Expected Value: Bidding for a House

[Master1022]

Summary:: The price of a house is uniformly distributed between 0 and 1000 but we do not know its exact value. If we place a bid higher than the value, then we obtain the house, but if our bid is lower then we get nothing. If we know we can sell the house on to another person (guaranteed) for $nV$ (e.g. $n = 1.5$) where $V$ is the value of the house, then should we bid for the house? If so, how much?

Question: The price of a house is uniformly distributed between 0 and 1000 but we do not know its exact value. If we place a bid higher than the value, then we obtain the house, but if our bid is lower then we get nothing. If we know we can sell the house on to another person (guaranteed) for $nV$ (e.g. $n = 1.5$) where $V$ is the value of the house, then should we bid for the house? If so, how much?

My attempt:
We can consider the expected value of our profit $P = nV - B$ from a deal. If we win the auction ($B > V$), then we have a guaranteed pay-off of $nV - B$, otherwise we gain nothing. I am struggling to represent this mathematically and extract a useful answer from this problem.

We can define indicator variables for when we win (1) or not (2). Then:
$$ P = X_1 + X_2 $$
$$ E[P] = E[X_1] + E[X_2] $$
we know that $E[X_2] = 0$ (the scenario where we don't get the house) and thus by linearity of expectation:
$$E[P] = E[X_1] = E[nV - B] = nE[V] - E[ B ]$$
We want to make a profit and therefore want $E[P] > 0$:
$$ nE[V] - E[ B ] > 0 $$
If the distribution of the house's value is uniform then $E[V] = 500$ and therefore, we want to bet less than $500 n$. In the case where $n = 1.5$, then we would bet (just) less than 750.

Is this correct logic? I have seen other versions of this problem on the internet and I have seen some responses to those variants say that we should not be betting, but don't provide much explanation to back up that claim.

Any help is greatly appreciated.

 

[PeroK]

What happens if $n = 1.5$ and you bid $750$? Can you just work out that one example?

 

[Dale]

We can consider the expected value of our profit P=nV−B from a deal.

OK, so here you have profit as a function of V and B. From there you can simply marginalize over V to get the expected profit as a function of B.

 

[Master1022]

What happens if $n = 1.5$ and you bid $750$? Can you just work out that one example?

Thanks for your reply. So if $n = 1.5$, then we want the value of the house to be $500 \leq V < 750$ to make a profit. 1/4 of the time we make a profit, 1/4 of the time we don't even win the auction, and the other 1/2 of the time we lose money.

Our expected winnings are: (I have used the average net profit and losses that we could face)
$$ = \frac{1}{4} \cdot (187.5) + \frac{1}{4} \cdot 0 + \frac{1}{2} \cdot (-375) $$
This yields a negative expected pay-out so this strategy actually turns out to be a bad strategy.

Was this reasoning correct? I got the averages for profit and losses by considering the values on the edges (0, 500, and 750) and saw what gains/losses those yielded.

Thank you for pointing that out. I will try and generalize the probabilities in terms of values and see whether I can make progress from there.

 

[Master1022]

OK, so here you have profit as a function of V and B. From there you can simply marginalize over V to get the expected profit as a function of B.

Thanks for responding. Could you perhaps explain what you mean by 'marginalize over V'? I am unfamiliar with that terminology.

 

[PeroK]

Thanks for your reply. So if $n = 1.5$, then we want the value of the house to be $500 \leq V < 750$ to make a profit. 1/4 of the time we make a profit, 1/4 of the time we don't even win the auction, and the other 1/2 of the time we lose money.

Our expected winnings are: (I have used the average net profit and losses that we could face)
$$ = \frac{1}{4} \cdot (187.5) + \frac{1}{4} \cdot 0 + \frac{1}{2} \cdot (-375) $$
This yields a negative expected pay-out so this strategy actually turns out to be a bad strategy.

Was this reasoning correct? I got the averages for profit and losses by considering the values on the edges (0, 500, and 750) and saw what gains/losses those yielded.

Thank you for pointing that out. I will try and generalize the probabilities in terms of values and see whether I can make progress from there.

