Expert: Scott A Wilson - 5/4/2009

Question
QUESTION: Using two six sided dice, numbered 0,2,4,6,8,10, any roll 10 and over gets a second roll, and if the second roll is a ten or over, a third roll and so on...
What are the probabilities for scores 0-20?
for scores 0-8 it is just the regular probability space of 36 possibilities, or really, um 15 of the 36.
0: 1/36
2: 2/36
4: 3/36
6: 4/36
8: 5/36
10: 6/36  and then for it to stay a ten it would be necessary to roll a 0,0 which has a probability of 1/36, so does that decrease the probability to 6/6^4 or 1/6^3?

The second event space( or third and fourth event ) has the remaining 21/36 of the probability space only now it goes into 1/6^4, right?
 I don't know how to calculate the scores without writing everything out.  Permuttions?  Combinations? Pascal's triangle?

ANSWER: The first sums are OK.
0: 1/36
2: 2/36
4: 3/36
6: 4/36
8: 5/36
10: 6/36, but you also need to know that
12: 5/36
14: 4/36
16: 3/36
18: 2/36

20 only has a chance of 1/36.

The next thing to do is to know when a total of 10 is rolled,
4/6 of the time there is no 10, but 2/6 = 1/3 there is.
Take the same distribution above and add 10 to it.  
Note that all of the probabilities were multiplied by 1/3.
Note that 36² = 1296.

While we are doing that, we also need to worry about 1/3 of the 12's having a 10 in them, 1/3 of the 14's having a 10 in them, 1/3 of the 16's having a then in them, and 1/3 of the 18's having a ten in them.

This would make the list into
The first sums are OK.
0: 1/36
2: 2/36
4: 3/36
6: 4/36
8: 5/36
10: (6/36)(2/3)
12: (5/36)(2/3)
14: (4/36)(2/3)
16: (3/36)(2/3)
18: (2/36)(2/3),
and now to add in the second roll.

When doing the following, only worry about the sums of 18 or less.
For 10's, you would add in the table at the top with each of the probabilities times (6/36)(1/3) = 6/108.
For 12's, you would add in the table at the top with each of the probabilities times (5/36)(1/3) = 5/108.
For 14's, you would add in the table at the top with each of the probabilities times (4/36)(1/3) = 4/108.
For 16's, you would add in the table at the top with each of the probabilities times (3/36)(1/3) = 3/108.
For 18's, you would add (2/36)(1/36) to 18.

To explain it in more detail,
10 would generate (6/36)(1/3)(1/36) of getting another 10,
(6/36) (1/3)(2/36) chance of getting a 12,
(6/36) (1/3)(3/36) chance of getting a 14,
(6/36) (1/3)(4/36) chance of getting a 16, and
(6/36) (1/3)(5/36) chance of getting a 18.

12 would generate (5/36) (1/3)(1/36) of getting another 12,
(5/36) (1/3)(2/36) chance of getting a 14,
(5/36) (1/3)(3/36) chance of getting a 16, and
(5/36) (1/3)(4/36) chance of getting a 18.


14 would generate (4/36) (1/3)(1/36) of getting another 14,
(4/36) (1/3)(2/36) chance of getting a 16, and
(4/36) (1/3)(3/36) chance of getting a 18.

16 would generate (3/36) (1/3)(1/36) of getting another 16, and
(3/36) (1/3)(2/36) chance of getting a 18.

18 would generate (2/36) (1/3)(1/36) of getting another 18.

Go through and generate the values, add them all together,
and there you have it.
All of the rest of the ways to roll would generate something over 20.



---------- FOLLOW-UP ----------

QUESTION: It seems that this amswer is to a different question; I don't understand the 1/3 probability part.  Let me rephrase it:  Any roll with a sum, of both dice, which is ten or larger than ten (e.g 4;6 ,  6;8 , 10;10) gets a second roll of both dice.  This rule applies to the second roll, and third, and so on presumably to infinity. So if a person rolls any one of the six combinations possible with two dice to get ten, then they will get a second roll.  At that point the chances of the ten score staying a ten score drop drastically:  chances of a ten with two dice are 6/36; chances of a ten with four dice are 6/36 X 1/36.  The 1/36 part is the probability of rolling a 0;0.
 Anyway I have mostly figured this out, but I don't know how to apply Pascal's triangle or any formula to it.  Calculus might help.  I found a pattern for probabilities in the second roll, yielding
10: 6/6^4
12: 12/6^4
14: 32/6^4
16; 50/6^4
18: 70/6^4
20: 91/6^4
This accounts for 801/1296 of the sample space, the remaining probability  being scores through 30, and scores resulting from further rolls.
 This solution seems inelegant, especially since it took me several hours to work it out, and it showed up on a test that allowed ten minutes for a solution.  Live and Learn.
 Thanks for your time!  RAndy

Answer
You probably have the answer, but here's my insight into the problem.
I believe I got it right this time if you meant you reroll if one of the two dice is a 10.

QUESTION: This question uses two six sided dice.
The dice are numbered 0,2,4,6,8, and 10.
Any roll 10 and over gets a second roll.

Note that if you rolled a 10 and then rolled another 10,
then the total is already 20, and that's too far to worry about.

The basic probabilities for the first roll:
0: 1/36
2: 2/36
4: 3/36
6: 4/36
8: 5/36
10: 6/36
12: 5/36
14: 4/36
16: 3/36
18: 2/36
20: 1/36.

