This question is from DEGROOT's PROBABILITY and STATISTICS.
Rolling Dice Repeatedly:
Suppose that two dice are to be rolled repeatedly and the sum T of the two numbers is to be observed for each roll. We shall determine the probability p that the value T=7 will be observed before the value T=8 is observed.
The desired probability p could be calculated directly as follows: We could assume that the sample space S contains all sequences of outcomes that terminate as soon as either the sum T=7 or the sum T=8 is obtained. Then we could find the sum of the probabilities of all the sequences that terminate when the value T=7 is obtained. However,there is a simpler approach in this example. We can consider the simple experiment in which two dice are rolled. If we repeat the experiment until either the sum T=7 or the sum T=8 is obtained, the effect is to restrict the outcome of the experiment to one of these two values. Hence,the problem can be restated as follows: Given that the outcome of the experiment is either T=7 or T=8, determine the probability p that the outcome is actually T=7. If we let A be the event that T=7 and let B be the event that the value of T is either 7 or 8, then A∩B=A and p=Pr(A|B)=Pr(A∩B)/Pr(B)=Pr(A)/Pr(B). But Pr(A)=6/36 and Pr(B)=(6/36)+(5/36)=11/36. Hence, p=6/11.
Now,my doubts are
1.) Why the author says "We could assume that the sample space S contains all sequences of outcomes that terminate as soon as either the sum T=7 or the sum T=8 is obtained. Then we could find the sum of the probabilities of all the sequences that terminate when the value T=7 is obtained." ?
2.)How can we go from lengthy sequences of outcomes that terminate as soon as either the sum T=7 or the sum T=8 is obtained to just the outcome of the experiment for which either T=7 or T=8 ?
Please just let me know why the author stops as soon as sums 7 or 8 are obtained.(I know that stopping before getting 7 or 8 does not give any useful info for this question but why stop at all? Why can't we extend the sequences after 7 or 8 too?)
If u go further u will be answering the question of the nature like: "probability of getting two 7s before u get one 8" OR "probability of getting three 7s before u get one 8" OR "probability of getting two 7s before you get two 8s"... so on..
you can make this question arbitrarily difficult by making it something like: "what is the probability of getting exactly n 7s before getting m-th 8"
"You don't have to believe in God, but you should believe in The Book." -Paul Erdős
Thans a lot ,VASUDHA, you are great as do not get frustrated!!
Actually,I asked the same question at math.se and stats.se.The stats.se answer is really great. But, Tyler starts without explaining why assume the seq. terminates as soon as 7 or 8 are obtained.
I think my doubt is best understood by Whuber on stats.se ,but he didn't answer at all! Tyler has helped a lot, but this is the last doubt with me for this question.
Let me elaborate my doubt. Suppose we have a sequence such that it terminates at 7: the seq. is (just for example) 234567 and if now we extend it as 2345677 or 23456778 or 234567952431523 or other ways, doesn't the probability get affected? And because these sequences are the required ones, why should we omit them?
Do you know some limit argument for it? As infinity is involved, I have to be very careful!!
And if you don't get my doubt, please let me know. I will ask my question with other reasoning.
I hope you are no bored!! But if you understand me, and answer me, I will be truly obliged.
I think that's because of independence of outcomes in successive throws.
Let's just add the probabilities of all sequences where 7 appears before 8.
What are the possibilities? 1) you get 7 followed by any sequence, 2) you get something other than 7 followed by 7 followed by any sequence, 3) you get something other than 7 twice followed by 7 followed by any sequence, and so on..
Probability of 1) is P(7)*P(anything|7 in first round)*P(anything|results of first 2 rounds)*...
Probability of 2) is [1-P(7)] *P(7)*1*1*1...
So basically you will end up adding probabilities of sequences that terminate at 7