I think that is right. If we bid $750$ and the house is worth more than that then nothing happens. So, we can just look at the case where we get the house. The average value in that case is $375$, so the average income is only $1.5 \times 375$, which is less than the bid.

 

[etotheipi]

I agree with the final answer, but in the working, are you sure we don't need to multiply by the probability of the bid being greater than the actual value? If $W$ and $L$ are the events in which the house is won or lost respectively (with indicator functions $I_W$ and $I_L$), and the random variable denoting the profit is $A = nV - b$, then then the law of iterated expectations is$$\mathbb{E}(A) = \mathbb{E}(A|I_W = 1) P(I_W = 1) + \mathbb{E}(A|I_L = 1) P(I_L = 1)$$$\mathbb{E}(A|I_L = 1)$ is just going to be zero, whilst $\mathbb{E}(A|I_W = 1) = nV - b$, if $b$ is the bid you chose. That would suggest$$\mathbb{E}(A) = \mathbb{E}(nV - b | I_W = 1)P(I_W = 1)$$and $P(I_W = 1) = P(V < b)$, for the fixed $b$ that we chose.

Of course the $P(I_W = 1)$ cancel in the end, but don't we still need it in there to get the correct form of $\mathbb{E}(A)$?

 

[Dale]

Thanks for responding. Could you perhaps explain what you mean by 'marginalize over V'? I am unfamiliar with that terminology.

Sure, sorry about that. Marginalizing $f$ over $V$ means to integrate $f$ with respect to $V$ in order to get rid of $V$. So here you would evaluate $f_{marginal}(B)=\int P(V) f(B,V) dV$. Since $V$ is uniformly distributed from 0 to 1000 then $P(V)=1/1000$, and you already found $f(B,V)=nV-B$. So you just evaluate the integral and that gives you the expected profit as a function of $B$. Then you maximize that and if it is positive then you bet that amount.

 

[Master1022]

Sure, sorry about that. Marginalizing $f$ over $V$ means to integrate $f$ with respect to $V$ in order to get rid of $V$. So here you would evaluate $f_{marginal}(B)=\int P(V) f(B,V) dV$. Since $V$ is uniformly distributed from 0 to 1000 then $P(V)=1/1000$, and you already found $f(B,V)=nV-B$. So you just evaluate the integral and that gives you the expected profit as a function of $B$. Then you maximize that and if it is positive then you bet that amount.

Not to worry, I am happy to learn. Thanks for your response. Okay, so if I follow this method:
$$ f_{marginal}(B)=\int P(V) f(B,V) dV = \frac{1}{1000} \int_0^{1000} (nV - b) dV $$
$$ f_{marginal}(B) = \frac{1}{1000}\left( \frac{n (10^3 )^2}{2} - (1000)b \right) $$

If we look for ranges of $b$ to make the expression positive, we once again end up with $b < 500n$, which we saw doesn't work (at least not for n = 1.5) with the help of @PeroK . This suggests that my $f(B,V)$ needs to be corrected. I will continue to see whether I can correct the function.

I agree with the final answer, but in the working, are you sure we don't need to multiply by the probability of the bid being greater than the actual value?

I agree, I had started working on a method with generalized variables and I have to consider these probabilities as well.

If $W$ and $L$ are the events in which the house is won or lost respectively (with indicator functions $I_W$ and $I_L$), and the random variable denoting the profit is $A = nV - b$, then then the law of iterated expectations is$$\mathbb{E}(A) = \mathbb{E}(A|I_W = 1) P(I_W = 1) + \mathbb{E}(A|I_L = 1) P(I_L = 1)$$$\mathbb{E}(A|I_L = 1)$ is just going to be zero, whilst $\mathbb{E}(A|I_W = 1) = nV - b$, if $b$ is the bid you chose. That would suggest$$\mathbb{E}(A) = (nV - b)P(I_W = 1)$$and $P(I_W = 1) = P(V < b)$, for the fixed $b$ that we chose.

Of course the $P(I_W = 1)$ cancel in the end, but don't we still need it in there to get the correct form of $\mathbb{E}(A)$?