Note that if a 20 is rolled the game is over for anything at 20 or more ends the game.

We need to look at when a 10, 12, 14, 16, 18, or 20 is rolled.
For 10, 12, 14, 16, or 18, there are 2 ways to have a ten included,
so the probability is 2/36.  To get a 20, the game is already over, but 2 10's were rolled and there is only 1 way to do that.

If a 20 was rolled, there was definitely a 10 rolled.  The chances of staying at 20 are the same as the chances of rolling two 0's, which ends the game.  The chance of rolling 2 0's is 1/36, so the odds of keeping that 20 is (1/36)(1/36) = 1/1296.
This updates the odds on a 20 to
20 1/36 - 1/36 + 1/1296 = 1/1296.

If the total of the first two dice is 18, there is at least one 10 that was rolled and you must roll again.  The only chance of keeping the 18 is to roll two 0's, which has chance 1/36.  So the new odds on an 18 is (2/36)(1/36) = 1/1296.  The chance that a 2 is gotten is 2/36, so the chance of a 20 is (2/36)(2/36) = 4/1,296.  Now the previous chances with these added on are
18: 2/36 - 2/36 + 1/1296 = 1/1296
20: 1/1296 + 4/1296 = 5/1296.

If the total was 16, there were 2 ways that had a 10 (6,10 or 10,6) out of 36 rolls.  The ways to keep a 16 are (2/36)(1/36) = 2/1296.
An 18 from here has chance (2/36)(2/36) = 4/1296.
A 20 from here has chance of (2/36)(3/46) = 6/1296.
Adjusting the previous odds, we now have
16: 4/36 - 2/36 + 2/1296 = 74/1296;
18: 1/1296 + 4/1296 = 5/1296; and
20: 5/1296 + 6/1296 = 11/1296.

If a 14 was rolled, there were 2 ways that had a 10 (4,10 or 10,4).
The chance of staying at 14 is (2/36)(1/36) = 2/1296
The chance of going a 16 is (2/36)(2/36) = 4/1296.
The chance of going an 18 is (2/36)(3/36) = 6/1296.
The chance of going a 20 is (2/36)(4/36) = 8/1296.
To update the old chances, we have
14: 4/36 - 2/36 + 2/1296 = 74/1296.
16: 74/1296 + 4/1296 = 78/1296;
18: 5/1296 + 6/1296 = 11/1296; and
20: 11/1296 + 8/1296 = 19/1296.

If a 12 was rolled, there were 2 ways that had a 10 (2,10 or 10,2).
The chance of staying at 12 is (2/36)(1/36) = 2/1296.
The chance of going to 14 is (2/36)(2/36) = 4/1296.
The chance of going to 16 is (2/36)(3/36) = 6/1296.
The chance of going to 20 is (2/36)(4/36) = 8/1296.
To update the old chances again, we have
12: 5/36 - 2/36 + 1/1296 = 109/1296;
14: 74/1296 + 2/1296 = 76/1296;
16: 78/1296 + 4/1296 = 82/1296;
18: 11/1296 + 6/1296 = 17/1296; and
20: 19/1296 + 8/1296 = 27/1296.

If a 10 was rolled, there are 2 ways that had a 10 (0,10 or 10,0).
The chance of staying at 10 is (2/36)(1/36) = 2/1296.
The chance of going to 12 is (2/36)(2/36) = 4/1296.
The chance of going to 14 is (2/36)(3/36) = 6/1296.
The chance of going to 16 is (2/36)(4/36) = 8/1296.
The chance of going to 18 is (2/36)(5/36) = 10/1296.
The chance of going to 20 is (2/36)(6/36) = 12/1296.
The last one is the only way there is of rolling a 10 on the second roll.  Out of 6/36 ways on the first roll, only 2/36 will let a second roll be made.  Out of these, only 1/6 have another 10, but it matters not since the total will now be at least 30 if you get a third 10.  The only way to get in on the third roll is if a 0,10 were rolled the first time, a 0,10 were rolled the second time, and a 0,0 were rolled the third time.  The odds of that happening are
(2/36)(2/36)(1/36) = 1/11664.

Anyway, the final table is
0: 1/36;
2: 2/36;
4: 3/36;
6: 4/36;
8: 5/36;
10: 6/36 - 2/36 + 1/1296 = 145/1296;
12: 109/1296 + 4/1296 = 113/1296;
14: 76/1296 + 6/1296 = 82/1296;
16: 82/1296 + 8/1296 = 90/1296;
18: 17/1206 + 10/1296 = 27/1296; and
20: 27/1296 + 12/1296 + 1/11664 = 39/1296 + 1/11664.

That seems a little odd that a 20 would be more likely that an 18,
but both include a 10 rolled, so the first time doesn't add in.
In that way, a 20 is easier to get to since its always 2 more than and 18 on the number rolled the second time.

Again, there is only one way to roll three times.
That is getting
1st roll) 10,0 or 0,10
2nd roll) 10,0 or 0,10
3rd roll) 10,0 or 0,10.


Now as I'm sitting here thinking, the way for this to not be right is if the previous roll and the current roll are not added together.
Now did you mean you only reroll if the total is 10?

Or maybe you already have the answer, so thanks for the wonderful problem.

That would put a whole new thought in on this problem.  But I believe they are, so here is what is known.