# probabilty

• Oct 13th 2012, 12:31 AM
nitin1
probabilty
When a dice is rolled the outcome of the face is summed up each time, and rolling is stopped when the sum becomes greater than 100. Which of the following have more probability to become sum?
a) 103
b) 102
c) 100
d) all have equal probability
e) 101

thanks if you can help me in this ;)
• Oct 13th 2012, 02:41 AM
chiro
Re: probabilty
Hey nitin1.

One approach I would take is to look at the conditional distribution of starting at 94-99 and then using a markov chain to do a transition matrix for all the states 94 to 105 and then use this to get the long term probabilities for each state.

Have you encountered Markov Chains?
• Oct 17th 2012, 04:39 PM
awkward
Re: probabilty
Quote:

Originally Posted by nitin1
When a dice is rolled the outcome of the face is summed up each time, and rolling is stopped when the sum becomes greater than 100. Which of the following have more probability to become sum?
a) 103
b) 102
c) 100
d) all have equal probability
e) 101

thanks if you can help me in this ;)

Let's suppose we write down the sum of the rolls after each roll, stopping when then sum is greater than 100, and let p(n) = the probability that the integer n appears in the list. We can show p(101) > p(102) > p(103) > p(104) > p(105) > p(106) without actually computing these probabilities (thank goodness, because I can't think of a way of computing them without a lot of computation).

To prove the claim, suppose that the final sum is 101. The previous sum must have been one of 95, 96, 97, 98, 99, or 100. If the previous sum was 95, the last roll must have been 6; if 96, it must have been 5, etc. In each case, the probability of rolling the required number is 1/6. So
$\displaystyle p(101) = \sum_{n=95}^{100} \frac{1}{6} \; p(n)$.

Now suppose the final sum was 102. The previous roll must have been one of 96, 97, 98, 99, or 100. 101 is not a possibility because we would have stopped earlier if that were the case. Again, the required roll in each case has probability 1/6. So
$\displaystyle p(102) = \sum_{n=96}^{100} \frac{1}{6} \; p(n)$.

Comparing the previous two equations, we see
$\displaystyle p(101) = p(102) + \frac{1}{6} \; p(95)$
so p(101) > p(102). The other inequalities, e.g. p(102) > p(103), are proved similarly.