I roll a six-sided die 200 times. What is the expected number of occasions on which
I will see two consecutive rolls that differ by 1?
For example, a sequence 1334546 would count as 3 such consecutive pairs.
The solution said that the probability of getting a single pair of this form is $\frac{10}{36} = \frac{5}{18}.$ I dont understand how this is obtained.
Rolling a D6 and getting consecutive rolls that differ by 1
probability
Related Question
- [Math] Expected value of rolling a $5$ followed by a $6$ different than rolling two consecutive sixes
- Probability Theory – Expected Rolls to Get 3 Consecutive Numbers
- Probability – Expected Number of Dice Rolls Before Rolling ‘1,2,3,4,5,6’
- Probability Theory – Expected Number of Rolls Until All Consecutive Differences Seen
- The probability that all faces have appeared in some order in some six consecutive rolls
- EV for rolling a consecutive 56, and why
Best Answer
One of the ten is '1 followed by 2'. Find the others and count them.