Risk & Reward

Three Rolls to Stop

Roll a die up to three times, taking a face value or rolling on, but forfeiting whatever you pass up. What is the game worth and when should you stop?

solvedmedium1 min

You may roll a fair die up to three times. After the first or second roll, if it shows xx you may either take xx dollars or roll again, but choosing to roll again forfeits the xx you just saw. On the third roll you simply take whatever it shows. What is the game worth, and what is your strategy?

Reveal solutionHide solution

#Solve from the last roll

On the third roll you take the face value, worth an expected 3.53.5.

#The second roll

On the second roll you keep a face only when it beats the value of rolling on, which is 3.53.5, so you stop at 44, 55, or 66 and continue otherwise. That makes the second roll worth

E[max⁡(x,3.5)]=12⋅5+12⋅3.5=174.(1)\E\big[\max(x, 3.5)\big] = \tfrac{1}{2}\cdot 5 + \tfrac{1}{2}\cdot 3.5 = \frac{17}{4}. \tag{1}

#The first roll

On the first roll you keep a face only when it beats 174=4.25\tfrac{17}{4} = 4.25, so you stop at 55 or 66. The whole game is then worth

13⋅112+23⋅174=116+176=143.(2)\tfrac{1}{3}\cdot\tfrac{11}{2} + \tfrac{2}{3}\cdot\tfrac{17}{4} = \frac{11}{6} + \frac{17}{6} = \frac{14}{3}. \tag{2}
roll 1123456
roll 2123456
roll 3123456
shaded = keep and stop
Solve from the last roll back. The third roll is worth 3.5, so on the second you keep any face above that, stopping at 4. That makes the second roll worth 17/4, so on the first you keep a 5 or 6 and the game is worth 14/3.

#Read it off

The game is worth 143≈\tfrac{14}{3} \approx $4.67. Stop on the first roll at a 55 or 66, on the second at a 44 or more, and take whatever the third roll gives. Each stage's threshold is just the value of the rolls still to come.