A gambling game involves a fair die with six faces. Each face is marked with a distinct number of dots from one and six. You pay a fee to play every round of the game. The fee for the first round is ₹0.01. The fee increases by ₹0.01 after each round is played. Thus, the fee to play the second round is ₹0.02, the fee to play the third round is ₹0.03, and so on. Once you have paid the fee, it won't be returned to you.

In each round, after you pay the fee, the die is rolled and you win ₹1.00 times the number of dots that appear on the face facing up. You may quit the game at any moment. How long should you play this game if you want to maximize your profit? What is the profit you expect to make?

[SOLVED]

2 comments

Tanny Libman solved this puzzle:

The expected winnings per turn is

\[ \frac{1}{6} \sum_{k = ₹1}^{₹ 6} k = ₹3.50. \]

Therefore, the expected winnings after \(n\) turns is \(3.5n\). The total fee to play \(n\) turns is \[ \frac{1}{100} \sum_{k=1}^n k = \frac{n (n + 1)}{200}. \] Therefore, the total profit is \[ 3.5n - \frac{n (n + 1)}{200} = \frac{699n - n^2}{200}. \]

This is a quadratic function, so it has exactly one maximum. We know that it's a maximum and not a minimum because the coefficient of \(n^2\) is negative. The maximum is at the average of its two roots, i.e. \(349.5\). But since we can't play half a turn, the maximum profit will occur after playing \(349\) turns. The expected profit is \[ \frac{₹699 \cdot 349 - ₹349^2}{200} = ₹610.75. \]

Raul Miller solved this puzzle:

In each turn, we earn an average of ₹3.50. Our cost for the 349th turn will be ₹3.49. The 350th game will provide an average profit of 0 and following games will, on average, cost more than they return. The expected profit in 349 games is \[ 349 \cdot ₹3.50 - ₹0.01 \sum_{k=1}^{349} k = ₹610.75. \]

Credit

This is an original puzzle from cotpi.

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