Here's a fun one

[LetYouDown]

Uncle asked me this. There's a brand of tea you can buy by the box. Every box has a figurine in it. Just a small porcelain figurine. There's 20 unique types of figurine. What is the expected number of boxes you'd need to buy before you had a 99.9% probability of having a complete set? Assume an infinite number of boxes are in existence and the figurines are evenly distributed.

[moomoocow]

I'll take a crack at this.

Let N be boxes you buy.
In order NOT to have a complete set, you need to be missing 1 of the figures -
Prob of missing 1 figure in N boxes = 19/20^N
but there are 20 figures - so chance of missing one of the 20 is roughly 20*(19/20)^N (this double counts situations where you are missing 2 or more figures - but that should be quite small -

so set 20*(19/20)^N <= 0.01
and we get N about 150

Seems low - maybe that approximation isn't fantastic - maybe the double counting is quite bad.

[BruceZ]

[ QUOTE ]
Uncle asked me this. There's a brand of tea you can buy by the box. Every box has a figurine in it. Just a small porcelain figurine. There's 20 unique types of figurine. What is the expected number of boxes you'd need to buy before you had a 99.9% probability of having a complete set? Assume an infinite number of boxes are in existence and the figurines are evenly distributed.

[/ QUOTE ]

Hey, this used to say expected value. [img]/images/graemlins/wink.gif[/img] This is just like the roullete question you asked a while back. We can solve it the same way:

Let p = the probability that a specific figurine fails to appear in N boxes such that the odds against not getting all 20 is 0.1%. We can bound p as:

1/20 * 1/1000 < p < 1 - (999/1000)^(1/20)

The left side of the inequality would be equal if the boxes were mutually exclusive, and the right side would be equal if the boxes were independent.

1/20,000 < p < 1/19,990...

p = (19/20)^N

N = ln(p)/ln(19/20)

193.066 < N < 193.076

194 boxes required to meet the objective.

[BruceZ]

[ QUOTE ]
so set 20*(19/20)^N <= 0.01
and we get N about 150

[/ QUOTE ]

Should be <= 0.001. Your method is the same as mine, where you used just the left-hand side of my inequality. That is, you used mutual exclusivity but not independence. Using both allows you to know if your result is accurate to the nearest integer. It wouldn't give you the same integer if the number of figurines were very large, like over 1000.

[Nalapoint1]

Answer is 20.You open the boxes to find the figurines you want,and just buy those boxes.

[RiverDood]

And there's a variant that can make time go faster (well maybe only a little faster) on long car drives.

Look to see what months are on passing cars' license plate registrations. See how many cars it takes before all 12 months are represented.

I haven't run the numbers yet, but I'm guessing that this problem will solve 50% of the time somewhere around 40 license plates. We ran it once and got a solution at about 65 license plates, but it took forever for August to show up.

[BruceZ]

[ QUOTE ]
And there's a variant that can make time go faster (well maybe only a little faster) on long car drives.

Look to see what months are on passing cars' license plate registrations. See how many cars it takes before all 12 months are represented.

I haven't run the numbers yet, but I'm guessing that this problem will solve 50% of the time somewhere around 40 license plates. We ran it once and got a solution at about 65 license plates, but it took forever for August to show up.

[/ QUOTE ]

No less than 33 and no more than 37. The number can be found exactly by inclusion-exclusion.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
And there's a variant that can make time go faster (well maybe only a little faster) on long car drives.

Look to see what months are on passing cars' license plate registrations. See how many cars it takes before all 12 months are represented.

I haven't run the numbers yet, but I'm guessing that this problem will solve 50% of the time somewhere around 40 license plates. We ran it once and got a solution at about 65 license plates, but it took forever for August to show up.

[/ QUOTE ]

No less than 33 and no more than 37. The number can be found exactly by inclusion-exclusion.

[/ QUOTE ]

And using inclusion-exclusion shows that 35 comes the closest to 50%, but the probability of not having them all is actually 50.07%. With 36, the probability of not having them all is 46.82%.

12*(11/12)^36 - C(12,2)*(10/12)^36 + C(12,3)*(9/12)^36 - C(12,4)*(8/12)^36 + C(12,5)*(7/12)^36

=~ 46.82%.

[LetYouDown]

Thanks. More obvious than I thought...in hindsight, and I didn't even recall the roulette problem I posted. So in theory, if you had purchased 194 boxes and didn't have a complete set, you could attribute it to either a statistical anomaly or some kind of distributional bias.