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Learning Bayes statistics from Allen Downey's Think Bayes

There are three dice, 6-sided, 8-sided and 12-sided. A randomly chosen dice is rolled and the outcome is "1". What's the probability it was the 6-sided dice?

Is it correct to setup the sample space made of 26 outcomes like so?

outcomes from 6-sided dice: [ 1 2 3 4 5 6
outcomes from 8-sided dice:   1 2 3 4 5 6 7 8
outcomes from 12-sided dice:   1 2 3 4 5 6 7 8 9 10 11 12 ]

Or is it just [1 2 3 4 5 6 7 8 9 10 11 12 ] ?

The prior is 1/3, the likelihood P(rolled-1|6-sided) is 1/6, but what is probability of evidence P(rolled-1)? Is it 3/26? Or 1/12?

I'm not getting the correct result using either.

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Based on Sextus Empiricus' answer https://stats.stackexchange.com/a/650221/313507, here's the sample space: enter image description here

And here are the probabilities of each outcome: enter image description here

The solution of the exercise is, using Bayes formula, $$ P(6-sided|rolled-1) = \frac{P(6-sided)*P(rolled-1|6-sided)}{P(rolled-1)} $$ $$ P(6-sided|rolled-1) = \frac{\frac{1}{3}\cdot\frac{1}{6}}{\frac{1}{8}}=\frac{4}{9} $$

The 1/8 probability of rolling one can be obtained by summing up the probabilities of the three outcomes of rolling a one $$ P(rolled-1) = 1/18 + 1/24 + 1/36 = 1/8 $$ It's the same as calculating it using the law of total probability

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    $\begingroup$ There are many possible correct answers, but all of them need to distinguish the six ordered pairs (die, outcome==1), which your second proposal does not do. $\endgroup$
    – whuber ♦
    Commented Jun 30 at 18:19
  • $\begingroup$ Ok, the correct probability of evidence (dice rolled 1) is 0.125 or 1/8 (because 1/3*1/6+1/3*1/8+1/3*1/12). Looking at the first sample space, why isn't just 3/26? $\endgroup$
    – yingele
    Commented Jun 30 at 18:46
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    $\begingroup$ the (natural) statespace is literally the 26 outcomes you mention (ie a table of which die and which result for that die) $\endgroup$
    – seanv507
    Commented Jun 30 at 18:54
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    $\begingroup$ do the same calculation for (dice = 6 sided and outcome =1) as for dice rolled =1 . what do you get? $\endgroup$
    – seanv507
    Commented Jun 30 at 19:04

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Is it correct to setup the sample space made of 26 outcomes like so?

outcomes from 6-sided dice:  [ 1 2 3 4 5 6
outcomes from 8-sided dice:    1 2 3 4 5 6 7 8
outcomes from 12-sided dice:   1 2 3 4 5 6 7 8 9 10 11 12 ]

This works. It looks like a joint distribution of 'the type of dice' (rows) and the 'number rolled' (columns).

You can fill in the probabilities of each of the 26 samples in this space,

outcomes from 6-sided dice:  [ a b c d e f
outcomes from 8-sided dice:    g h i j k l m n
outcomes from 12-sided dice:   o p q r s t u v w x y z ]

then consider the probability of the type of dice, conditional on the outcome being 1.

I will fill in the first row for you, e.g. the probability that the dice is 6-sided (1/3) and the roll is 1 (1/6) is the product 1/3 x 1/6 = 1/18, or with a different denominator 4/72 (72 is gonna be useful for comparison with the other cases)

outcomes from 6-sided dice:  [ 4/72 4/72 4/72 4/72 4/72 4/72
outcomes from 8-sided dice:    g    h    i    j    k    l    m    n
outcomes from 12-sided dice:   o    p    q    r    s    t    u    v    w    x    y    z ]
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    $\begingroup$ 26 letters, how convenient. $\endgroup$
    – qwr
    Commented Jul 1 at 4:54
  • $\begingroup$ This is helpful. My mistake was to assume that each outcome in the sample space has equal probability. It doesn’t: rolling a 1 with a 6-side dice is more likely than rolling a 1 with an 8-side dice. As a consequence, I used the wrong denominator in the Bayes formula, i.e., the probability of rolling a one. It is the sum of the probabilities 1/18+1/24+1/36=0.125. This can be also obtained by applying the law of total probability, P(rolled-1) = P(6-dice)*P(rolled-1|6-dice) + P(8-dice)*P(rolled-1|8-dice)+P(12-dice)*P(rolled-1|12-dice)=1/3*1/6+1/3*1/24+1/3*1/36=0.125. $\endgroup$
    – yingele
    Commented Jul 1 at 22:03
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    $\begingroup$ @DavidMayer you can also express the probabilities as 4/72, 3/72 and 2/72 and the odds of 4:3:2 for the occurrence of the different dice when the roll is 1 are more directly visible. $\endgroup$
    – Sextus Empiricus
    Commented Jul 2 at 6:48

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