this occurs in k attempts is modeled by:
[math] p(k) = (\frac{5}{6})^{k-1}(\frac{1}{6}) \text{ for } k=1,2,3,...[/math]
Compute the expected value of this distribution, or look it up.
>>7717678
I need to correct this I think..
I believe [math] k = 0,1,2,3,... [/math]..
Sorry.. I'm just waking up.
>>7717685
Damnit.. Another correction.. Depending on the whether we start with k = 0, or k=1 changes the distribution. Just check the wiki for geometric distributions..
[math] p(k) = (\frac{5}{6})^{k}(\frac{1}{6}) \text{ for } k = 0,1,2,3,4,... [/math]
>>7717678
No correction was needed.. This is all right.
The expected value, if [math] Y \text{ follows } geo(\frac{1}{6}) \text{ is } \mathbb{E}[Y] = 6 [/math]
>>7717636
By on average, I'm assuming you want to know how many times you have to roll a six sided die to have a 50% of getting a number twice. If that's what you want, then this is how I would try and calculate it.
For the first roll, it doesn't matter what you get. For the second roll, you have a [math]\frac{5}{6}[/math] possible rolls that won't have a doubled number. For the third roll, you have a [math]\frac{5}{6} \cdot \frac{4}{6}[/math] chance of not having a doubled number. That's 55.56% of it not happening or a 44.44% of it happening. For the fourth roll, you have a [math]\frac{5}{6} \cdot \frac{4}{6} \cdot \frac{3}{6}[/math] chance of it not happening, so you have a 27.78% chance of it not happening, or a 72.22% chance of it happening. So, the fourth roll is the roll where it becomes more likely to have rolled at least one number twice than to have no repeats.
As of now, I'm not sure how you would calculate how many rolls it would take to have a 50% chance of getting 1 twice.
>>7717676
This replyer's hint is the solution to the most naive assumption: that you simply want to get some number twice, and it doesn't matter which. Throw seven dice, and there has to be at least two of the same by the pigeon-hole principle.
>>7717676
What'd I do to deserve that kind of response?
>>7717678
Ok, I'll try and work with that, I've genuinely not done any serious math for 10 years, artfag here.
>>7717701
Maybe I should have explained my scenario a bit more.
The hope is to roll as many dice as possible, as soon as the 2nd one is rolled you have to stop rolling.
> How many dice can I expect to roll.
>>7717769
OP-- you hopefully noticed that the statement of the problem greatly changes the outcome. You should try to state it as precisely as possible to ensure that you compute the thing you're actually looking for.
>not using a stochastic process
