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I am trying to define a probabilistic model for an experiment involving the following steps:

  1. Choose at random 1 dice from a bag containing 5 die (1 each of 4 sided, 6 sided 8 sided 12 sided and 20 sided)
  2. Roll chosen die once
  3. Observe the outcome.

So far, I've come up with 2 models through which probabilities can be assigned to the outcomes. Each model assigns a different probability value to each outcome.

The first method:

I assign equal probabilities to each trial. Then, I assign probabilities to outcomes by summing over trials.

Example: Probability that outcome is 1 = $\frac{5}{50}$= 0.1

(Here I assume that the sample space is composed of 4+6+8+12+20=50 equally likely outcomes and the number of favorable outcomes is 5)

The second method:

I assign probabilities to outcomes assuming that a given outcome is a succession of two 'sub'-outcomes- "choosing a die" and "rolling it once". I treat these 'sub'-outcomes as independent and calculate probability based on this assumption.

Example:Probability that outcome is 1 = $\frac{1}{5}$$\frac{1}{4}$+$\frac{1}{5}$$\frac{1}{6}$+$\frac{1}{5}$$\frac{1}{8}$+$\frac{1}{5}$$\frac{1}{12}$+$\frac{1}{5}$$\frac{1}{20}$ = 0.135

(Here I assume that the probability of choosing any one of the five dice is $\frac{1}{5}$; and having chosen a given die, the probability of outcome being 1 is $\frac{1}{4}$ for die with 4 sides, $\frac{1}{6}$ for die with 6 sides and so on)

Which method is correct?

Summing probabilities over all outcomes in both cases gives 1.

Using method 1:

P(1)=P(2)=P(3)=P(4)=0.1

P(5)=P(6)=0.08

P(7)=P(8)=0.06

P(9)=P(10)=P(11)=P(12)=0.04

P(13)=P(14)=...=P(20)=0.02

Using method 2:

P(1)=P(2)=P(3)=P(4)=0.135

P(5)=P(6)=0.085

P(7)=P(8)=0.0517

P(9)=P(10)=P(11)=P(12)=0.0267

P(13)=P(14)=...=P(20)=0.01

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  • $\begingroup$ Using Baye's conditional probability, P(1) = P(1|4 sided die chosen) + P(1|6 sided die chosen) + P(1|8 sided die chosen) + P(1|12 sided die chosen) + P(1|20 sided die chosen) = $\frac{1}{4}$+$\frac{1}{6}$+$\frac{1}{8}$+$\frac{1}{12}$+$\frac{1}{20}$ = 0.675. I cannot understand how the weight $\frac{1}{5}$ comes into play here; unless the premise is flawed. In which case, how? $\endgroup$
    – Rob Plant
    Apr 27, 2018 at 3:45

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The second method is correct. In the first method, assigning each possibility of "occurrence of 1" equal probability would mean that you'd have the same chances to get 1 from a 6 faced and a 20 faced die. But you know the probability of getting 1 from a 20 faced die would obviously be less. That's one way of looking at it. Hope it helps.

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