dimanche 19 janvier 2014

Nash Equilbrium Always at Saddle Points?

I have a question....



I was reading through this really interesting article on the mathematics of bluffing - basically it uses a tree diagram to work out the best strategies for bluffing and calling in a very simple gambling game. As this is a zero sum game, I would have thought that the Nash equilibrium is always at the saddle point of the graph - and this works for the example given in the article. However I then tried it with different numbers (with an initial entry of 1 pound as before, but then a bet of 2 pounds) to get the following 2 equations:



Expected Value for Player 2 : 2/6 C (2B-1)

Expected Value for Player 1: 2/6 C (1-2B)



Player 1 is the one that has the choice of bet and Player 2 is the one that has the choice of call.



Which if you plot using Wolfram Alpha you get a saddle point of (1/2, 0)



which would imply that the optimal strategy is to bet 1/2 the time and to call 0 of the time. But clearly Player 1 would choose to bet 0 of the time to maximise his expected value. So what's gone wrong? Why is the saddle point not the Nash equilibrium?



Many thanks :)





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