Calculating probability distribution for rolling 4 dice plus reroll lowest die

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

he he :biggrin:
 


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
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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