MHB How can I maximize my points when rolling multiple dice with different options?

[zweilinkehaende]

Hi,

i was wondering how to calculate this problem:

Rolling a 5 or a 6 on a dice means you score a point. You are given x dices. You have 2 options: Either you add y dices to your dicepool and can roll dices that show a 6 again (still gaining a point) or you can reroll your dices not showing a 5 or 6.
At which values of y and x should i choose which method to gain the most points (on average).

I made multiple attempts but always failed because i could in theory score infinite points using the first method.

I hope i chose the right subforum.

Ty for your help.

PS.: Please excuse my poor english, it's not my first language.

 

[I like Serena]

Hi,

i was wondering how to calculate this problem:

Rolling a 5 or a 6 on a dice means you score a point. You are given x dices. You have 2 options: Either you add y dices to your dicepool and can roll dices that show a 6 again (still gaining a point) or you can reroll your dices not showing a 5 or 6.
At which values of y and x should i choose which method to gain the most points (on average).

I made multiple attempts but always failed because i could in theory score infinite points using the first method.

I hope i chose the right subforum.

Ty for your help.

PS.: Please excuse my poor english, it's not my first language.

Hi zweilinkehaende! Welcome to MHB! :)

If you can only score points and not lose them, adding an infinite number of dice will indeed gain you an infinite number of points.

Is there perhaps a condition that you lose points?
Perhaps you lose a point if you roll 1-4?

 

[awkward]

zweilinkehaende,

I am not sure I understand the rules. Here is what I think you are saying. Please let us know if this is correct or not.

Positive integers $x$ and $y$ are fixed in advance. You have two options:

1. Roll $x$ dice, then re-roll any that do not come up 5 or 6. Your score is the total numbers of 5s and 6s.
OR
2. Roll $x+y$ dice, then re-roll any that come up 6. Your score is the total numbers of 5s and 6s, including the dice that came up 6 on the first roll.

Is this a correct interpretation?

If so, in the first option the expected number of 5s or 6s on the first role is $ (2/6) x$. The expected number of 5s or 6s on the second role is $(2/6) (x - (2/6)x) = (8/36) x$. So the expected total score is $(2/6) x + (8/36)x = (5/9) x$.

In the second option, the expected number of 5s on the first role is $(1/6) (x+y)$, and the expected number of 6s is also $(1/6)(x+y)$. The expected number of 5s or 6s on the second role is $(2/6)(1/6)(x+y)$. So the expected total score is $((2)(1/6)(x+y) + (2/6)(1/6)(x+y) = (7/18)(x+y)$.

 

[zweilinkehaende]

Hi zweilinkehaende! Welcome to MHB! :)

If you can only score points and not lose them, adding an infinite number of dice will indeed gain you an infinite number of points.

Is there perhaps a condition that you lose points?
Perhaps you lose a point if you roll 1-4?

No there is no condition under which points are lost. I'm searching for the optimal strategy depending on the x and y values. For example if x is 60 and y is 1:
Strategy 1 (adding y and rerolling 6s) would give me approximatly 24 on average. (61*1/3 + 2/6*1/6*61 + 2/6*1/6*1/6*61+...)
Strategy 2 (rerolling 1-4s) would give me approximatly 33 (60*1/3+(60-(60*1/3))*1/3= 33.33).

Obviously i would choose Strategy 2 in this case. But how do i calculate when to switch strategies? (with x and y values reversed it would be quite obvious)

The player doesn't get to choose the x or y values, only strategies.

 

[zweilinkehaende]

zweilinkehaende,

I am not sure I understand the rules. Here is what I think you are saying. Please let us know if this is correct or not.

Positive integers $x$ and $y$ are fixed in advance. You have two options:

1. Roll $x$ dice, then re-roll any that do not come up 5 or 6. Your score is the total numbers of 5s and 6s.
OR
2. Roll $x+y$ dice, then re-roll any that come up 6. Your score is the total numbers of 5s and 6s, including the dice that came up 6 on the first roll.

Is this a correct interpretation?

If so, in the first option the expected number of 5s or 6s on the first role is $ (2/6) x$. The expected number of 5s or 6s on the second role is $(2/6) (x - (2/6)x) = (8/36) x$. So the expected total score is $(2/6) x + (8/36)x = (5/9) x$.

In the second option, the expected number of 5s on the first role is $(1/6) (x+y)$, and the expected number of 6s is also $(1/6)(x+y)$. The expected number of 5s or 6s on the second role is $(2/6)(1/6)(x+y)$. So the expected total score is $((2)(1/6)(x+y) + (2/6)(1/6)(x+y) = (7/18)(x+y)$.

The first one is correct, but the second one can be infinetly continued. The 6s on the second role can be reroled again. And the 6s that come up again, and again... .

My main problem is that i don't know how to express 2/6*(x+y) + 2/6*1/6*(x+y) + 2/6*1/6*1/6*(x+y) + ... correctly.

 

[I like Serena]

The first one is correct, but the second one can be infinetly continued. The 6s on the second role can be reroled again. And the 6s that come up again, and again... .

My main problem is that i don't know how to express 2/6*(x+y) + 2/6*1/6*(x+y) + 2/6*1/6*1/6*(x+y) + ... correctly.

It seems to me that you're only supposed to reroll once...
Anyway, what you have there is a so called geometric series.
In general we have:
$$a + ar + ar^2 + ... = \frac{a}{1-r}$$
So:
$$\frac 26 \cdot (x+y) + \frac 26\cdot \frac 16\cdot (x+y) + \frac 26\cdot \frac 16\cdot \frac 16\cdot (x+y) + ...
= \frac{\frac 26\cdot(x+y)}{1-\frac 16} = \frac 25 (x+y)$$