Need a math geek to help me out with a probability question:
You have a range of 20 cards each consisting of a number and a letter, labeled as follows:
1a 1b
2a 2b
3a 3b
4a 4b
5a 5b
6a 6b
7a 7b
8a 8b
9a 9b
10a 10b
One at a time, you draw a number/letter card. You can't draw any of the above cards twice and you do not return the card to the deck.
What formula would you apply to figure out the average amount of unique numbers drawn after x draws. (ie: after x draws, on average you would have Y% of unique numbers.)
On the first draw, you have a 20/20 chance of drawing a unique number.
On the second draw, you have an 18/19 chance of drawing a unique number.
etc etc
>pic unrelated
Wouldn't it always be 100% because you don't return the card to the deck?
>>8066146
Rather, 1a / 2a are both unique numbers, 1a / 1b are not unique numbers.
>>8066146
Please accept photos of my bike as payment for any help you can offer.
>>8066122
(18x+(x+2)(1-x)/2)/19
>>8066184
So to find the average amount of drawn cards after say, 15 draws would it be (18*15+(15+2)(1-15)/2))/19
I'm a bit of an idiot, so might need a hand D:
>>8066202
Sorry average amount of unique numbers in the draw*
>>8066202
Yes
>>8066202
No, that's completely wrong
>>8066207
So after 15 draws you'd have an average of 6.84 uniques? That sounds a tad low.
>>8066202
Sorry made a mistake. It should be
x(39-x)/38
The formula will be long as fuck and will contain many sigmas
anything else is wrong
Suppose you draw x cards, for some fixed x=0,1,...,20.
For i=1,...,10 let [math]I_{x,i}[/math] be the random variable that = 1 if the number i appears at least once, and = 0 otherwise. Then the average number of unique numbers is [eqn]E[ \sum_{i=1}^{10} I_{x,i} ] = \sum_{i=1}^{10} E[I_{x,i}] = \sum_{i=1}^{10} P(I_{x,i}=1) = 10\times P(I_{x,1} = 1)[/eqn]
because $E[I_{x,i}] = 1 \times P(I_{x,i}=1) + 0 \times P(I_{x,i} = 0)$, and the probability of any particular number i appearing is the same regardless of i.
Anyway, the probability of the number 1 being drawn at least once in x cards is [eqn]P(I_{x,1} = 1) = 1 - P(I_{x,1} = 0) = 1 - \frac{ 18!}{20!} \frac{(20 - x)!}{(18 - x)!} [/eqn] which is well-defined for all x=0,1,...,20 provided you cancel the factorials out properly, and the final number you're after is simply 10 times that.
>>8066226
Lol no. The problem isn't really that complicated. If you would like to prove me wrong go ahead.
>>8066237
*fixing Latex
The expectation of an indicator variable is the probability:
[math]E[I_{x,i}] = 1 \times P(I_{x,i}=1) + 0 \times P(I_{x,i} = 0)[/math].
Also thanks to the other thread on the front page I realized that if you have access to a table of binomial coefficients, the final formula simplifies to:
[eqn] 10 \times (1 - \binom{18}{x} / \binom{20}{x})[/eqn]
which holds for all x under a suitable definition of the binomial coefficients.
>>8066217
How would the formula work if the deck was like 1a 1b 1c etc.?
>>8066254
True.
You can tell I considered the problem over once I derived the formula, and the algebraic simplification was merely an after thought
(especially since that part can be outsourced to WolframAlpha).
>>8066281
Yes but there is an easier way to solve the whole thing without simplifying all that out. I want to see if you can figure it out.
>>8066280
Well, I was going to generalize this anyway.
Suppose the deck contains numbers from 1 to n, where number i appears on [math]t_i[/math] cards. Let [math]T = \sum_{i=1}^nt_i[/math] denote the total number of cards in the deck.
Then by the argument in the earlier thread, the average number of unique numbers in x draws is [math]\sum_{i=1}^n1 - P(I_{x,i} = 0)[/math]
For each i, that probability is the number of ways to draw x cards from [math]T-t_i[/math] (containing no i's), divided by the total number of ways to draw x cards from T. Putting it all together, the formula is
[eqn]n - \frac{ \sum_{i=1}^n (T-t_i) (T-t_i-1) \cdots (T-t_i-x+1) }{ T (T-1) \cdots (T-x+1) }[/eqn]
>>8066301
thanks mate :)
