POLL: Pirate map :)

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There is a box which holds infinite number of balls and when you pick one, the balls possibility of being one certain color is 1/8. After how many tries our chance of having balls from all 8 colors is above or equal to %50.

This is a hard one as a math problem, but in the real world I agree with Kitrax

so it is easy to solve, the answer is over 70 (and maybe it is also over 1 million)

 

I would think that the result i that case would be infinite -4, Argohir.

 

Thinking back to high school statistics assuming there are an equal number of each map piece for this to be possible to work out without some very high level of mathmatics and heaps of pages of working.

After 8 purchases there is only a 0.24033% (yes I did the calcuations here, this is not some figure out of the air) chance of having all the pieces assuming the chance of finding each piece is exactly the same. You would have a 1.92261375% chance of finding all pieces on the 9th purchase. As for the 10th purchase it would be higher still by the previous figure + two other figures yet I honestly can't be bothered working out what they would be . On the 11th it would be the figure from the 10th try + 3 other figures. 12th would be the 11th figure + 4 other figures... there's a pattern forming obviously.

The 'other figures' are from the other ways you could come to 8/8 pieces. The first way is finding a new piece every time, the 2nd way is finding a new piece every time bar the last time then finding it on your next attempt. The 3rd way would be to finding a new piece every time except the 6th time but then finding new pieces every time. There are so many possibilities of what could happen each time the mind boggles.

 

Abomination (or anyone else who knows the answer), how did you calculate this thing (if all the chances on getting pieces are equal)?

The chance of getting all 8 pieces with 8 purchases is P = 40320 * ((1/8)^8) = 0.00240. (Btw, how do you call the thing which you need to get the 40320? In Dutch we say '8 above 8', but my calculator writes it as 8P8). But how does it work when you do more purchases?

My guess of the chance of all eight pieces with 9 purchases would be P = 362880 * ((1/8)^8) * ((7/8)^1) = 0.01893. But this isn't the same your answer... what am I doing wrong here?

 

Now, why *wouldn't* you Harbs?

I remember having to make an equasion out of this exact problem...well, it was for the Monopoly game...but out teacher likened it to the ball pen theory. If you have a ball pen (those lard netted areas in a McD's play area with hundread of 4 inch plastic balls) and you fill it with with 200 balls for each of 8 colors, then add in a single "black" ball, mix all the balls around, and put on a blind fold:

• What is the chance that you will pull out 9 balls in a row, all of a differnt color, therefor winning the game with the minimum required tries?
• What is a chance that you will pull out 9 of the same color of balls?
• Given all the info, how many times will you have to take from the ball pen before you have at least one of each colored ball?

I don't remember how to do it, but after seeing the number, no one wanted to waste the time/money at McD's to play their game.

The problem here is that we don't know several factors, including:
How many *total* pieces of each part of the map have been distrubituted?
Does each location have an equal number of *all* the pieces?
Is a treasure map even worth this much effort, even for half the actual map?

So as you can see...there is no real answer to the question without more information.

 

Damnit Mes! Now I have to explain it.

You're right about the formulae used to discover the 8th purchase. The reason the next purchase is different is that once you find the map you won't keep buying and it is something that could have happened. Thus you need to ADD that probability to the next set of probabilities. I actually made a mistake, the 9th purchase % would be higher as I forgot that there would be more than just one extra chance of finding the map on the 9th purchase. Let me explain...

It is possible to only miss getting a new piece once, yet it could happen on a different purchase. For example it could happen on your second purchase, you got the same piece you got the first time, yet on the 3rd, 4th, 5th etc. you find a new piece each time till you find the map complete. Another situation is you find a new piece the first and second times, the third time you find one of the two pieces you already have yet from the 4th onwards you find a new piece each time till you have the map. Repeat the same situation for the 4th time you find the same piece yet every other time you find a new one and so on. Basically, on the 9th purchase you would be 'adding' 6 more probabilities of finding the map to the chance you would have found the map by then, as well as the chance that you would find the map on the 8th chance if you found a new piece every time. Thus it'll be 0.24003% + (miss 2nd time probability) + (miss 3rd time probability) + (miss 4th time probability) ... + (miss 7th time probability).

