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There are 4 boxes. One box has $100. The other 3 has nothing. You pay \$X to choose any box. If it doesn't have the money, you can choose another box. How many should \$X be to ensure a fair game if the person plays optimally? The solution is \$40.

The solution is as follows:

You can define $E_1, E_2, E_3, E_4$ to be the events that the first, second, third, fourth boxes you choose has the money, respectively. So we want the expected payment for playing the game to be equal to $100. So we have

$$ 100 = XP(E_1) + 2XP(E_2) + 3XP(E_3) + 4XP(E_4) \\ = 0.25X + 0.50X + 0.75X + 1.0X \implies X = 40 $$

Now suppose if you choose 2 boxes and haven't gotten the money, the price changes to $Z$. What would $X$ and $Z$ be to ensure a fair game? I am stuck on this problem. According to the above equation, the equation becomes

$$ 100 = XP(E_1) + 2XP(E_2) + 3XP(E_3) + 4XP(E_4) \\ = 0.25X + 0.50X + 0.75(2X + Z) + 1.0(2X + 2Z) $$

So we have 1 equation and 2 unknowns and if we bound $X,Z \in [0, 100]$, then we have an infinite number of solutions. Is this the answer, or am I missing something here?

roulette01
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  • Perhaps in the second case, the fair amount of $Z$ should two-thirds of $$100$ making the fair amount of $X$ two-sevenths of $$100$. I really do not know what the question is aiming for. – Henry Mar 29 '21 at 01:08
  • @Henry Why do you believe it should be those numbers in particular? The initial question was just asking what should $X$ be to ensure the expected payoff for the player is zero. The second question asks the same thing but with the condition that $X$ changes to $Z$ if the first two choices are not correct. – roulette01 Mar 29 '21 at 01:10
  • My idea was that when raising the price, it might be sensible to make the player indifferent between playing on or giving up. But it is just an idea – Henry Mar 29 '21 at 01:17

2 Answers2

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The key is the condition "a fair game if the person plays optimally".

You have analyzed only one strategy: playing the game until you win the money. In the first version of the problem, that is necessarily the optimal strategy: if it was worth playing the game with $4$ boxes, it becomes worth playing the game with $3$ boxes, as well.

In general, we can consider $4$ strategies; the $k^{\text{th}}$ strategy is to stop after $k$ boxes have been opened (if you haven't yet won). In the second version of the problem:

  • The $k=1$ strategy can be disregarded: if it was worth opening the first box, it is worth opening the second box, which has the same price but better chances.
  • The $k=3$ strategy can be disregarded: if it was worth opening the third box, paying $Z$ for a chance at $\$100$, then it's worth opening the fourth box and paying $Z$ for a certain $\$100$.

But the $k=2$ strategy and $k=4$ strategy are both valid. So the constraints on fairness are that:

  • The $k=2$ strategy does not earn free money: $$\frac14(100-X) + \frac14(100-2X) + \frac12(-2X) \le 0.$$ Equivalently, $7X \ge 200$.
  • The $k=4$ strategy does not earn free money: $$\frac14(100-X) + \frac14(100-2X) + \frac14(100-2X-Z) + \frac14(100-2X-2Z) \le 0.$$ Equivalently, $7X + 3Z \ge 400$.
  • There is a way to break even: equality holds in at least one of the cases above.

I would consider any strategy that satisfies all three conditions "fair, if the person plays optimally". This means that either $X = \frac{200}{7}$ and $Z \ge \frac{200}{3}$, or $X \ge \frac{200}{7}$ and $Z = \frac{400-7X}{3}$. This includes:

  • The case $X=40$ and $Z=40$, which is unsurprisingly fair, because it's the same as the first version of the game.
  • The case $X=\frac{200}{7}$ and $Z = 100000$, which is fair if the person plays optimally and stops after two boxes. (In this game, it is a bad idea to open the third box, since you pay more money to open it that you could possibly win.)

Possibly the intended solution, though, is $X = \frac{200}{7}$ and $Z = \frac{200}{3}$, which is fair if the person plays either the $k=2$ or the $k=4$ strategy.

Alternatively, you could argue that even if equality does not hold in either case, we have "a fair game if the person plays optimally". The game with $X=100000$ and $Z=100000$ is a fair game if the person plays optimally: if the person walks away and does not play at all, they break even.

