Another Warframe-Mathematics Discovery

[Renegade343]

Continuing on from this thread (see post #1), I asked myself a question in my dream: 

 

If a player were to press the Random Colours button continuously for a Warframe, what is the least number of presses he/she has to make in order for his/her one favourite colour combination to appear at least 90% of the time? It takes only one hit to succeed.

 

Thus, the calculation is below: 

 

P(F) = 1.82488E-13, P(F') = ([5.47981E12]-1)/(5.47981E12)

X = No. of times player presses Random Colours button to get his/her one favourite colour

X ~ B(n, 1.82488E-13)

 

P(X = r) = nCr(n, r) * ([1.82488E-13]^r) * (([5.47981E12]-1)/(5.47981E12))^(n-r)

P(X = 0) < 0.1

Assume P(X = 0) = 0.1,

0.1 = nC0(n, 0) * 1 * (([5.47981E12]-1)/(5.47981E12))^n

n = 1.27921E13

 

That means for a player to hit the Random Colours button repeatedly and hoping to have at least a 90% chance of getting the colour combination he/she wants, he/she will have to hit the button 1.27921E13 times. Assuming if it takes around 5 seconds for a click, the colours to load and to see if the colour combinations are correct, then the player must expect to spend: 

 

1.27921E13 * 5s = 6.39605E13 seconds = 2,026,785.9 years just to get his/her one favourite colour combination at lesat 90% of the time.

 

Have fun with that Random Colours button.

Edited July 25, 2014 by Renegade343

[KenseiSeraph]

I'm impressed.

[pooptaco]

Did you know that if you shuffle a full deck of cards, the overwhelming probablilty is that the resulting shuffled deck has never once in the history of humanity been shuffled exactly so.

 

IIRC: the formula is something like: 52^52 + 52^51 + 52^50... etc, etc, etc.  That's how many possible deck-shuffles there are.

 

[edit] I looked it up and I was way off.  It's 52 factorial.  Aka: 52 * 51 * 50 * 49 * etc.  Still, it's a huge number.  Google say:

80658175170943878571660636856403766975289505440883277824000000000000

Edited July 23, 2014 by pooptaco

[CY13ERPUNK]

 i <3 math

 

+1 OP

[Renegade343]

Did you know that if you shuffle a full deck of cards, the overwhelming probablilty is that the resulting shuffled deck has never once in the history of humanity been shuffled exactly so.

 

IIRC: the formula is something like: 52^52 + 52^51 + 52^50... etc, etc, etc.  That's how many possible deck-shuffles there are.

The possible number of ways a 52-card deck can be shuffled is (using permutations, as each order of the card matters): 

 

52P52 = 52!/(52-52)! = 8.06582E67

 

So, to get a duplicate shuffle (getting the same shuffle out of all the shuffles done) at least (1.2398E-66)% of the time, with one try being enough, the number of shuffles needed is: 

 

P(S) = 1.2398E-68, P(S') = ([8.06582E67]-1)/(8.06582E67)

X = No. of times shuffled to get duplicate shuffle

X ~ B(n, 1.2398E-68)

 

P(X = r) = nCr * ([1.2398E-68]^n) * (([8.06582E67]-1)/(8.06582E67))^(n-r)

P(X = 0) < (1 - 1.2398E-68)

Assume P(X = 0) = 1 - 1.2398E-68

1 - 1.2398E-68 = 1 * 1 * ([8.06582E67]-1)/(8.06582E67)^n

n = undef (my calculator just broke down from that)

 

So yes. Quite a lot of shuffles.