As to what those probabilities are I'm not sure how you could find those out since I forget the formulae since I did stats such a long time ago. However I knwo what figures you will need to discover the correct answer.

I know what you're talking about but I forget the name for it myself. You're trying to find the number of possibilities and it is 8x7x6x5x4x3x2x1. I think to find the 9th number of possibilities it would be (8P8 - 1) x 9 = 362871[[ The -1 to get rid of the time when you find all the pieces on the 8th go that will not have any more possibilities stemming from it]]. And for the 10th time it would be ((8P8 -1) x 9 - 6) x 10 = 3628710. As for the 11th time, uh, I really can't be bothered working out the possible outcomes where you collect a piece you already have twice yet every other time you collect a new piece. It will NOT be 12, it will possibly be 30 or close to? I know what answers I seek I just don't know how to find them.

That is why we would make an assumption that there are an equal number of pieces yet an infinite number of pieces. Every time you attempt to take a piece the chance of that piece being any piece is 1/8 and your previous purchase will not affect that in any way.

 

Yes there are several factors but to simplify the problem we can assume that the pieces are distributed equally, each has a ratio of 1/8. And we can also assume that even after you get 1 piece, the proportion doesn't change, it means the number of pieces is infinite; because we don't know the total number of the pieces. Maybe there is a complicated formula for the situations like this in advanced math, but I don't know such a formula, so I should think all the possibilities one by one and there are so many possibilities.
But as Abomination says, your chance of finding all the pieces at the first try is 0,240325927734375% (it is 7/8 * 6/8 * 5/8 *4 /8 * 3/8 * 2/8 *1 /8) and your chance is higher for 9 tries, and even higher for 10 tries and it goes on like this. If your chances were even, the answer would be 209 (because 50/0,240325927734375 is 208,05079365079365079365079365079) but your chances are getting higher as you try more, so the answer is less than 209. That's all I can say for now.

 

According to McDonald's website, the odds of winning a car in their latest promotion are dependant on the number of entries, but are estimated at 1:6,706,697.

 

You guys are right about the chances of getting all eight pieces after eight purchases; it is (1x2x3x4x5x6x7x8)/(8^8)= 40,320/16x10^6 = 0.25% (by the way, the combination above the fraction is called the factorial of 8, or 8!). But then the chances increase pretty fast with every extra piece you get.

Actually, I think the simplest way to solve the problem is to evaluate the chances of not getting one piece. If we are speaking of the chances of not getting one particular piece, it is pretty simple; each time you make a purchase, you have a chance of 7/8 of not getting that piece. So, after you made N purchases, the chances of not getting that piece are (7/8)^N. Now, if after N purchases you don't have the whole map, this means that you are missing at least a piece, so what you have to compute are the chances of not getting either the first piece, or the second, or the third, all the way to the eight. But this is not so trivial, since these do not add up directly.

For simplicity, one could try solving for the case when the whole map has only 4 pieces first.

Okay, so here it goes for a 4 pieces map:

Probability of not getting all 4 pieces after N purchases =
P(not 1)+ P(not 2) + P(not 3) + P(not 4) -
- P(not 1&2)- P(not 1&3)- P(not 1&4)- P(not 2&3)- P(not 2&4)- P(not 3&4) +
+ P(not 1&2&3) + P(not 1&2&4) +P(not 1&3&4) +P(not 2&3&4)

where
P(not i) = (3/4)^N (probability of not getting piece i)
P(not i&j) = (2/4)^N (this is the probability of not getting either piece i or piece j)
P(not i&j&k) = (1/4)^N (i,j,k can be different numbers from 1 to 4)

If you use this, I believe that for 4 purchases, you will obtain that the probability of not getting all 4 pieces is 232/256, which means that the probability of getting them is 1- (232/256) = 24/256 = 4!/4^4

[ July 25, 2006, 01:45: Message edited by: khaavern ]

 

We are such geeks...

 

Wow man. This was all too long ago. Nice to see there's other stuff than spam on the boards. (Other words: Glad to see this is not an other Arrr-thread )
A pity I'm not good at both, seemingly.

