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[GuyP]

[META] The stats so far!

[GuyP]

OK, so now you can answer a maths question for me.

Let's assume that there is one silver card in every ten packs, and you have ten packs in front of you.

By the time you've opened the fifth pack, you've opened half the packs, and hence have a 50% chance of having got the silver card. This is where my revised "on average" comes from - just because one in ten cards have a silver, it's just as likely that the first pack you buy will contain a Riemann as the tenth pack.

On the other hand, why doesn't that extrapolate out? It seems fairly clear that you'd need to buy about 100 packs to get 10 silvers, so we're back to the same ratio and needing 120 packs (i.e: £300 worth) to get 12 silver cards.

I suspect this might have something to do with the fact that nothing is "guaranteed." For instance, there's a 1/12 chance of a silver card being Riemann, but just because you've found 11 cards doesn't mean the next one will definitely be a Riemann. I don't know how to express that in terms of probability, however - "probability over time", if you will. I think my Maths teacher stopped at the point where they explain that such a thing doesn't really exist, in that a coin doesn't know which way up to land based on the way it landed previously.

Anyway, I'm curious about this because it's a bit headmessy. Your thoughts? (Jeb... calling Jeb...)

[Juxta]

Sorry, wrong Jamie, but the calculations for amounts spent thus far are based upon the costs of cards which have been solved. Not counting duplicates...which...uh...I have a few of. Just a few. Ahem. A stack more than six inches in height is still classed as "a few", right? Plus...the couple of hundred I've given away/traded in bulk for higher value cards...oh dear...

J

[GuyP]

Too true, too true. In which case the big, shiny reward is probably already paid for, hurrah! Now, if you could work out the probability of various cards been duplicated based on the cards already on the leaderboard and thereby estimate how many packs must have really been bought...

[idlemichael]

Holy monkey, my brain seems to be leaking out of my ears...

[cassandra]

I love the hours spent solving stat Though I'm not sure about the 5 min per solve Or the years of work put in by people who don't have the card (see Silver). I think we all must have managed a good decade so far.

[JebJoya]

Right, just to warn you all, this is only part 1 of the post, I can't work out part 2 at the moment, but rest assured it will be coming:

Right, dredging this out again, now that I'm here (only 9 days late ), let's have a look at this.

Basically, you need to use a binomial distribution (Wiki Linki).

So, we set our random variable X~B(n,p), where n is our number of trials (10 in your example) and p the probability of success (prob of a silver - 1/10 in your example).

Just to add a note at this point - I assume here that you cannot get 2 silvers in 1 pack - if this isn't the case, you can edit using 6 cards in place of 1 pack and changing the probabilities respectively.

Right, so using the simple formula, we can find that the probability of you getting k silver cards from 10 packs is:

nCk * (p)^k * (1-p)^(n-k)

So, for example, the probability of getting 0 silvers is:

10C0 * (0.1)^0 * (0.9)^(10) = 0.34

So about 34% chance of getting none from 10 packs, using the made up probability provided.

Moreover:

P(X==1) is 38%
P(X==2) is 19%
P(X==3) is 5.5%
...
P(X==10) is 0.00000001%

Moreover, to cover GuyP's particular point, the Expected result E(X) is equal to np for the binomial distribution, so to expect to get EXACTLY 1 card, you need 10 packs.

However, I'm now starting to get confused... Perhaps by taking 7 packs, that's the first time you drop the P(X==0) below 50% so you expect at that point to get at least a silver... Yes.. Sounds about right...

Right, onto the next bit...

The probability of getting an exact card... It's a bit late, so apologies if this is wrong, it shouldn't be though... We need BD's within BD's or something...

So if we define X(n)~B(n,1/10) , Y(m)~B(m,1/12), then we get

P(Getting a particular silver in n packs) = ### P(X(n)==m) * P(Y(m) != 0)

Where ### is read as "the sum over all m of:"

Sooo... Where are we... Let me write this down, hold on a sec...

...

The next day...

The internet went down last night, and I thought about this, and maybe this isn't the most sensible way to do this, so how else could we?

I will stop there in this post, and have a think about things, there must be a way to simplify things... In conclusion of part 1, you need to buy 7 cards to have a >50% chance of getting 1 or more silvers


Jeb (exhausted)

[fitzyfitz]