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
Parker Mitchell
Wouldn't it always be 100% because you don't return the card to the deck?
Christian Scott
Rather, 1a / 2a are both unique numbers, 1a / 1b are not unique numbers.
Landon Rodriguez
Please accept photos of my bike as payment for any help you can offer.
Anthony Cooper
...
Robert Thompson
(18x+(x+2)(1-x)/2)/19
Colton Brooks
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:
Jaxon Davis
Sorry average amount of unique numbers in the draw*
Hunter Flores
Yes
Justin Price
No, that's completely wrong
Jace Hall
So after 15 draws you'd have an average of 6.84 uniques? That sounds a tad low.
Wyatt Williams
Sorry made a mistake. It should be
x(39-x)/38
Xavier Baker
The formula will be long as fuck and will contain many sigmas
anything else is wrong
Nathaniel Reed
Seems to work?
Nolan Scott
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.
Sebastian Cook
Lol no. The problem isn't really that complicated. If you would like to prove me wrong go ahead.
Eli Hall
*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.
James Parker
Not to burst your bubble but that simplifies to x(39-x)/38. The factorials can be reduced out.
Asher Barnes
How would the formula work if the deck was like 1a 1b 1c etc.?
Samuel Evans
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).
Eli Lopez
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.
Kevin Hughes
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]
