Dice Problem: Chance of throwing three doubles in a row.

[genepeer]

Something I just learnt and thought I'd share. Here's the game: you throw a pair of dice and whenever you get three doubles in a row, you earn a point. You will have to throw six doubles in a row to earn two consecutive points. So after throwing for a countless number of times, what is the probability of earning a point on your next throw?

Hint: Stochastic matrix

Edited January 26, 2011 by Gene-Peer

[Drake Glau]

With your last question can we assume you already rolled 2 doubles? making the chance of the next one being a double still 1/36?

For 3 doubles in a row wouldn't it be (1/36)^3? Since each throw is independent? I don't know what these Stochastic matrices are so I'm just throwing stuff around in my head.

[Daedalus]

Some thing I just learnt and thought I'd share. Here's the game. You throw a pair of dice and whenever you get three doubles in a row, you earn a point. For instance, you will have to throw six doubles in a row to earn two consecutive points. So having throwing for a countless number of times, what is the probability of earning a point on your next throw?

Hint: Stochastic matrix

Ahh interesting. Basically you have to consider the probability that you have already thrown two doubles plus the probability that you will throw one now. Right? Double meaning two of the same number, I guess? I don't know what a stochastic matrix is, but basically it's a 1 in 6 chance of getting a double in any throw, so there's a 1 in 36 chance the previous two throws have both been doubles. Then the chance of you getting another double to get the point is 1 in 6 again, so I guess that's a simple multiplication and results in 1/216 or (1/6)^3. I think.

[Drake Glau]

To get a double there'd be two dice. 1/6 of getting x and then the other is 1/6 chance of getting x. So getting double is already (1/6)^2 or 1/36, right? I hate probability. Then getting three of them in a row would be independent and (1/36)^3?

[Daedalus]

To get a double there'd be two dice. 1/6 of getting x and then the other is 1/6 chance of getting x. So getting double is already (1/6)^2 or 1/36, right? I hate probability. Then getting three of them in a row would be independent and (1/36)^3?

No, you get anything with the first dice and then there's a 1/6 chance of getting the same anything with the second...

[Drake Glau]

Right...mine would be if it had to be a specific number, sorry...

[genepeer]

(1/6)^3 would be the probability of rolling 3 doubles in a row at the start of the game. The probability of getting a point on the fourth roll is 5/6*(1/6)^3 because you'll have to "miss" on the first throw then get your 3 doubles in a row. When I say countless number of times, I mean a lot not just four. So there are also countless number of possibilities to consider if you wanted to use this approach. The probability matrix is really the only way to solve this. Wikipidea "Stochastic matrix" and they'll give a similar example to explain; it's a really interesting way of doing probabilities. I'll try to write the solution for this tomorrow.

Edited January 23, 2011 by Gene-Peer

[genepeer]

When you play this game, there are 4 possible states:

Rolled 0 doubles in a row (St1)

Rolled 1 double in a row (St2)

Rolled 2 doubles in a row (St3)

Rolled 3 doubles in a row (St4)

Whenever you roll, you transition between these states. When you start the game, you are in St1. After your first roll, you can obviously only be in St1 or S2. The probability of moving to St2, i.e. the probability of rolling a double, will be 1/6, while the probability of staying in St1 is 5/6. Probability of moving to S3 or S4 both remain 0. This can be represented by a stochastic vector

(5/6 1/6 0 0)

The i-th element in the vector represents the probability of moving from St0 to St-i.

Similar vectors can be made for the other states. For St2,

(5/6 0 1/6 0)

Note that the second element is zero, because after rolling, you’ll either miss a double then go back to St1 (0 doubles in a row) or get a double then move to St3 (2 doubles in a row).

For St3,

(5/6 0 0 1/6)

And St4,

(5/6 1/6 0 0)

This is to show that whenever you throw 4 doubles in a row, it is counted as a 3 in a row, then 1 double in a row. So you move from St4 to St2. These vectors can be combine to produce a matrix, M.

(5/6 1/6 0 0)

(5/6 0 1/6 0)

(5/6 0 0 1/6)

(5/6 1/6 0 0)

The 3rd element in the 2nd row is the probability of moving from St2 to St3. Since the game always starts at St1, we are only interested in the 1st row.

When we want to consider the probability after n throws/turns, then we use the probability matrix, Mⁿ. So the probability of being in St4 (i.e. getting a point) after 4 rows will be the 4th element in the 1st row of M⁴. This will turn out to be the value I calculated in my last post.

My question asked what the probability is after countless number of throws. This means we want to see what happens to Mⁿ as n tends to infinity. So raise M to a large number like 1000. The probability of earning a point will be the 4th element in the 1st row of this new matrix. Question answered.

Interesting enough, as n tends to infinity, all row become identical. This means that it won’t matter if you cheat by starting the game saying you’ve rolled 2 doubles in a row already. Winning a point on the 1000th roll will still have the same probability as calculated.

NB: Ok, now I have a serious problem! I can't attach pictures at all. I'm getting "upload failed" I had pictures for the matrix and vectors.

[Austin Glau]

NB: Ok, now I have a serious problem! I can't attach pictures at all. I'm getting "upload failed" I had pictures for the matrix and vectors.

Servers capped as of now, try again later.