How can you quantified the cost of an improvement in success probability over time?

Question

The game Trivia Crack asks a question and gives four possible answers. Assume that I do not know the answer, so my attempt would be a pure guess at a 25% chance of success.

I have two optional tools at my disposal:

• A Bomb removes two wrong answers, leaving only two possible answers This improves my odds to 50%.
• A Double Guess lets me guess twice, and my first wrong guess is not counted. Assuming my first guess is wrong, this would reduce it to three possible answers, improving my odds to 33%.

The Bomb costs 6 coins. The Double Guess only costs 3 coins.

Over time, which is the better strategy, assuming a finite supply of coins? Is it more cost effective to improve from 25% to 50% odds at 6 coins, or from 25% to 33% odds at 3 coins?

Answer 1

The double guess:

\begin{align} \mathbb{P}(right \; answer|Bomb) &= 0.5 \\ \mathbb{P}(right \; answer|double \; guess) &= (1/4)*1 + (3/4)*(1/3) = 0.25 + 0.25 = 0.5 \end{align}

So you get the same odds for half as much money. The second derivation follows (denoting $R$ as 'right answer', $FS$ as 'first shot', $SS$ as 'second shot') because the first and the second shot are independent if we don't have any knowledge. The result is then that

\begin{align} \mathbb{P}(R|DG) &= \mathbb{P}(R|FS) + \mathbb{P}(R|SS \text{ & not } FS) \\ &= \mathbb{P}(R|FS) + \mathbb{P}(R|SS)\mathbb{P}(R|\text{ not } FS) \\ &= (1/4)*1 + (3/4)*(1/3) \end{align}