Misha Lavrov
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  • In the first version of the game, why is that necessarily the optimal strategy? Is it because one should be playing the game in a memoryless way? I think I have trouble justifying that step. – roulette01 Mar 29 '21 at 13:54
  • In the first version of the game, your chances of winning increase at every step; therefore if it is worth opening the first box, it should also be worth opening the second box. This can be verified mathematically; the expected value of the $k=1, 2, 3$ strategies is always either less than the expected value of the $k=4$ strategy, or less than the expected value of not playing. – Misha Lavrov Mar 29 '21 at 15:49
  • Makes sense. Another question I have is you said "This means either $X = 200/7$ and $Z \geq 200/3$, or $X \geq 200/7$ and $Z = (400-7X)/3)$. But you could also have $X =200/7$ and $Z = (400-7X)/3)$, i.e., strict equalities, right? – roulette01 Mar 29 '21 at 16:04
  • Yes: that falls under both cases at the same time, so it's certainly allowed. It is the $X=\frac{200}{7}, Z = \frac{200}{3}$ solution I mention at the end, which is fair under both strategies. – Misha Lavrov Mar 29 '21 at 16:10
  • When you say "fair" here, you don't actually mean "break even" right? It seems you distinguish between the two where break even is if equality holds and fair is when the inequality holds. – 24n8 May 10 '21 at 20:45
  • @24n8 If the inequality were strict in all cases, it would not be a fair game. But it's okay for the inequality to be strict in some cases - that just means that those cases aren't the best strategy. – Misha Lavrov May 10 '21 at 20:53
  • For the case where $X=Z=40$, that wouldn't be fair for the player if using the $k = 2$ strategy right? Since it would result in the player losing money, on average? – 24n8 May 10 '21 at 21:09
  • @24n8 Yes, but the $k=2$ strategy is suboptimal for $X=Z=40$. I'm fine with still saying that the $X=Z=40$ game is fair, because the player can always use the $k=4$ strategy. If we want "fair" to mean "fair for all strategies", we can't get what we want: fair for the $k=3$ strategy would mean giving away money to the $k=4$ strategy, for example. – Misha Lavrov May 10 '21 at 21:12
  • Right. I agree There is no $X, Z$ that ensures the game is fair for all strategies, not even for just the combined $k =2$ and $k=4$ strategies. – 24n8 May 10 '21 at 21:14
  • Actually, I think the equality for both cases is fair for both $k=2$ and $k=4$, which is what you concluded. I just misread that part earlier. – 24n8 May 10 '21 at 21:15
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Suppose you have two boxes. One has nothing and one has $100. Otherwise, the rules are the same as in your first game.

The fair price $Z$ for each play is now given by $$100 = \frac{1}{2}Z + \frac{1}{2}(2Z) = \frac{3}{2}Z \implies Z = \frac{200}{3}.$$

Now, with your equation, solve for $X$ in $$100 = 0.25X + 0.25(2X) + 0.25(2X+Z) + 0.25(2X+2Z)$$ to find $$X = \frac{400-3Z}{7}$$

and then use our value of $Z$ to find $$X = \frac{200}{7}$$

Brian Moehring
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  • Hmm. I am not sure why $100 = 0.5Z + 0.5 * 2Z$ is valid. When we are left with two boxes, we would have already exhausted $2X$, but this is not accounted for when determining $Z$. Shouldn't it be accounted for? – roulette01 Mar 29 '21 at 01:20
  • This seems to assume the player plays in a "memoryless" way. – roulette01 Mar 29 '21 at 01:23
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    @roulette01 Is the price of a lottery ticket today affected by the price of all the lottery losers you had before? It's not a perfect analogy, but it's the principle. You can choose to play again or stop, but the price to continuing shouldn't be affected by how much you've already lost. It should be set by the chance of winning as of that point. – Brian Moehring Mar 29 '21 at 01:23
  • The player can choose whatever strategy he/she wants. The price is set from the perspective of the game-creator, and there's no reason to set a price for the third and fourth boxes that will have a surely positive payout to the player. (such as with $40, when the price is fixed for all four) – Brian Moehring Mar 29 '21 at 01:25
  • If instead of the price changing after choosing 2 incorrect boxes, the price changes for each incorrect choice, then it seems the series of prices for the box would be $25, 33\frac{1}{3}, 50, 100$, as this would ensure that the player would net $0 at every stage? – roulette01 Mar 29 '21 at 14:49