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BonJoviJones
11-21-2004, 02:59 PM
Roll a die. If you roll a 6, roll again. Repeat until you don't roll a six. Your score is the sum of all rolls.

What is the average score in this game?

kyro
11-21-2004, 03:21 PM
(1/6)(1+2+3+4+5) + (1/6)(6) + (1/6)(1/6)(1+2+3+4+5)+ (1/6)(1/6)(6) + (1/6)(1/6)(1/6)(1+2+3+4+5) + ...

15/6 + 1 + 15/36 + 1/6 + 15/216

15/6(1 + 1/6 + 1/36 + ... 1/6^n) + 1(1 + 1/6 + 1/36 + ... + 1/6^n)

(15/6)(6/5) + (1)(6/5)

3 + 1.2

4.2

I think, someone correct me if I'm wrong.

BruceZ
11-21-2004, 07:22 PM
[ QUOTE ]
(1/6)(1+2+3+4+5) + (1/6)(6) + (1/6)(1/6)(1+2+3+4+5)+ (1/6)(1/6)(6) + (1/6)(1/6)(1/6)(1+2+3+4+5) + ...

15/6 + 1 + 15/36 + 1/6 + 15/216

15/6(1 + 1/6 + 1/36 + ... 1/6^n) + 1(1 + 1/6 + 1/36 + ... + 1/6^n)

(15/6)(6/5) + (1)(6/5)

3 + 1.2

4.2

I think, someone correct me if I'm wrong.

[/ QUOTE ]

Right. Or you could just note that since the probability of not rolling a 6 is 5/6, it will take an average of 6/5 rolls to not roll one, and the average score per roll is 3.5. Then 3.5 * 6/5 = 4.2.

gaming_mouse
11-21-2004, 07:34 PM
[ QUOTE ]

Right. Or you could just note that since the probability of not rolling a 6 is 5/6, it will take an average of 6/5 rolls to not roll one, and the average score per roll is 3.5. Then 3.5 * 6/5 = 4.2.

[/ QUOTE ]

Clever.

kyro
11-21-2004, 10:06 PM
i have no idea what you just did. i'm going to spend my night figuring that out.