## More dice rolling

Posted by Chris on September 30, 2011 – 5:56 pm

1. What is the average sum of the rolls when you roll two dice until the sum is 6?

e.g. (1,2),(3,5),(2,4) gives a sum of 11.

2. What is the average sum of the rolls when you roll two dice until at least one is a 6?

e.g. (1,2),(3,5),(2,6) gives a sum of 11.

3. What is the average sum of the rolls when you roll two dice until exactly one is a 6?

e.g. (1,2),(3,5),(6,6),(1,6) gives a sum of 23.

4. As I’ve not seen it before: What is the average number of rolls of a die to have got each number (at least) once?Сайт знакомств

October 1st, 2011 at 5:12 pm

#1 44.4

#2 13.6

#3 16.2

#4 14.7

October 2nd, 2011 at 6:41 am

I’ll confirm that ragknot has the right numbers. Now can someone solve the problems using logic/probability?

October 3rd, 2011 at 2:45 am

1: 5/36 rolls have a total of 6. So the expected number of rolls including the six total is 36/5. Excluding the six gives 31/5. The average non six total roll is 222/31. So the answer is (222/31)*(31/5)=44.4

2: 11/36 rolls include at least one six. So the expected number of rolls including the the last is 36/11. Excluding the last is 25/11. The average roll is 150/25. So the answer is (25/11)*(150/25)=13.6

3: 10/36 rolls include exactly one six. So the expected number of rolls including the last is 36/10. Excluding the last is 26/10. The average roll is 162/26. So the answer is (26/10)*(162/26)=16.2

4: Suppose there are 7 states: x_6…x_0. The i in x_i stands for the amount of numbers you still have to throw.

x_6 -> x_5 takes us 1 throw.

x_5 -> x_4 takes us 6/5 throw on average.

this goes on and gives us, 6/4, 2, 3, 6.

This leads to a total of 14.7

October 3rd, 2011 at 4:32 am

Hi Jan. You got it.

For 2, the 13.6 should be 13.636363…

A bit more info: If you draw a table showing the various rolls, you’ll see that for the 25 rolls that don’t involve a 6, that the total sum of the rolls is 150. The sum of the rolls that include exactly one 6 is 2*45 = 90, and the total sum of the rolls is 252 (=150+90+12). Jan’s 222 is 252-5*6, i.e. the total sum less that due to rolling a sum of 6.

My prepared answer for problem 4 is: The probability of getting any one of the six numbers on the first roll is 6/6. The probability of getting one of the remaining five numbers on the second roll is 5/6, for the third number it’s 4/6, for the fourth number it’s 3/6, for the fifth number it’s 2/6 and for the sixth number it’s 1/6.

For each case the number of rolls is 1/p, so altogether that gives:

6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7