# A complex dice game (at least to me)

1. Jul 21, 2008

### bobze

Hello all,

I was thinking about how a certain probability would be calculated and the answer is eluding me. It has been a number of years since I have taken a statistics and probabilities class so I am a little rusty.

Anyway, to jump right into my question: How would one figure out the probability of rolling 3 dice and getting a 3 6's. The twist is, dice rolled on a 6 remain the 6 for the next round.

So for example, if I roll 3 dice and get a 1/3/6 my next roll would be the first two dice as I already got a 6 and that is removed/remember by the game.

Also, how would one go about figuring out how many turns of the game one would expect before they attained 3 6's.

I know that for rolling 3 sixes at once, I could call the first roll X and the second roll x(1-x) such that on the nth round I would have x(1-x)(n-1)

Knowing that I came up with $$E$$ix(1-x)(i-1) when i=1 to +$$\infty$$

Then solve the limit to get the expectation of rounds =x(1/x2)

This wont work however for a game where a roll of 6 is removed, just the expectation of rolling all 6's. I am confused on how to start to setup the equation for this new game.

Any help would be much appreciated:rofl:.

2. Jul 23, 2008

### bobze

Did I explain this poorly?

3. Jul 23, 2008

### CRGreathouse

If you have only one die, there's a 1/6 chance of rolling a 6 each time and a 5/6 chance of rolling nothing. Thus the expectation is
e(1) = 1 + 5/6 * e(1)
e(1)/6 = 1
e(1) = 6

If you have two dice, you have a 1/36 chance of finishing in one step, a 10/36 chance of rolling one 6, and a 25/36 chance of rolling none. The expectation is
e(2) = 1 + 10/36 * e(1) + 25/36 * e(2)
11/36 e(2) = 1 + 10/36 * 6
e(2) = 36/11 * 8/3 = 96/11

Can you finish the problem from here?

4. Jul 23, 2008

### bobze

I think so, though I am not sure if I understand why or if it is correct. Here is what I came up with;

With 3 dice you have a 1/216 chance to win the game in 1 roll, a 90/216 chance of rolling at least one 6 and 125/216 chance of rolling no 6 at all.

So...

e(3)= 1 + 90/216 * e(2) + 125/216 * e(3)
91/216e(3)= 1 + 90/216 * 96/11
e(3)=216/91*51/11
e(3)=11016/1001 or 11.00

So I would expect to win the game (on average) in 11 rounds?

Can anyone explain the formula? I don't understand how you came by it.

5. Jul 23, 2008

### bobze

Also, I think I made a mistake in my other calculation. For determining the expected number of rolls to get all 6's on 3 dice would it just be (1-p)/p

So;

$$1-(1/216)$$ = 215
$$\overline{(1/216)}$$

6. Jul 23, 2008

### CRGreathouse

I just came up with it as I was trying to find a way to explain this to you. It avoids infinite series and other things that I thought might be tricky.

The "1" at the beginning is your first roll, which you have to make in all cases. The fractional parts are just each different outcome. You solve the simple ones first, then move on to the complex ones.

Not quite. You need to distinguish between rolling 1 six and 2 sixes. This formula gives the expected number of turns if you can keep only one six per round if you roll two. You want

e(3) = 1 + blah * e(1) + blah * e(2) + 125/216 * e(3)

7. Jul 23, 2008

### bobze

Ok let me try again then:

there are 25 ways you can get one 6 (from (5/6)2*1/6) so there are 65 ways to get two 6's, from (25-90)

So

e(3) = 1 + 25/216 * e(1) + 65/216 * e(2) + 125/216 * e(3)
91/216e(3) = 1 + 25/216 * 6 + 65/216 * 96/11
e(3) = 1711/396 * 216/91 = 10266/1001 or 10.26 rounds

Correct?

8. Jul 23, 2008

### CRGreathouse

Heh, getting there. There are 25 ways of getting 6-other-other, but also 25 ways of getting other-6-other and 25 ways of getting other-other-6. This should make sense -- you're more likely to roll 1 six out of three dice then 2.

9. Jul 25, 2008

### bobze

Ahh, that makes sense so it should be:

e(3) = 1 + 75/216 * e(1) + 15/216 * e(2) + 125/216 *e(3)
91/216*e(3) = 1 + 75/216 * 6 + 15/216 * 96/11
e(3) = 216/91 * 487/132
e(3) = 8766/1001 or 8.76 rounds

correct?