Collect all three!

Probability Level 4

You play a new online game called, "Pokey Mango!" It's a game where you go around collecting strange cartoonish beasts.

Suppose there are only three such beasts in the game, and each beast has the following probability of being found:

  • Snorlax 1 6 \frac{1}{6}
  • Drowzee 1 3 \frac{1}{3}
  • Magikarp 1 2 \frac{1}{2}

The expected total number of beasts you need to collect before you have "Collected them All" (In this case there are only 3 3 ) is a b \frac{a}{b} where a a and b b are coprime positive integers. What is a + b a+ b ?

Image credit: http://forums.smitegame.com/ , http://nintendo.wikia.com/ , http://www.dltk-kids.com/


The answer is 83.

Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
Sep 22, 2016

Define the following:

  • E = E = Expected number of catches for getting all three before you have collected any
  • E n = E_n = Expected number of catches for getting all three once you have the digits in n n

(e.g. E 23 = E_{23} = Expected value of getting all three once you have 2 2 and 3 3 )

Finally, let 6 6 represent the Snorlax, 3 3 represent the Drowzee, and 2 2 represent the Magikarp.

This leads to the following set of linear equations:

  • E = 1 + 1 2 E 2 + 1 3 E 3 + 1 6 E 6 E = 1 + \frac{1}{2}E_2 + \frac{1}{3}E_3 + \frac{1}{6}E_6
  • E 2 = 1 + 1 3 E 23 + 1 6 E 26 + 1 2 E 2 E_2 = 1 + \frac{1}{3}E_{23} + \frac{1}{6}E_{26} + \frac{1}{2}E_2
  • E 3 = 1 + 1 3 E 3 + 1 6 E 36 + 1 2 E 23 E_3 = 1 + \frac{1}{3}E_3 + \frac{1}{6}E_{36} + \frac{1}{2}E_{23}
  • E 6 = 1 + 1 2 E 26 + 1 3 E 36 + 1 6 E 6 E_6 = 1 + \frac{1}{2}E_{26} + \frac{1}{3}E_{36} + \frac{1}{6}E_6
  • E 23 = 1 + 5 6 E 23 E_{23} = 1 + \frac{5}{6}E_{23}
  • E 26 = 1 + 2 3 E 26 E_{26} = 1 + \frac{2}{3}E_{26}
  • E 36 = 1 + 1 2 E 36 E_{36} = 1 + \frac{1}{2}E_{36}

Solving these, we find that E = 73 10 E=\frac{73}{10}

73 + 10 = 83 73+10 = \boxed{83}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Great problem! I haven't seen a coupon collectors problem with unequal probabilities posted on Brilliant before. There is a nice general solution for n n "beasts", (or coupons, etc.), with probabilities p 1 , p 2 , . . . , p n p_{1}, p_{2}, ... , p_{n} , such that

E = ∫ 0 ∞ ( 1 − ∏ k = 1 n ( 1 − e − p k x ) ) d x \displaystyle E = \int_{0}^{\infty} \left(1 - \prod_{k=1}^{n}(1 - e^{-p_{k}x})\right) dx .

Brian Charlesworth - 4 years, 8 months ago

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Whoa.... Cool answer @Brian Charlesworth ! I hadn't seen that one before!

Geoff Pilling - 4 years, 8 months ago

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