When do I have 80% chance of 6 rolls of either 1, 2 or 3?

Question

Problem: Every roll has 4 outcomes: 1, 2, 3 and 4 There are 50% chance for 1-3 at every roll. (And that means 50% for 4) If one roll did not give 1-3 at one roll, then it is 100% chance at the roll after. (So every second roll is guaranteed to be either 1, 2 or 3) If rolled 1, 2 or 3 then there is equal chance of either 1, 2 or 3.

So: 1/6 chance of 1, 1/6 chance of 2, 1/6 chance of 3, 1/2 chance of 4, If last roll were 4 or it is the first roll.

Success: six 1's, six 2's or six 3's. So I need 6 rolls of either 1, 2 or 3.

Example: 1,4,1,2,3,4,1,1,1,1 (now there is 6 1's)

Notes: I know that after 32 rolls there are 100% chance of six rolls of either 1, 2 or 3.

Question: When do I have 80% chance of 6 rolls of either 1, 2 or 3?

Edit 1: If we ignore that I cannot roll two fours in a row then:

I could work backwards: there is 1-((1/6)^5)*(1/2)=0,9999356996 of not getting it in the first 6 rolls.

Then what about 7 rolls? There is 5/6 chance of not hitting a number that matches. (1-((1/6)^5)(1/2))(5/6)=0.8332797497

So with 9 extra rolls: (1-((1/6)^5)(1/2))(5/6)^9=0,1937942376 I hit below 20% probability of not hitting 6 equals. That is 15 rolls.

If I did consider that I could not get two fours in a row, then it would take even fewer extra rolls; Because the probability would be even better. I just dont know exactly how much.

Right?

Edit 2 I see some great answers below that I will look at soon. Here is just an idea I were thinking about.

Initially we forget the 4's. That would give:

(1-((1/3)^5)*(1/2))*(5/3)^4=0,1975308642

Meaning 6+4=10 rolls if there is never rolled a four.

If there were a 100% chance of rolling a four, but not two fours in a row; then we would just multiply with two: 20 rolls.

But now that the probability is 50%, then I must ask how many 4's is there "in between" non-four rolls after 10 rolls. Again it must be a range, so I would like to know the 80% chance again.

(1-((1/2)^8))*(1/2)^(10-8)=0,2490234375

But that does not seem right, and maybe my whole method is wrong then.

I will take a look at the others answers now :)

Answer 1

To take $r$ rolls, you have to get

• $6$ rolls of one of the three numbers $1,2,3$,
• and $a$ rolls of the next with $0\le a \le 5$,
• and $b$ rolls of the other with $0\le b \le 5$,

being able to do that ${5+a+b \choose 5}{a+b \choose a}$ different ways since the final attempt must be the desired number,

• and $r-6-a-b$ rolls of the number $4$

out of the $6+a+b$ places you might have done it so in ${6+a+b \choose r-6-a-b}$ ways $\big($or $0$ ways if $r <6+a+b$ or if $6+a+b < \frac{r}{2}\big)$.

So I think the probability you take $r$ rolls is $$\sum\limits_{a=0}^{5} \sum\limits_{b=0}^{5} \frac{3}{6^{6+a+b}} {5+a+b \choose 5}{a+b \choose a}{6+a+b \choose r-6-a-b}$$

Using R to calculate this

probs <- function(r){
  f <- function(a,b){
    3 * choose(5+a+b, 5) * choose(a+b, a) *
        choose(6+a+b, r-6-a-b)  / 6^(6+a+b)
    }
  sum(outer(0:5,0:5,f))
  } 
probrolls <- numeric(32)
for (n in 1:32){
  probrolls[n] <- probs(n)
  }
sum(probrolls) # check adds up to 1
# 1
sum(probrolls*(1:32))  # expected number 
# 18.27188

The probabilities and cumulative probabilities are as follows, so to have at least an $80\%$ chance of $6$ rolls of either $1$, $2$ or $3$ appears to need $22$ rolls.

cbind(probrolls, cumprob=cumsum(probrolls))
         probrolls      cumprob
 [1,] 0.000000e+00 0.000000e+00
 [2,] 0.000000e+00 0.000000e+00
 [3,] 0.000000e+00 0.000000e+00
 [4,] 0.000000e+00 0.000000e+00
 [5,] 0.000000e+00 0.000000e+00
 [6,] 6.430041e-05 6.430041e-05
 [7,] 5.144033e-04 5.787037e-04
 [8,] 2.014746e-03 2.593450e-03
 [9,] 5.320264e-03 7.913714e-03
[10,] 1.096679e-02 1.888051e-02
[11,] 1.915676e-02 3.803727e-02
[12,] 2.974389e-02 6.778115e-02
[13,] 4.227031e-02 1.100515e-01
[14,] 5.602002e-02 1.660715e-01
[15,] 7.001649e-02 2.360880e-01
[16,] 8.299626e-02 3.190842e-01
[17,] 9.345859e-02 4.125428e-01
[18,] 9.987578e-02 5.124186e-01
[19,] 1.010285e-01 6.134471e-01
[20,] 9.633151e-02 7.097786e-01
[21,] 8.604229e-02 7.958209e-01
[22,] 7.132705e-02 8.671480e-01
[23,] 5.417352e-02 9.213215e-01
[24,] 3.707282e-02 9.583943e-01
[25,] 2.240174e-02 9.807961e-01
[26,] 1.167466e-02 9.924707e-01
[27,] 5.105289e-03 9.975760e-01
[28,] 1.811591e-03 9.993876e-01
[29,] 4.989406e-04 9.998865e-01
[30,] 9.978811e-05 9.999863e-01
[31,] 1.287589e-05 9.999992e-01
[32,] 8.047428e-07 1.000000e+00

---

As a check, here is a simulation of $10^5$ cases that comes close to that result and those probabilities

runofrolls <- function(){
   appearances <- numeric(4)
   while (max(appearances[1:3]) < 6){
     roll <- sample(1:4, 1, prob=c(1/6,1/6,1/6,1/2))
     appearances[roll] <- appearances[roll] + 1
     if (roll == 4){
       nextroll <- sample(1:3, 1, prob=c(1/3,1/3,1/3))
       appearances[nextroll] <- appearances[nextroll] + 1
       }
     }
   return(sum(appearances))
   }

set.seed(2021)    
sims <- replicate(10^5, runofrolls()) 
mean(sims)
# 18.26183 
quantile(sims, 0.8)
# 80% 
#   22 
mean(sims <= 21)  
# 0.79601
mean(sims <= 22)  
# 0.86814