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There's a treasure box with \$100. There's a 0.5 probability that the box contains a bomb. The bomb has a probability of exploding on the i-th (for $i \leq 100$) day given by a uniform distribution. The bomb will eventually explode if there is a bomb. If it does not explode in the end, you can take all the money inside the treasure box. On the $i$-th day, how much are you willing to pay to play the game?

I am a little confused by the question, but I proceeded with finding the conditional probability that the treasure box doesn't have a bomb given that it doesn't explode the first $i$ days.

Let $A$ denote the event that it doesn't have a bomb. Let $D_i$ denote the event that it went the first $i$ days without exploding. We want to derive $P(A | D_i)$.

\begin{align} P(A|D_i) = \frac{P(D_i|A)P(A)}{P(D_i)} \\ P(A) = 0.5 \\ P(D_i | A) = 1 \\ P(D_i) = P(D_i | A) P(A) + P(D_i|A^c) P(A^c) \\ = 1 * 0.5 + P(D_i|A^c) * 0.5 \end{align}

For $P(D_i|A^c)$, which is given that there's a bomb in the box, what is the probability that the bomb doesn't explode in the first $i$ days. Finding the complement is easier. The complement is the probability that it explodes on any given day in the first i days, which is $i/100$. So then the complement of this is $1 - i/100$. So we have

\begin{align} P(D_i) = (2 - i/100) * 0.5 \\ P(A|D_i) = \frac{0.5}{(2 - i/100) * 0.5} \\ = \frac{1}{(2-i/100)} \end{align}

This probability gets smaller and smaller as the number of days increases without the box exploding.

After finding this, I am not sure how to answer the definitively answer the question of how much I'd pay to play the game on a given day. Because the conditional probability that the box has a bomb decreases with the number of days where the bomb doesn't explode, I would be inclined to say that I'd pay less and less, i.e., the amount I'd pay to play the game decreases as days go by.

Is there an actual amount that is optimal to offer to play such a game?

student010101
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  • Surely the box becomes more attractive as time goes by. Say there were $10^{10}$ instants at which the bomb could go off (if there were one). Then, by the $10^{10}-1$ instant you would be really, really sure that there wasn't a bomb, so you should pay nearly $100$. Of course, at the first instant, you should pay only $50$. – lulu Oct 06 '20 at 15:44
  • @lulu Oh wait, I said it backwards in my OP. I would pay more as time goes on. Is there a numerical relationship between how much I'd pay and the number of days gone by? Should it be proportional to the derived conditional probability? What are the conditions? You mentioned at the beginning it's 50, but what justifies the use of 50, instead of, say, 45, 55, etc..? – student010101 Oct 06 '20 at 15:47
  • At any given moment you should pay $100\times P(\text {no bomb})$. At time $0$ the probability that there is no bomb is $.5$ so the value of the box at time $0$ is $50$. – lulu Oct 06 '20 at 15:48
  • But why should I pay $100 \times P(no \ bomb)$? Why couldn't it be offset by some constant? – student010101 Oct 06 '20 at 15:50
  • Are you worrying about the phrasing "how much are you willing to pay..."? If so, then I agree, that's confusing. I took it to mean "what is the expected value of the box?" – lulu Oct 06 '20 at 15:50
  • @lulu Yeah that's the part I'm confused about. Honestly, at first I was like, if I'm not competing with anyone, why don't I just wait until the end? – student010101 Oct 06 '20 at 15:51
  • One could make an economics problem out of it by defining a utility function for the would be buyer. But none of that is mentioned, and there's no way to guess at it, so I think Expected Value is the way to go. – lulu Oct 06 '20 at 15:52
  • @lulu If we're trying to maximize the expected value from the view of the buyer of the treasure box, then this method assume the seller plays optimally right? – student010101 Oct 06 '20 at 15:56
  • I agree with lulu's interpretation. The way I see it, you can think of it as though time has passed and it is currently the $i$'th day. The current owner of the box explains everything in the problem about the box, the 100 dollars, and the possibility of there being a bomb. The owner then offers to sell you the box right there and then and will not offer again in the future. Given $i$ and given the other stated information, what is a "fair" price to pay for the box at that moment? (fair, meaning that if you were to buy the box your expected monetary gain from the transaction is zero) – JMoravitz Oct 06 '20 at 16:12
  • That is to say, there is no "waiting until the end"... and the seller isn't "playing" so there is no "optimal choice" for the seller. – JMoravitz Oct 06 '20 at 16:13
  • @JMoravitz The expected value of the box on the first day is 50 right? and this assumes that the box may explode after we purchase it but before we open it? Because otherwise if the box explodes before we purchase it, then we wouldn't want to make the purchase, in which case, I'm not sure what the expectation would end up being. – student010101 Oct 06 '20 at 16:16
  • Oh, the way I interpret it, the box that is offered to us to purchase has not exploded yet. Given that the box has not exploded by the $i$'th day and it is currently the $i$'th day, how much are we willing to pay. Yes, the fair cost on the initial day before the bomb has had any opportunity to explode would be $50$. – JMoravitz Oct 06 '20 at 16:20
  • @JMoravitz Ah okay, so the expectation (or perhaps conditional expectation I should say) of the box value for the $i + 1$-th day is $E[N_{i+1}] = 100 * P(A|D_i) + 0$, where notice the RHS uses the information for the first $i$ days? – student010101 Oct 06 '20 at 16:33

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