Game A
Roll a fair dice repeatedly until we land on a 1.
Let A denote the expected sum of all the values (including the final 1).
Game B
Roll a fair dice repeatedly until we land on a 6.
Let B denote the expected sum of all the values (including the final 6).
Which is larger, A or B?
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Nice solution (+1)
There's another solution that uses Doob's optional stopping theorem. It's a fairly advanced tool, but it yields the answer almost immediately.
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I hadn't heard of that theorem before, so thanks for mentioning it. From what I've just read, as this is a martingale the expected sum will be the same regardless of the designated "stopping time", or in this case stopping number, so we can conclude A and B are equal without making any calculations.
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We do have to be a little careful in defining the martingale, but if you let S n denote the sum after n rolls of the die, then M n = S n − 2 7 n is a martingale with bounded increments. If you then let the N th roll be the first time any chosen face shows up, then the optional stopping theorem tells you E [ M N ] = E [ M 0 ] = 0 ⟹ E [ S N ] = 2 7 E [ N ] = 2 7 ( 6 ) = 2 1
There is an easier way to do the calculation using law of iterated expectation .
If we let X be the expected value of the sum before landing on a 1, we have X = 6 X + 2 + 6 X + 3 + 6 X + 4 + 6 X + 5 + 6 X + 6 . Then, A = X + 1 .
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In general, suppose we keep rolling until we get the number N . Then the expected value of each roll up until we roll N is A N = 5 2 1 − N . As the die is fair, the probability that we roll N on any given roll is 6 1 , and the probability we don't is 6 5 . If we take m ≥ 1 rolls to get N then the final sum will be N + ( m − 1 ) A N . The expected sum E N is then
6 1 N + 6 1 ∗ 6 5 ( N + A N ) + 6 1 ( 6 5 ) 2 ( N + 2 A N ) + . . . + 6 1 ( 6 5 ) m − 1 ( N + ( m − 1 ) A N ) + . . . . =
6 N k = 0 ∑ ∞ ( 6 5 ) k + 6 A N k = 1 ∑ ∞ k ( 6 5 ) k = 6 N × 6 + 6 A N × ( 1 − 6 5 ) 2 6 5 = N + 5 A N = N + ( 2 1 − N ) = 2 1 .
So regardless of which number we roll until, the expected (inclusive) final sum is invariant, and thus the answer is A and B are equal .
(Note: A shortcut could have been to note that the expected number of rolls to get any given number is 6 , so the expected sum would be N + 5 A N = 2 1 as found above, but I wasn't certain that this was airtight so I went with the long version to be sure.)