Ok that means the spinning wheel draws 5 cards and it will not draw the first 10 pillar and only 4 of the 5 drawn cards will spin?
Just a question to be able to calculate it:
Is it sampling with or without replacement?
....
3 different cards = 23.2% winning chance
2 different cards = 52% winning chance
...
I assume its not correct without calculating it. Just 3 different cards can be 3/1/1 and 2/2/1 and 2 different cards can be 4/1 and 3/2.
And I dont know whether its so easy to calculate.
The combined with the original post in this thread seemed like a good starting point. So I started with a 24/6 deck cuz the math should be the simplest.
With 10 chances per variable of finding the rare, there should be a (24/30)^10 = 10.7% chance of a pillar and a 89.3% chance of finding a rare.
The fact that there are four variables rather than three or five is irrelevant, I think. There's a 89.3%^3 = 71.1% chance of winning a rare per spin. Chriskang says 33.8% so obviously my math is wrong. Little help, anyone?
@Kevkev
Its not so easy to calculate or rather its complex.
1.
At least I would calculate first, how high your chance for each possible "shortlist"(5cards) is(18pillar/6rare1/6rare2):
1. 5pillar/0rare1/0rare2 = ohh small
0.000526%(without replacement)
2. 4pillar/1rare1/0rare2 = ....
3. 4pillar/0rare1/1rare2 = ....
4. ...
2.
After that I would calculate the chance for each "small shortlist"(4cards):
For example 3/1/1 can be 2/1/1, 3/0/1 and 3/1/0 after taking 1 card away.
3.
Then I would calculate the chance for every "small shortlist to win one of each card"
For example 4/0/0 = 100% to win pillar
4.
That means 5/0/0 has an 0.000526% to appear and a 100% to win a pillar, thus you have a 0.000526% to win a pillar. If you sum all the cases up you get the total chance, but there are much cases....
I need to know whether its sampling with or without replacement to answer this question.
And the spinning wheel only has "10 fails" in total?
In this calculating I assume its without replacement, thus if the spinning wheel draws a pillar first there are only 23/6 left when drawing the next card:
1.
rares | pillars(pillars really in spin) | case | chance |
0 | 15(5) | 5/0 | 0,8429% |
1 | 14(4) | 4/1 | 7,5862% |
2 | 13(3) | 3/2 | 24,1379% |
3 | 12(2) | 2/3 | 34,8659 |
4 | 11(1) | 1/4 | 24,1379 |
5 | 10(0) | 0/5 | 7,5862 |
| | | =99,157% |
Its hypergeometric: 15 cards(10 fails + 5) drawn and chance to draw 15/0; 14/1, .....
Its only 99,157% total, because 9/6 isnt possible.
Now you just need to look what happens if you take one card away:
shortlist | 2. small shortlist1 | 3. chance to win rare | 3. chance to win pillar | 2. small shortlist 2 | 3. chance to win rare | 3. chance to win pillar | 4. chance to win pillar in total | 4. chance to win rare in total |
5/0(0,8429%) | 4/0 | 0% | 100% | | | | 0,8249% | |
4/1(7,5862%) | 4/0(20% of 7,5682%= 1,51%) | 0% | 100% | 3/1(80% of 7,5862%) | .... | ... | +1,51%+.... | ... |
.... | ... | .. | ... | .. | .. | ... | ... | ... |
... | ... | ... | .. | .. | .. | .. | =total chance pillar | =total chance rare |
I think its just too much work.