I agree (I think). I have a quick follow-up question: what about the region where we lose money (i.e. below $\frac{b}{n}$)? My first reaction was to say that we need to include it as another term, but a second look at the term $b - nV$ seems to incoporate the losses for the case where $b > nV$.

My current attempt:
These were my steps (please do let me know if you spot any errors):

We can see that we need to bet $V \leq b \leq nV$ and therefore the size of the profitable region is $V(n - 1)$, in probability space, for any given $n$. Thus the probability of the value being within that region is $\frac{V(n - 1)}{1000}$. If the value is within that region, then our profit is $nV - b$.

I think we can ignore the case where we lose the auction as the pay-off is 0 so that term disappears.

Now I am trying to think about the probability where we lose money. Although I think I should understand the follow-up question I had because maybe this term is already included in the above... If we were to include it, then we would lose money whenever the value is less than $\frac{b}{n}$, thus occurring with $\frac{b}{1000n}$ chance. Am I correct in saying that the loss would be $b - nV$ or have I mixed up the signs?

Then I would write the expected value down and try to find the values of b such that:
$$ \frac{V(n - 1)}{1000} \cdot (nV - b) - \frac{b}{1000n} \cdot (b - nV) > 0 $$

Does that look like the correct expression? I will go on to solve it in the meanwhile, but just want to check whether it is right or wrong from a conceptual standpoint.

Thanks for all the continued help.

 

[PeroK]

I think you can shortcut the whole process by setting the problem up in the right way before you do any serious calculations.

 

[Master1022]

I think you can shortcut the whole process by setting the problem up in the right way before you do any serious calculations.

Yes, I will try to start back at the basics and see whether any simpler arguments come to mind. I think I am over-complicating this, but I cannot see where.

$$ \frac{V(n - 1)}{1000} \cdot (nV - b) - \frac{b}{1000n} \cdot (b - nV) > 0 $$

One error I am thinking about is whether the pay-off terms have been used correctly here? For example, when we did the calculation above, we took the averages of the edge cases to find the average pay-offs/losses. I wonder if an integral is needed for the gain and loss amounts (i.e. integrating over V as @Dale had suggested). I also wonder if I have used incorrect limits when attempting to use the integration above? I will take some time to see what others respond before posting random ideas again.

 

[Dale]

This suggests that my f(B,V) needs to be corrected.

Oh, yes, you are right. If $B<V$ then $f(B,V)=0$ because you lose the bid. That means you do not get the house so you do not gain $nV$ from selling it but you also do not lose $B$ from buying the house. So it is a piecewise function which is a little more difficult to evaluate.

 

[Master1022]

I was just thinking along the lines of $P(I_W = 1) = P(V < b) = \frac{b}{1000}$. Then you have$$\mathbb{E}(A) = \mathbb{E}(nV - b)P(I_W = 1) = \frac{(n\mathbb{E}(V) - b)b}{1000}$$i.e. your expectation is an upside down quadratic in $b$, with roots at $0$ and $750$. If that's correct, then $b=350$ gives you maximum expectation of profit, even if you are more likely to win the house at $b = 750$.

This makes more sense! Thank you very much.

 

[PeroK]

Perhaps I'm over-simplifying this, but if the odds are against you, then don't bet; and, if the odds are in your favour, then maximise your bet. That's what I'd do.

 

[jbriggs444]

Perhaps I'm over-simplifying this, but if the odds are against you, then don't bet; and, if the odds are in your favour, then maximise your bet. That's what I'd do.

If I follow the logic, the first observation is that no matter how much you bid, if you win you get a house with a value that is uniformly distributed between zero and your bid. You can sell this house for n times its value. The expected value of this sale is going to be an easily calculated multiple of the bid. This fixed multiple will depend only on n.

Either the expected result will exceed the bid or it won't. If it does, you bid the maximum. If it does not then you bid nothing or simply walk away.

 

[Master1022]

If I follow the logic, the first observation is that no matter how much you bid, if you win you get a house with a value that is uniformly distributed between zero and your bid. You can sell this house for n times its value. The expected value of this sale is going to be an easily calculated multiple of the bid. This fixed multiple will depend only on n.

