Probability – Analyzing Probability of an Outcome with Potential for Retries

probability

I’ve been trying to figure out how to solve a particular scenario. Even after taking an Intro Statistics course, I haven’t found a solution. I apologize for the complexity of the following analogy.

Suppose, for example, you play a dice game with the following rules:

If you roll a
1-2: You get a point,
3-5: Nothing happens,
6: You are provided two more dice rolls with the same rules.

Rolling a 1 or 2 will add 1 point to the point total and the die will be discarded from the game.

Rolling a 3, 4, or 5 has no effect. The die is discarded without providing more dice to be rolled or adding to the point total.

Rolling a 6 results in the die being discarded from the game without awarding a point; following which, the player is provided two die.

One die is provided to roll at the start of the game. As stated, after a die is rolled it is discarded from the game.

It is only possible to roll more than one die in a given game if the original die lands on 6 – in which case the die will be discarded and the player will be provided two dice to roll.

Retries (6: roll) are awarded regardless of whether or not a 6 was previously rolled. Theoretically, the game can consist of dozens of rolls, albeit very unlikely.

Die are rolled individually, regardless of how many the player has acquired.

The game ends when all die have been rolled and discarded. The number of times a point was rewarded (1-2: roll) will result in the final score.

The most efficient way to win with one point is by rolling a 1 or 2 (1/3 odds). The most efficient way to win with two points is by rolling the sequence 6, 1 or 2, 1 or 2 (1/54 odds).

What I don’t understand is the function that provides for the possibility of being rewarded multiple reroll opportunities, allowing for a game that doesn’t have a predetermined number of rolls.

What is a formula I can use to discover the probability of completing the game with $n$ points? I would like to see how this is done.

If this analogy doesn’t make enough sense, I will do my best to provide clarification.

Best Answer

One approach is to use generating functions, using the fact that with a single die you have a probability $\frac12$ of zero points, a probability $\frac13$ of one point and a probability of $\frac16$ of two new dice each with the same generating function.

So I think you have $$g(x)=\frac12+\frac13x + \frac16 g(x)^2$$ which as a quadratic has solutions $$g_1(x)=3 + \sqrt{2} \sqrt{3-x} \text { or } g_2(x)=3 - \sqrt{2} \sqrt{3-x}.$$

The first of these will end up generating a probability of zero points of $g_1(0)=3+ \sqrt{6}>1$, so can be rejected as spurious, while the second has a Taylor series which looks like $$g_2(x)=\left( 3-\sqrt{6}\right)+\sqrt{\frac{1}{6}}x +\sqrt{\frac{1}{864}}x^2 +\sqrt{\frac{1}{31104}}x^3 +\sqrt{\frac{25}{17915904}}x^4+\cdots \\ \approx 0.5505 + 0.4082x+0.0340x^2+0.0057x^3+0.0012x^4+\cdots$$

and that suggests the probability of zero points at the end of the game is about $0.5505$, of one point about $0.4082$, of two about $0.0340$, of three about $0.0057$, of four about $0.0012$, falling rapidly.

Added Here is a simulation in R, suggesting this is plausible:

set.seed(1)
maxn <- 100000
score <- numeric(maxn)
for (n in 1:maxn){
  dice <- 1
  points <- 0
  while (dice > 0){
    roll <- sample(6,1)
    if (roll <= 2){ points <- points + 1 }
    if (roll == 6){ dice <- dice + 2 }
    dice <- dice - 1
    }
  score[n] <- points
  }
table(score)/maxn
# score
#       0       1       2       3       4       5       6       7 
# 0.55086 0.40832 0.03373 0.00562 0.00104 0.00034 0.00008 0.00001
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