Dice Duel With a Pay-to-Boost Option

Game Theory · Medium · Free problem

You and an opponent each roll a fair six-sided die independently. If your number is strictly greater than your opponent's, you win $\

$; on a tie or a loss you get nothing.

Part 1: What is your probability of winning, with no options?

Part 2: You (and only you) may pay $\$0.25$ to add

$ to your roll. You decide whether to pay after seeing your own roll but before learning your opponent's. For which of your rolls should you pay, and what is your net expected value playing optimally?

Hints

  1. First nail the no-option win probability by counting the $a > b$ pairs out of $36$.
  2. The $+2$ boost normally lets you beat two more opponent faces, a /6 = 1/3$ gain in win probability -- compare that to the $\$0.25$ cost.
  3. The boost is only partly useful at the top: on a $6$ you can beat just one more face, so the gain is
    /6 < 1/4$. Pay on
    $-$5$, decline on $6$.

Worked Solution

How to Think About It: Without the option this is the standard "strictly greater" dice comparison. With the option, paying $\$0.25$ buys you a jump of $+2$, which raises your win probability by however many extra opponent values you now beat -- but only up to the cap, since you can never beat more than all six.

Part 1: Of the $36$ equally likely pairs, you win on

5$ (the count of $a > b$). $P(\text{win}) = \frac{15}{36} = \frac{5}{12} \approx 0.417$.

Part 2 (Quick Estimate): Adding

$ normally lets you beat $ additional opponent faces, raising win probability by /6 = 1/3 \approx 0.333$ -- worth far more than the $\$0.25$ cost. The exception is when you are near the top, where the boost is partly wasted.

Formal: Let your roll be $r$. - For $r = 1,\dots,5$: boosting to $r+2$ beats exactly $ more faces, so $\Delta P(\text{win}) = 2/6 = 1/3$, and net gain $= 1/3 - 1/4 = 1/12 > 0$. Pay. - For $r = 6$: you already beat $5$ faces; boosting to $8$ beats only

$ more (you cannot beat your own value), so $\Delta P = 1/6$, net $= 1/6 - 1/4 = -1/12 < 0$. Do not pay.

Net EV: baseline $5/12$ plus the value of optimal boosting on rolls

$-$5$ (each occurs with prob
/6$): $\frac{5}{12} + \frac{5}{6}\left(\frac13 - \frac14\right) = \frac{5}{12} + \frac{5}{72} = \frac{35}{72} \approx 0.486.$

Answer: Part 1: $5/12$. Part 2: pay on rolls

$-$5$, not on $6$; net EV $= 35/72 \approx 0.486$.

Intuition

The lesson is marginal-value-of-an-edge: the boost is worth buying exactly when the extra win probability it unlocks exceeds its price, and that extra probability shrinks once you are already near the top of the distribution. This is the same reasoning a trader uses when deciding whether to pay for an incremental improvement (tighter data, faster execution) -- it pays off in the meat of the distribution but has diminishing returns at the tails. Note the decision is made conditional on your own roll, which is what makes the policy state-dependent rather than a blanket yes/no.

Open the full interactive solver →