Calculating probability distribution for rolling 4 dice plus reroll lowest die

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 5K views
Paraxis
Messages
2
Reaction score
0
Hi,

I am trying to figure out what the probability distribution is for the following:

1. Roll 4 dice.
2. Take the lowest die and re-roll it.
3. Take the sum of the three highest dice.

The result will be between 3 and 18.

I know how to figure out the probability distribution for rolling 4 dice and taking the three highest, but how can we calculate the distribution when re-rolling the lowest die?

My thought process goes as follows:
When you roll the 4 dice and keep the three highest, the result from re-rolling the lowest remaining die is only kept if the roll is higher than any of the three dice that were kept.

e.g.

Say you roll a 6,4,2 and 1. Keep the 6,4 and 2 and re-roll the 1.
The new roll is only kept if and only if it exceeds 2.
Therefore there is a 2 in 6 probability for the sum to be 12 and there is a 1 in 6 probability for the sum to be each of 13, 14, 15 or 16.

I can calculate this long hand by writing up all 1296 possible combinations for 4 dice, but if you could supply me with a formula, it would be much quicker...

Thanks,

Paraxis
 
Physics news on Phys.org
Welcome to PF!

Hi Paraxis! Welcome to PF! :wink:
Paraxis said:
1. Roll 4 dice.
2. Take the lowest die and re-roll it.
3. Take the sum of the three highest dice.

Isn't that the same as
1. Roll 5 dice.
2. Take the sum of the three highest dice?​
 


tiny-tim said:
Isn't that the same as
1. Roll 5 dice.
2. Take the sum of the three highest dice?​

...

I've racked my brain and can't see how it is NOT!

Wow... Just roll 5 dice... I guess I over-thought things!

Unless someone demonstrates that it is not the same as rolling 5 dice...

Thanks! Much MUCH easier to calculate...

So the distribution is:

3 = 1 (0.01%)
4 = 5 (0.06%)
5 = 15 (0.17%)
6 = 41 (0.45%)
7 = 90 (0.99%)
8 = 171 (1.88%)
9 = 300 (3.31%)
10 = 485 (5.35%)
11 = 697 (7.68%)
12 = 946 (10.43%)
13 = 1163 (12.82%)
14 = 1326 (14.62%)
15 = 1335 (14.72%)
16 = 1196 (13.18%)
17 = 854 (9.41%)
18 = 447 (4.93%)
 


Paraxis said:
So the distribution is:

3 = 1 (0.01%)
4 = 5 (0.06%)
5 = 15 (0.17%)
6 = 41 (0.45%)
7 = 90 (0.99%)
8 = 171 (1.88%)
9 = 300 (3.31%)
10 = 485 (5.35%)
11 = 697 (7.68%)
12 = 946 (10.43%)
13 = 1163 (12.82%)
14 = 1326 (14.62%)
15 = 1335 (14.72%)
16 = 1196 (13.18%)
17 = 854 (9.41%)
18 = 447 (4.93%)

That totals 9072 permutations but 65 = 7776

I make the correct frequencies 1, 5, 15, 41, 90, 170, 296, 470, 665, 881, 1055, 1155, 1111, 935, 610, 276