Either the expected result will exceed the bid or it won't. If it does, you bid the maximum. If it does not then you bid nothing or simply walk away.

Thanks for the post - I am keen to develop my intuition here as well so I apologize if my question comes across as trivial. I might be misunderstanding this, but I don't think we will know the value $V$ until after we have either won or 'lost' the house. Can you explain what you mean by "If it [the expected result] exceeds the bid, then we bid the maximum"? Does this just mean if $nV$ is expected to be greater than the bid, then bid the maximum? I only wrote an arbitrary $n$ so I wasn't tied down to any specific value (like 1.5). I think that once $n$ starts to increase slightly (even to, let's say, 5), the problem starts to break down as it very quickly becomes a good deal (which I think is what your point is stating and therefore you want to secure the win by bidding the maximum? - is this a correct interpretation). I don't mean for this post to come across in a rude way.

 

[jbriggs444]

Thanks for the post - I am keen to develop my intuition here as well so I apologize if my question comes across as trivial. I might be misunderstanding this, but I don't think we will know the value $V$ until after we have either won or 'lost' the house.

Yes. We do not know the value of V. But if we bid $B$ then the conditional distribution of $V$ under the condition that we win is something that we do know.

Can you explain what you mean by "If it [the expected result] exceeds the bid, then we bid the maximum"? Does this just mean if $nV$ is expected to be greater than the bid, then bid the maximum?

Yes, exactly. Fortunately for us the answer to the question: "is $nV$ expected to be greater than the bid" is an easily determined function of n.

 

[Master1022]

Yes. We do not know the value of V. But if we bid $B$ then the conditional distribution of $V$ under the condition that we win is something that we do know.

Okay, I think I see what you mean. Does this suggest that we shouldn't be making bids for $n < 2$? For example, we know that if we bid 600 and win, then the distribution of $V$ is uniform between 0 and 600 (this is how I interpret the conditional distribution of $V$). Therefore, it has an expected value of 300. Thus, $300 \cdot 1.5 = 450$ which is less than 600 and therefore we know that 600 isn't a good bet to make for n = 1.5.

 

[jbriggs444]

Okay, I think I see what you mean. Does this suggest that we shouldn't be making bids for $n < 2$? For example, we know that if we bid 600 and win, then the distribution of $V$ is uniform between 0 and 600 (this is how I interpret the conditional distribution of $V$). Therefore, it has an expected value of 300. Thus, $300 \cdot 1.5 = 450$ which is less than 600 and therefore we know that 600 isn't a good bet to make for n = 1.5.

Right. 600 is not a good bet to make for n = 1.5.

Now keep following the logic. Is 500 a good bet to make for n = 1.5. Is any bet a good bet to make for n = 1.5?

 

[Master1022]

Now keep following the logic. Is 500 a good bet to make for n = 1.5. Is any bet a good bet to make for n = 1.5?

No!

I think the use of this conditional distribution of $V$ might have been the explanation I saw earlier, but it made no sense with only one line of math and no explanation. The expected payoff is as before:
$$ E[P] = n \cdot E[V] - B $$ where we can use the conditional distribution to say $E[V] = B/2$ and therefore any $n$ below 2 suggests that we shouldn't be placing any bids as $E[P] < 0$.

Okay, thank you very much for helping me see this much simpler approach.

 

[Master1022]

The only question I still have about this is whether I ought to have included a probability term in the expected value calculation as suggested earlier by @etotheipi ? I feel like the conditional distribution accounts for this to some extent (the fact that it is conditioned upon the successful purchase of the house) but am not completely sure. This wouldn't change the answer by including a factor of $B/1000$, but just wanted to see whether that was 'technically' required in the expression?

 

[jbriggs444]

The only question I still have about this is whether I ought to have included a probability term in the expected value calculation as suggested earlier by @etotheipi ? I feel like the conditional distribution accounts for this to some extent (the fact that it is conditioned upon the successful purchase of the house) but am not completely sure. This wouldn't change the answer by including a factor of $B/1000$, but just wanted to see whether that was 'technically' required in the expression?

