Forum > Tuber's secrets > thread

Went to check the forum, noticed the site wouldn't load, and had to leave the game paused on Level 102 for the rest of the day. XD
Wow, you got lucky. You were able to avoid having to deal with the trouble of saving. Well done

Apparently the chances of getting 2 of the same co.5 items, one after the other, is about 1 in 1,383,000.
Where did you get this number ? It depends on whether the item is an effect or a point, because there is a different number of each.

There are 8 coef.5 effect items.
700 (chances of getting one) * 8 (chances of getting the right one) squared (chances of getting it twice in a row) = 1 in 31,360,000

There are 13 coef.5 point items.
700 (chances of getting one) * 13 (chances of getting the right one) squared (chances of getting it twice in a row) = 1 in 82,810,000

Now, i got my 2 in the same game, anyone care to run those odds?
If you played 100 levels:
(5600 / 100) * (5600 / 99) = 56 * 56.57 = 1 in 3,167.68 (for an effect item)
(9100 / 100) * (9100 / 99) = 91 * 91.92 = 1 in 8,364.65 (for a point item)

I think. I am not a statistician, so forgive me if I have done something wrong.

It was diamond of aces. I'm not complaining, but i just haven't got my BoS yet...
Me neither (after six months). Join the club.

Vanilleon: Well... I would appreciate Terry Pratchett being right in this point But maybe that which he describes is only true on the Discworld? I'd guess so because there are several weird things that appear only there (like... sunlight flowing slowly?)


Apparently the chances of getting 2 of the same co.5 items, one after the other, is about 1 in 1,383,000.
Where did you get this number ? It depends on whether the item is an effect or a point, because there is a different number of each.

There are 8 coef.5 effect items.
700 (chances of getting one) * 8 (chances of getting the right one) squared (chances of getting
it twice in a row) = 1 in 31,360,000
But you don't care which one you get. So it is 700 * (8*700) - the first one is any coeff 5 and the second one is this specific coeff 5. This equals 3.920.000

Furtermore, I would say that the effects and points items should be counted together. The 1/700 chance counts for both together; because of the varying numbers of items the respective chances for effects and points items might be different.

So in my opinion it is 700 * (21*700) * 2
700: chance for the first coeff 5
21*700: chance for the same coeff 5 again
and the 2 because there are two items in the first of the two levels (and because you want a specific item in the second level, there is only one chance to get it unless you have )
700*21*700*2=20.580.000

*sighs* I guess I am wrong too. I don't know too much about statistics.

But I think two items in the same game should be easily calculated. Assuming you played 100 levels with items, you can just take any two of these levels, so it is "100 nCr 2" (this is what my calculator calls it... can't use the correct writing here); 100 nCr 2 = 100!/(98!*2!) = 100*99/2 = 4950
So if my idea is a good one, you get 20.580.000/4950=4157... So you should get two of the same kind of coeff 5 item every 4000 games. Looks like a quite small number here... But for me I can say that I had this situation twice so far - once and once

Ohhh, I almost had 1 234 567 points, and I only needed to kick one bomb, when I had to take 500 crystal or die

But you don't care which one you get. So it is 700 * (8*700) - the first one is any coeff 5 and the second one is this specific coeff 5. This equals 3.920.000

That is an excellent point, and you're right. I concede this mistake.

But other than that, we are basically saying the same thing. It's just that yours is a general form for E and P, and mine assumes that the probabilities for coef.5-E and coef.5-P are equal (which has never been adequately tested). I think this is a safe assumption, but it is important to note that it is still an assumption.

Either way, the numbers are similar, and they seem reasonable. (And for the record, I could never remember the difference between combinations and permutations. I knew it at one point ...)

I think it's funny that we're discussing statistics in the DV thread ...

How come you have the chance of getting the same Coef.5 item in 2 levels in a row is 1:1+mil and having the same Coef.5 item through the whole game is 1:3+mil?

How come you have the chance of getting the same Coef.5 item in 2 levels in a row is 1:1+mil and having the same Coef.5 item through the whole game is 1:3+mil?

The 3+mil figure is actually for two in a row. Through the whole game is in the thousands (4157). I don't know where that 1+mil figure came from ... either we're doing something wrong or whoever came up with that did something wrong.

Started game.

Breased through, at lev 10 i had + lives (new record for me)

at 50 i had 1m

Missed the rigor dangerous key since it was my first time in that PD, but i got the crystals.

Made it to lev 69.

Got 1.277M points

Died and saw a spanish game over screen...

and mine assumes that the probabilities for coef.5-E and coef.5-P are equal (which has never been adequately tested). I think this is a safe assumption, but it is important to note that it is still an assumption.
I don't think so (I am sorry ). Basically it depends on how the items you get are calculated.
I can think of two possible ways.
First way: The calculation is based on single items: Every item has its own chance for appearing. So if you get an effect item, each of the coeff 6s has a certain "starting chance" of being this item; each of the coeff 5s has a higher chance, and so on. There are a certain number of effect items, and the chances of getting one of them depend on this number and on the item's "starting chance". Since there is a higher number of point items, the same "starting chance" leads to a lower real chance for every single item.
Second way: The calculation is based on coefficients. If an effect item appears, first the coefficient is chosen, and then every item within this group has the same chance of appearing. Same with point items, so the chances of getting a coeff 5 item are the same, no matter what kind of item you get.

I don't know which one is right - or even if there is a third way of calculating! But I think it might be true. You can see it quite well if you look at the coeff 6s: the french forum has a thread where the highest numbers of coeff 6s of one kind are listed. For example someone got 7 golden mushrooms so far, another player got 5 purple pearls, and so on. The effect item numbers are somewhere around 5-7 each while the point item numbers are always around 2 or 3. This is a quite big difference, and there are four effect items and four point items, so I would say the first way (see above) is right.

Either way, the numbers are similar, and they seem reasonable.
Yes

I think it's funny that we're discussing statistics in the DV thread ...
I think this thread is just the right one for statistics xD

Why do I always die right after I get ?

Hmm... Here's my two cents on the whole Coeff 5 probability whatnot.

In 200 items on 100 levels every item has a 0.1% chance of being coeff 5. This could be any of either 8 items, or 13 items, and it will only be one of them at a time. 0.2 coeff 5 items are to appear in one 200 items game, so it would take 5 100 level games to get one. for a certain coeff 5 effect item:

0.1% = .001 = 1/1000

1/21 x .001 ~ 0.0000476 or 1/21000

1/8 x .001 = 0.000125 or 1/8000 items you get a certain effect coeff 5.

1/13 x .001 ~ 0.0000769 or 1/13000 items you get a certain effect coeff 5.

40 games to see 8 coeff 5 effect items, not necessarily one of each kind, I think.
65 games to see 13 coeff 5 point items, again not necessarily one of each kind.

Which totally works out when you think of me, with 23 coeff 5 items found and having played something over 100 games. But that's just me.

As for figuring out the likelyhood of getting two of the same? My head is failing but what I see is something like 1 in 441,000,000. Eh. Who knows?