telekinetic

So as I sit hear eating a starburst 2-pack (of which there is an entire drawer in the office), I started wondering about the probabilities they represented (clearly a busy day at work).

For people not familiar, there are four flavors, pink red orange and yellow, and each pack has two candies. The packs do have a left and right orientation, so in theory you could have 16 different candy combinations, if you consider 'right pink/left yellow' different from 'right yellow/left pink'.

Easy question: What is the probability that if you gathered 16 packs from a large population, you would have one of each combination?

Harder question: What is the probability that you would get one pack of each possible combination, if (like a sane person) you count 'right pink/left yellow' and 'right yellow/left pink' as being the same?

Harder question abstracted to dice: What is the probability of rolling 11 times and getting each number, 2 through 12, exactly once?

I may just be super tired, but the only answer I've come up with to second problem seems excessively fiddly and counterintuitive to me, though for some reason the third seems easier, since I'm used to thinking of statistics and dice.

Please spoilerify your answers for a few days so people who want to try it have a chance...I'll try to mull over a better answer to the second one (and third, by extension) and post it up when I get home.

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Daniel_

You are evil and should be punished for puting this problem in my head.

I have to get a pad now. Grrrr!

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PonyPotato

I'll be thinking about this for the next couple of hours. Thanks. :P

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Rekna

Are you assuming the starbursts colors and orientation have a completely equal distribution? Without that assumption your first 2 questions are impossible to answer.

This one is pretty easy is you generalize it for order.

Spoiler:
First you need to calculate the odds of rolling this in order.

To do this you multiply the odds of rolling 2-12 individually and then multiply by the number of orders you can roll it.

odds for 7 6/36 =0.167
odds for 6 and 8 5/36 =0.139
odds for 5 and 9 4/36 =0.111
odds for 4 and 10 3/36 =0.083
odds for 3 and 11 2/36 =0.056
odds for 2 and 12 1/36 =0.028

Multiply them together and you get:

(6*5^2*4^2*3^2*2^2)/36^11=86400/131621703842267136

this is the probability of rolling in a specific order. Now you have one of these for each possible way to order the rolls.

To compute the number of orders you will need to use a partial factorial.
The first roll has 11 possibilities (2-12). After each roll the number of possibilities drops by one. Thus you have a partial factorial. 11*10*9*8*7*6*5*4*3*2=39916800

Multiplying the odds for one order above yields a probability of: 86400*39916800/131621703842267136=2.62024531 × 10-5
or 1 in 38,164.3656.

It has been a few years since I have done any probability & stats how did I do?

Tully Mars

I was going to start in on this and suddenly realized I've always considered pink, red orange and yellow to be colors and not flavors. I've decided to slowly back away from this problem as it is clearly beyond my comprehension.

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telekinetic

Yes, definitely...sorry, I meant to state it in the setup...

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Rekna

Spoiler:
The odds of getting any one of the individual packs in a single pick is 1/16. To verify this look at probability of a the color of a single candy being color a) which is 1 in 4. Then look at the probability of the color of the second candy which is also 1/4 (though technically it is slightly smaller since the population has been lowered by the first pick, I'm assuming the population is essentially infinite and an earlier pick doesn't affect the probability of a future event, thus each pick is an independent event). Multiply them together and you get 1/16.

Now we just have to compute the successive probabilities. The first pick has perfect odds (16/16) of being unique. The second pick has the odds of 15/16 of being unique and so on and each successive pick is 1 less on the numerator. Thus your odds are 16!/16^16 is equal to 1 in 881,657.952.

I would try to simplify my expression but i don't have a pen and paper to do it and typing it on a computer would be a pain so you will have to live with the decimal representation.

I'll leave the middle one for someone else thought if you follow the way I did this one and the hardest one it shouldn't be to hard.

Deltona Couple

Just from basic logic of probability (Forgive me some for it has been several years since I took Probability and Statistics in school) But....

Spoiler:
Basic numbers show that the odds of doing the die roll in sequence is MUCH higher than approx 1 in 36,000. just to have the chance of rolling a 2 then a 3 immediately after is 1:2916, then the odds of a 4 following THAT is 1:145800...etc... I dont have the time to go on much further, but as you can tell, it is exponentionally growing, so I am going to estimate that the odds of rolling 2,3,4,5,6,7,8,9,10,11,12 would be in the neighborhood of 1: "billions"


"Statistics of Dice Throw
The probababilities of different numbers obtained by the throw of two dice offer a good introduction to the ideas of probability. For the throw of a single die, all outcomes are equally probable. But in the throw of two dice, the different possibilities for the total of the two dice are not equally probable because there are more ways to get some numbers than others. There are six ways to get a total of 7, but only one way to get 2, so the "odds" of getting a 7 are six times those for getting "snake eyes". This simple example raises the idea of distinguishable states. For example, throwing a 3 is twice as likely as throwing a 2 because there are two distinguishable ways to get a 3. "
Taken from
Statistics of Dice Throw

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