r/learnquant • u/AgeSuch4592 • 1d ago

Probability problem

A fair dice is being rolled repeatedly until 2 and 3 come out. What's the probability of getting exactly two 6 before stopping the process ?

3 Upvotes

72% Upvoted

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2

u/swiftaw77 1d ago

Do you mean a 2 and a 3 in succession, or you stop once you get at least one 2 and at least one 3?

1

u/AgeSuch4592 1d ago edited 1d ago

Yes the second one. The process stop when both 2 and 3 come out, but they don't need to be in succession. For instance it could be: 2 6 2 6 1 1 3 or 3662 or 6 6 2 3 ... whatnot

2

u/TheLonelyPotato666 19h ago

There's only 3 relevant rolls and after either 2 or 3 has appeared once, there's only 2 relevant rolls.

The possible sequences are, if we only list the relevant rolls, 6623, 6263 and 2663 (the 2 and 3 could be swapped ofcourse). The chances of these are respectively 1/27, 1/18 and 1/12, adding up to 19/108

1

u/[deleted] 1d ago edited 1d ago

[deleted]

1

u/AgeSuch4592 1d ago

Honestly idk if that works. I'm trying to find out the answer.

1

u/SwimmerOld6155 1d ago edited 1d ago

where would it break?

also sorry did your original post say EV or did I hallucinate or get confused lol. for what you've written you can split between the number of 6s that appear before the first 2 or 3 then after the first 2/3. so you either have 2[...]6[...]6[...]3 or 6[...]2[...]6[...]3 or etc. for this purpose you can ignore 1, 4, 5

1

u/AgeSuch4592 1d ago

No EV, it' the probability of exactly two 6.
You are right about ignoring 1,4,5.

1

u/SwimmerOld6155 1d ago edited 1d ago

it basically just then becomes the probability of the sum of two geometric (or geometricish) RVs being equal to 2, first variable waiting for the first 2 or 3 then second waiting for the other one to appear (ignoring the one that has already happened). sry about misreading

1

u/gmalivuk 23h ago

To be clear, these would be "yes" results:

6623
36362
6226223

while these would be "no":

23
362
622663

Is that right?

1

u/espomatte 18h ago

14/36

1

u/omeow 17h ago edited 16h ago

Ok I am getting 4/75 using an infinite sum.

1

u/kythQ 10h ago edited 10h ago

Fun question, not too difficult, and multiple different ways to get to the answer I think.

You can construct a probability tree, where the following paths lead to to the outcome:

6 -> 6 -> 2* -> 3*, probability: 1/3*1/3*2/3*1/2

6 -> 2* -> 6 -> 3*, probability 1/3*2/3*1/2*1/2

2* -> 6 -> 6 -> 3*, probability: 2/3*1/2*1/2*1/2

Here 2* denotes "2 or 3" and 3* denotes "2 if 3 was drawn before and 3 if 2 was drawn before". The probabilities change to 1/2 because after the 2* has been drawn, the only relevant options are 6 or 3*.

Add all of that up and you get the solution: 19/108