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This is Problem 1.25 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.

You are handed two envelopes, and you know that each contains a positive integer dollar amount and that the two amounts are different. The values of these two amounts are modeled as constants that are unknown. Without knowing what the amounts are, you select at random one of the two envelopes, and after looking at the amount inside, you may switch envelopes if you wish. A friend claims that the following strategy will increase above 1/2 your probability of ending up with the envelope with the larger amount:

Toss a coin repeatedly. Let X be equal to 1/2 plus the number of tosses required to obtain heads for the first time, and switch if the amount in the envelope you selected is less than the value of X . Is your friend correct?

The answer is listed in another question.

My question is about one of the steps in the solution to the question.

answer first part

answer second part

I do not understand how they made this step

${1 \over 2} (P(A) + P(B) + P(C)) + {1 \over 2} P(B)$

${1 \over 2} + {1 \over 2} P(B)$

Woah. I just got it as I was typing. They know that A B and C are the only options, so they can sum them to 1 due to the normalization axiom.

If that is right, feel free to delete this.

Sebastian
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  • Your conclusion seems right to me. – David K Oct 12 '18 at 16:12
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    Some times all it takes to understand the answer to a question is to slow down and understand what the question is. Trying to explain the question to someone else helps you to slow down long enough to think about it yourself too. When writing papers or reports and the like, I find I can organize my thoughts best after trying to explain the content of what I want to write about to a friend or classmate. – JMoravitz Oct 12 '18 at 16:15
  • You are welcome to delete your Question yourself, since it has no upvoted or Accepted Answer. On the other hand you might consider a self-Answer if you think the solution will be of interest to future Readers. – hardmath Oct 12 '18 at 20:41

1 Answers1

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The answer to that particular part of the question is that A, B and C collectively represent all possible disjoint outcomes.

A is defined as the event where X (${1\over2}$ + the number of coin tosses) is less than the envelope with a smaller event.

B is where X is greater than the envelope with a smaller amount and less than the envelope with a larger amount.

And C is where X is greater than the envelope with a larger amount.

We have the normalization axiom, which says P($\Omega$) = 1, and we know A, B and C collectively exhaust all possible outcomes, so we can convert (P(A) + P(B) + P(C) to 1.

Sebastian
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