You are cordially invited to match wits with some of the best minds in IBM.

Seems some of us can’t see a problem without wanting to take a crack at solving it. Does that sound like you? Good. Forge ahead and ponder this month’s problem.

**Ponder This Challenge: **

Inspired by the viral 2048 game, this month’s question is a simpler one-dimensional version of it. Assume that random independent numbers, either 2 or 4 with a 50% chance each, come in from the left side of a bar with N slots. The numbers are always squeezed to the left and every time two equal numbers are the same – they are replaced by their sum. The game ends when all the N slots are occupied – and therefore there is no room for a new number.

What is the expected maximum number in the array when the game ends?

When N=1, the game ends after one round, leaving either 2 or 4 with the same probability and hence the average is 3.

When N=2, the game can reach 8 (for example, when the input is 2, 2, 4, 2, and the result is 8, 2), but can also be stuck with a 4 (such as when the input is 2, 2, 2, and the result is 4, 2).

Computing the average we get 5.5, or 11/2.

What is the average when N=5? Express this value as a quotient of two coprime integers.

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IBM Research will post the names of those who submit a correct, original solution on their website! If you don’t want your name posted then please include such a statement in your submission. Send your submission to the webmaster.

*If you have any problems you think we might enjoy, please send them in. All replies should be sent to:* webmster@us.ibm.com

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