Just a weird thought

[Ncoe]

Say there's a game where u get to pick one of four options, A,B,C or D where one of them gets you a point.

For each time u play you get exact probabilities for the separate options being the correct one.
Such as, A=50%, B=35%, C=10% and D=5%.
Obviously it's an easy choice to go with the option that has the highest probability for being correct.

But say you instead use some sord of randomizer to choose your pick, but with the corresponding probabilities.

So that in this example, you would on average pick option A 50% of the time, B 35% of the time etc.

Would this be equal resultwise to always picking the one with the highest probability?

[sixhigh]

<font color="#FF4444">[*] </font> Let's make this easy. Two options, A and B. A wins with probability p and B with 1-p. wolog p&gt;0.5. If you always choose A you win with probability p.

if you choose A with probaility p and B with probability (1-p) we have four cases:
1. (A, A, p*p) you choose A and A wins: probability p*p
2. (A, B, p*(1-p))
3. (B, A, (1-p)*p)
4. (B, B, (1-p)(1-p))

You win in case 1 and 4 - which is gonna happen with the probability p*p + (1-p)(1-p) = 1-2p+2p^2.

And 1-2p+2p^2&lt;p for every p in (0.5, 1).

This also holds true in a general case with more options.