Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**UrgentHelp****Guest**

So we have three functions: Choosing 'a' gives payoff: 3-q-2p.

Choosing 'b' gives: 2

Choosing 'c' gives: 4-2q-4p.

p and q are both less than 0 and do not necessarily add to one (because when deriving the above 3 payoffs, I expanded the (1-p-q))

'Derive the conditions under which 'a' is chosen'. Given the work we have done, what this means is, for what 'p' and 'q' is the payoff from 'a' above, i.e. 3-q-2p larger than 'b' and 'c'.

So for this I did: 3-q-2p>2 therefore 1>q+2p and

3-q-2p>4-2q-4p therefore q+2p>1 THEREFORE contradition. This CAN be a result, this isn't where I am stuck I am just showing what I am up to. (What that result means is that the player is risk averse), I.e. doesn't like risk, it comes from the utility functions at the start of the question and is a valid result. My question comes from, finding the conditions for which 'b' is chosen:

So; 2>3-q-2p is needed (i.e. payoff from 'b' is larger than payoff from 'a'), therefore q+2p>1

and we also need payoff from 'b' larger than payoff from 'c', i.e. 2>4-2q-4p, therefore 2q+4p>2. Now how can we manipulate/transform/ get conditions from these two inequalities which must hold: i.e. q+2p>1 and 2p+4p>2??

Please help me!!

**scientia****Member**- Registered: 2009-11-13
- Posts: 224

UrgentHelp wrote:

p and q are both less than 0

Do you mean **more** than 0?

UrgentHelp wrote:

The two inequalities are the same. Any permitted values of Now how can we manipulate/transform/ get conditions from these two inequalities which must hold: i.e. q+2p>1 and 2p+4p>2??

Offline

Pages: **1**