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Did I calculate the correct probability for these simple scenarios:

1: What is the probability of rolling 3 and 4 with two dice in two rolls?

If you roll either 3 or 4 in the first roll, you put that die aside and roll the second die again.

I would break the probability down like this:

  • 1/18 chance of getting both 3 and 4 (or 4 and 3) in the first roll.
  • 1/4 chance of rolling 3 (31, 32, 33, 35, 36, 13, 23, 53 and 63), and then 1/6 chance of rolling the 4 in the second roll.
  • 1/4 chance of rolling 4 (41, 42, 44, 45, 46, 14, 24, 54 and 64), and then 1/6 chance of rolling the 3 in the second roll.
  • 1/18 of rolling both 3 and 4 on the second roll (having rolled neither in the first).

This adds up to:

$$\frac{1}{18} + (\frac{1}{4} \times \frac{1}{6}) + (\frac{1}{4} \times \frac{1}{6}) + \frac{1}{18} = \frac{7}{36}$$

Is that correct?

2: What is the probability of rolling double 6 in two rolls?

Slightly different breakdown:

  • 1/36 chance of getting 66 in the first roll.
  • 10/36 chance of getting a single 6 (all combinations of 6, except 66), then 1/6 of getting the second 6 in the second roll.
  • 1/36 chance of getting 66 in the second roll.

$$\frac{1}{36} + (\frac{5}{18} \times \frac{1}{6}) + \frac{1}{36} = \frac{11}{108}$$

Correct?

I've watched several videos on probability calculations, the usual "pick a coin in a bag with some unfair coins", and this is the first time I'm actually trying to apply what I've learned.

Update: I was wrong

Correct answers:

$$\frac{1}{18} + (\frac{1}{4} \times \frac{1}{6}) + (\frac{1}{4} \times \frac{1}{6}) + (\frac{4}{9} \times \frac{1}{18}) = \frac{53}{324} = 16.4\%$$

and

$$\frac{1}{36} + (\frac{10}{36} \times \frac{1}{6}) + (\frac{25}{36} \times \frac{1}{36}) = \frac{121}{1296} = 9.3\% $$

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They are not correct: you do not include the probability of getting to the second roll without having made progress in the overall probability of getting the exact needed roll on the second roll alone.

Which is to say: The probability of making no progress on the first roll and then getting 34 or 43 on the second is $\frac{4}{9} \times \frac{1}{18}$, because you only have a 4-in-9 chance of making no progress.

Similarly, since there's a $\frac{25}{36}$ chance of making no progress on the first roll when the goal is to get two sixes, the probability of getting double six on the second roll after making no progress on the first is $\frac{25}{36}\times\frac{1}{36}$

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  • $\begingroup$ Thank you. I suspected something was wrong.... :) $\endgroup$ – Jakob Gade Mar 23 '16 at 1:39

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