How to calculate expected value of a game that doesn't need to be played to completion e.g. Blackjack and stopping when aces are gone.

Question

Consider a simple example where I have a prize behind one of two doors.

The prize is $1.90.

I will let you choose a door for $1.00.

If you choose the winning door first, you can walk away. If you choose the losing door first, you can pay me another dollar to open the other door.

So you'll either win on the first door and walk away with \$0.90 profit, or you'll lose on the first door, pay another dollar to open the second, and walk away with a -$0.10 loss.

1/2 * \$0.90 + 1/2 * -\$0.10 = $0.40 expected value.

How can this be extrapolated to a larger number of doors and a wider variety of prizes?

For example, 10 doors with 1 prize of \$7 and 2 prizes of \$1 and the rest \$0.

Edit with image for clarification: 3 Doors with values [3, 1, 0]

In this particular case I could just stop on "3" because it's obvious that's I can't win more. But how could I somehow programmatically or mathematically determine when the best place to stop is and what the expected value of the game is when I stop there?

Answer 1

First, you need to specify the conditions (whether deterministic or probabilistic) that a person will stop on. IE, they'll stop as soon as they profit, or they'll stop with probability (Profit/\$1).

Then, this just becomes a probability tree. In your given example, you had two branches: 50% chance they took the door with the prize and 50% they didn't. If you expand to your 10 door version, you start with 3 branches: .1 probability of \$7, .2 probability of \$1 and .7 probability of \$0.

After each branch, depending on your stop conditions, you'll either terminate at that point, or have further branches, which will be different based on which branch you're on.

Let's say that your stop condition is that you have to have profited (deterministic) to make things simpler. Then your \$7 branch and your \$1 branch are both immediate stops, and only the \$0 branch expands. The next step branches as follows: (1/9) chance of \$7, (2/9) chance of \$1, (6/9) chance of \$0.

You can repeat until your tree is complete with every end node being a point at which your player would stop.

Then, each end node has a probability equal to the product of every branch probability along the way.

For example, the end node of \$0 followed by \$7 would have a probability of 7/10 * 1/9 = 7/90.

Once you have a probability for each end node, you multiply by that node's value to calculate your expected value in the usual way.