Everyone sufficiently competent in probability knows that in The Monty Hall problem as most commonly presented, switching doors wins you the car $\frac23$ of the time. I have come up with this variation:

  • There is a pool of cars and goats, initially containing one car and two goats. You always know the pool's exact contents.
  • The game is played in an infinite sequence of rounds. Each round begins by sampling three objects uniformly at random without replacement from the pool to place behind the doors, again uniformly at random; if the sample does not contain at least two goats the pool is resampled. (Thus there might be three goats and no car, but never one goat and two cars.)
  • The standard Monty Hall game is then played: you pick a door, another door is opened revealing a goat, you decide whether to stay or switch.
  • After you win a car or goat one of the opposite object is added to the pool – a car if you won a goat and a goat if you won a car. The objects behind the doors are returned to the pool and the next round begins. This is similar to a "pity timer" mechanism found in certain loot boxes, hence this question's title.

If you play optimally every round:

  • What is the limiting proportion of your wins that are cars?
  • What is the limiting proportion of cars in the pool?

1 Answer 1


Let w and p ("win" and "pool") the two limiting proportions. Because of the pity timer we have

$p + w = 1$.

If three goats are drawn we are guaranteed to lose. Otherwise our chances are 2:1 in favour of winning.

The odds for three goats g3 vs two goats one car g2 are (in the limit replacement or not is irrelevant)

$g3 : g2 = w^3 : 3w^2p = w : 3p$.


$p:w = (g3 + \frac {g2} 3):\frac {2\ g2} 3 = (w+p):2p$


$p = w = \frac 1 2$.


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