Download Midterm Exercise 2: If you want to use iterative elimination of strictly

Midterm Exercise 2: If you want to use iterative elimination of strictly dominated
strategies in this game, you can use the following:
First we show that 200 is strictly dominated by the mixed strategy that puts
• probability 0 on 200
• probability 0.99 on 199
• probability (1-0.99)*0.99 on 198
• probability (1-(1-0.99)*0.99) on 197
• probaility [1-(1-(1-0.99)*0.99)] on 196
• and so on...
In short, each action (apart from 100) is 99 times as likely as all smaller actions together.
For example, 199 is 99 times as likely as all the other actions together or 198 is 99 more
likely than all the action 100,101,102. . . 197 together.
This mixed strategy does better than the pure strategy 200 if the other player says
100: With some very small probability (in fact 0.01100 ), you say 100 as well and then
you save the 5$ penalty. With the counter probability (that is 1 − 0.01100 ) our mixed
strategy leads to the same outcome for you as if you had said 200: You get 100$ and
have to pay 5$ penalty. But that means that in expectation our mixed strategy does
strictly better than saying 200.
This mixed strategy also does better than saying 200 if the other player says 100:
With probability 0.99 you say 199 and then your payoff is 204 (instead of 200 if you
said 200). If you say a number k with 195 ≤ k ≤ 198, you still do at least as good
as when saying 200. Only when you say a number k < 195 you do worse than saying
200. However, this happens only with the small probability 0.015 . Your maximal loss
(compared to saying 200) is when you say 100 which would give you a payoff of 105
which is 95 lower than when saying 200. Hence, your expected losses from saying a
number less than 195 are smaller than 95 ∗ 0.015 . Your expected gain from saying 199
with probability 0.99 is 4 ∗ 0.99 which is much higher than 95 ∗ 0.015 . Hence, the mixed
strategy does better than saying 200 if the othe player says 200.
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Does our mixed strategy also do better if the other player says anything else (not
100 or 200)? Let’s say he says n ∈ {101, 102, . . . 199}. With probability 1 − 0.01199−n
you say a number that is strictly greater than n. In this case, the mixed strategy gives
you the same payoff as the action 200. With both you get n − 5.
With probability 0.99 ∗ (0.01199−n ) you also say n. In this case, you save the 5$
penalty and therefore your payoff is 5$ higher than if you had said 200. The expected
gain (compared to playing 200) from this case is 5 ∗ 0.99 ∗ (0.01199−n ).
With probability 0.01200−n you say a number k strictly below n. In this case, you
get k + 5. This could be worse than what your payoff if you had said 200 (which
would have given you n − 5). That is, the mixed strategy does worse than saying 200
if k < n − 10. However, this happens with such small probability that the gain that
occurs if you say k = n (or k ∈ (n − 10, n)) more than outweighs this possible loss.
Remember that saying k = n is 100 times as likely as saying a number k < n (and
k = n is 10010 more likely than choosing a number below n − 10). The biggest loss you
have is when you say k = 100. In this case, you do n − 100 − 10 < 90 worse than if you
said 200. Hence, your expected loss from saying a number less than n is strictly less
than 90 ∗ prob(k < n − 10) = 90 ∗ 0.01200−n+10 .
Because 5 ∗ 0.99 ∗ (0.01199−n ) > 90 ∗ 0.01200−n+10 the gain from playing the mixed
strategy (compared to playing 200) are much higher than the expected losses for any
n ∈ {101, 102, . . . , 199}. Hence, the mixed strategy does strictly better than 200 no
matter which number hte other player chooses. This is equivalent to saying the mixed
strategy dominates the action 200.
Hence, we can eliminate 200 in the first round of elimination. In a second round,
we can then eliminate 199 which is strictly dominated by the mixed strategy that puts
• probability 0 on 200 and 199
• probability 0.99 on 198
• probaility 0.99*(1-0.99) on 197
• probability 0.99*[1-0.99*(1-0.99)] on 196
• and so on.
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This can be shown with the same arguments as above. One can proceed this way
to iteratively eliminate all actions until 100.
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