 

Aye, me matey MoN. Arrrr, I used to be good at this stuff in high school, but now I can't remember most of it. Thanks for taking the time to explain Abomination, you're now officialy my mathematical hero.

I'm just curious about one other thing: is this an actual McD thing and what can you win, or is this something you made up. If the second is the case, you should have used dice instead of treasure maps... it's like an unwritten rule that all probability questions must include dice

 

Well, me three! I took statistics an eon ago in school, and I liked the probability section most of all. (lottery drawing; 6 balls from a pool of 49, odds of matching all six are a little over 1:13million. B) )

This thread has been fascinating; I was struggling to remember the formulae to work all this out (assuming an even distribution, even though I think we all agree McD would weight the quanties of the pieces to make some prizes harder to get than others)

But yes, this is a real McDonalds contest in North America, tied into the release of Pirates of the Carribean: Dead Man's Chest (c'mon, the thread is "Pirate map : )" - din't ye see dat comin'?

Great work on the #s Abomination, khaavern and Argohir - this is keeping me following raptly!

And thanks for the 'offical' odds, Gawain, that may help in working backwards to find the actual distribution (Oh, and I've sat in your airport a couple of times waiting for a bus - you still got that cool taxidermed grizzly in the lounge? )

 

There is a dutch warehouse consern that expresses its prices with exclamation-marks iso euros. So 5 euro would be 5! Once I learned what factorials (faculteit) was, I thought these prices to be exhorbitantly high.
Only the stuff that costed 2 euros (2!) made sense to me.

 

Heh, prices expressed with factorials

Mesmero: one can rephrase this puzzle in terms familiar to a MMORPG addict let's say that some dragon drops at random one piece of a very rare armor each time you kill him. The whole armor has eight pieces; how many times you have to 'farm' that dragon till you get 50% chances of getting the whole armor

 

And here are the chances for 8, 9 and 10 tries

8 tries: 0,240325927734375% 40320 winning combinations out of 16777216 (8^8) combinations

9 tries: 1,051425933837890625% 1451520 - 40320 winning combinations out of 134217728 (8^9) combinations

10 tries: 2,68113613128662109375% 30240000 - 1451520 winning combinations out of 1073741824 (8^10) combinations

I do that by writing a program in Java, calculating them by hand is impossible. And I have to do a lot of changes to find the other results, because the numbers in 11 and more tries are out of Java's integer bounds. So, your total chance for 8, 9 and 10 tries is 3,97288799285888671875%, near 4% and the number of total tries is smaller than 28.

Edit: 11 tries: 5,22427260875701904296875% 479001600 - 30240000 winning combinations out of 8589934592 (8^11) combinations. My computer calculated it in approximately half an hour. So,the total chance in 11 tries is 9,19716060161590576171875% I will try to calculate the others but I don't know if I can do it.

[ July 27, 2006, 11:44: Message edited by: Argohir ]

 

Holy cow, that is a lot of work

Anyhow, I think the problem becomes simpler if the number of tries is larger.

I mentioned a general formula above, which contains a number of terms. To recap

P(not getting all) = sum_i P(not i) - sum_(i,j) P (not i&j) + ...
where sum_i means sum over all different single numbers (so a sum from 1 to 8), sum_(i,j) means a sum over different combinations of 2 numbers ((1,2), (1,3),... there are 28 such combinations).

The thing is, if you make enough tries, the first term becomes larger. For N tries

sum_i P(not i) = 8*(7/8)^N
sum_(i,j) P(not i&j) = 28*(6/8)^N

so because N is in the exponent, the second term decreases quite faster with N then the first one.

So let's say that only the first one matters. then, we would have
P(not getting all) ~ 8*(7/8)^N =0.5
or (7/8)^N = 1/16
which, if one uses logarithms, gives N about 22 or 23 (I think so, at least, since I do not have a calculator with logarithms handy). Since I neglected the second term, the result is probably a bit lower (I'd say 20 or 21 tries).

This will not work with a small number of tries (10 or 11), so I do not know how to verify Argohir's result.

[ July 28, 2006, 02:52: Message edited by: khaavern ]

 

Any number from 8 to infinity is quite possible, due to the happy laws of probability. But, screwing maths, I say, eh, about 12 times.