When I have watched Deal or No Deal (I try not to make a habit of it) I always do little sums in my head to work out if the banker is offering a good deal. Where odds drop below "evens" it's easy to see it's a bad deal, but what would be the correct mathematical way to decide if you're getting a good deal?
[Math] Optimal Strategy for Deal or No Deal
probabilityrecreational-mathematics
Related Solutions
We must ignore the "cannot lose on first click" rule as it severely complicates things.
In this answer, I will be using a notation similar to chess's FEN (Forsyth-Edwards Notation) to describe minesweeper boards. m is a mine and empty spaces are denoted by numbers. We start at the top of the board and move from left to right, returning to the left at the end of each row. To describe a specific square, the columns are numbered from a to h, left to right, and the rows are numbered from 8 to 1, top to bottom.
On a minesweeper board, all mines are adjacent to numbered squares that say how many mines are next to them (including diagonally). If there is ever a numbered square surrounded only by mines and other numbered squares, new squares will stop being revealed at that square. Therefore, the question is actually:
How many 9 × 9 minesweeper boards with 10 mines exist such that every blank square adjacent to a mine touches a square that is neither a mine nor adjacent to one?
I like to approach problems like these by placing mines down one by one. There are 81 squares to place the first mine. If we place it in a corner, say a1, then the three diagonal squares adjacent to the corner (in this case a3, b2, and c1) are no longer valid (either a2 or b1 is now "trapped"). If we place it on any edge square except the eight squares adjacent to the corners, the squares two horizontal or vertical spaces away become invalid. On edge squares adjacent to the corners (say b1) three squares also become unavailable. On centre squares, either 4 or 3 squares become unavailable.
The problem is that invalid squares can be fixed at any time. For example, placing mines first on a1 and then c1 may be initially invalid, but a mine on b1 solves that.
This is my preliminary analysis. I conclude that there is no way to calculate this number of boards without brute force. However, anyone with sufficient karma is welcome to improve this answer.
The key is: Monty knows where the car is (and will never open that door). We don't know where the million dollar is so we MIGHT open that door. For an illustration, we look at how the tree diagram differs for the two cases.
Suppose we have 3 doors, A, B and C and our car/million is in door A. We further assume we will always switch. (Once we understand this, we can extend it to $n$ doors and see that the situation will be similar.)
Case 1: Monty Hall Problem
If we switch, $P($Win$) = \frac{2}{3}$.
Case 2: Deal or No Deal scenario
Notice our assumption in the question is we only look at the situation if the million has not been opened. So we are in essence calculating a conditional probability. If we switch,
$P($Win $|$ Million not opened$) = \displaystyle \frac{P(\textrm{Win}\cap \textrm{Million not opened})}{P(\textrm{Million not opened})} = \frac{\frac{1}{6}+\frac{1}{6}}{\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}}=\frac{1}{2}$.
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Best Answer
There are (at least) two factors that mean that simply calculating the average of the remaining options is not enough to describe how someone should play.
Risk aversion
Someone's utility is not a predictable function of the amount of money that they win. For instance my utility from winning $\$$5 is more than 100 times my utility from winning 5 cents. However, my utility from winning $\$$100 million is less than 100 times my utility from winning $\$$1 million.