Shevek (![[info]](http://p-stat.livejournal.com/img/userinfo.gif){kind=small} shevek) wrote,@ 2007-02-04 21:14:00 |
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Deal or No Deal - a mathematicians's perspective.
For a long time, I thought the optimal strategy for Deal or No Deal was always to play the game to the end. The (statistical) expectation is easy to compute, and the banker rarely if ever offers more than about 60% of the expectation. So the expectation for 'no deal' is always higher than that for 'deal'. At the start of the game, the expectation is about 40,000. But this doesn't seem right, so I was looking for the theory which explains actual behaviour and the design of the game.
The fallacy is in the assumption that the player's valuation is linear in money. It isn't. Given the choice between 40,000 guaranteed, or a 50/50 choice between 100,000 and 0, almost everyone would pick the 40,000 guaranteed because the valuation tails off above a certain point.
And that is why the game works.
If it was a repeated game, not a single-play, then it wouldn't work this way, because the averaging strategy would work. But the answer above doesn't quite fit into classical game theory. Can anyone provide a better framework?
For a long time, I thought the optimal strategy for Deal or No Deal was always to play the game to the end. The (statistical) expectation is easy to compute, and the banker rarely if ever offers more than about 60% of the expectation. So the expectation for 'no deal' is always higher than that for 'deal'. At the start of the game, the expectation is about 40,000. But this doesn't seem right, so I was looking for the theory which explains actual behaviour and the design of the game.
The fallacy is in the assumption that the player's valuation is linear in money. It isn't. Given the choice between 40,000 guaranteed, or a 50/50 choice between 100,000 and 0, almost everyone would pick the 40,000 guaranteed because the valuation tails off above a certain point.
And that is why the game works.
If it was a repeated game, not a single-play, then it wouldn't work this way, because the averaging strategy would work. But the answer above doesn't quite fit into classical game theory. Can anyone provide a better framework?