I don't know about @PeroK, but my intuition was busy sweeping irrelevant details under the carpet in order to simplify things to a point where a decision could be made.

The expected value is zero if we lose the bid -- so ignore that possibility. It doesn't contribute.

It is a winning proposition? Had to sweat for a few seconds seeing that this was a function of n. And you've reasoned your way to seeing that it is a winning proposition if and only if n > 2.

If it's a losing proposition then we won't play -- so ignore that possibility. It doesn't contribute.

The more we bet, the higher the likelihood we win. So we should bet more.

The more we bet, the more we win. So we should bet more.

You do not need many numbers to make a decision.

If you want to go back and calculate your expected value for a particular value of n and of B then yes, you'll have to include a probability of $B/1000$, sure.

 

[PeroK]

If you want to do the maths, then first I'd scale the 1000 down to 1. Then, with a bid of $b$ you get the house with probability $b$, its average value is $b/2$ and you sell it for $nb/2$. The average gain/loss on the transation is $(\frac n 2 - 1)b$, which is losing for $n < 2$.

The expected loss/gain is simply $(\frac n 2 - 1)b^2$. Maximising $b^2$ is the same as maximising $b$, when $n > 2$.

 

[Master1022]

You do not need many numbers to make a decision.

If you want to go back and calculate your expected value for a particular value of n and of B then yes, you'll have to include a probability of $B/1000$, sure.

Completely agree and I do appreciate the help. Was just asking to confirm a mathematical technicality (just helping to practice e.v. calculations that show up elsewhere)

 

[etotheipi]

I was just thinking along the lines of $P(I_W = 1) = P(V < b) = \frac{b}{1000}$. Then you have$$\mathbb{E}(A) = \mathbb{E}(nV - b)P(I_W = 1) = \frac{(n\mathbb{E}(V) - b)b}{1000}$$i.e. your expectation is an upside down quadratic in $b$, with roots at $0$ and $750$. If that's correct, then $b=350$ gives you maximum expectation of profit, even if you are more likely to win the house at $b = 750$.

Nice discussion above, I did make mistake here, the conditional expectation $\mathbb{E}(nV - b | I_W = 1)$ should actually be$$\mathbb{E}(nV - B | I_W = 1) = n\mathbb{E}(V| I_W = 1) - b = n\frac{b}{2} - b = b \left(\frac{n}{2} -1 \right)$$i.e. not just using $\mathbb{E}(V) = 500$, but rather $\mathbb{E}(V | I_W = 1) = b/2$. then I get @PeroK's formula for $\mathbb{E}(P)$. so it's absolutely correct that if $n>2$, you should bid as much as possible. Sorry for this mistake!

 

[Dale]

Right. 600 is not a good bet to make for n = 1.5.

So I did the marginal calculation I described above and it is negative for all $B$. So this agrees with your analysis at least qualitatively

 

[kuruman]

I thought it might be worthwhile to extend this discussion by replacing the clueless buyer, who bids sight-unseen, with a savvy buyer who is allowed to evaluate the property and put in a bid randomly selected within the range $\pm f V$ ($f<1$). The same filter $B > V$ and profit formula $\text{Profit}=nV -B$ apply. The results are shown below. The two datasets correspond to two levels of savvy buyer: in red a real estate broker who can evaluate a property to $\pm 5\%$ of its value and in blue an amateur whose accuracy is $\pm 30\%$. The dotted lines through each set of points represent the results of linear fits with the equations shown next to each line. I also did a simulation for the original question (clueless buyer) that produced the linear fit $$\langle \text{Profit} \rangle_{\text{clueless}}=\frac{V_m}{3}n-\frac{2V_m}{3}$$where $V_m$ is the maximum value of the property, here equal to 1000. I do not show this on the same plot because it will crowd out the other two plots and does not say anything new. It confirms what has been already stated that the breakeven point is at $n=2$.

Now for the math. The probability that the value of the property be between $V$ and $V+dV$ is $dP_V=\dfrac{dV}{V_m}.$ The probability of the bid being between $B$ and $B+dB$ is $dP_B=\dfrac{dB}{2fV}.$

The expectation value for the profit is given by $$\langle \text{Profit}\rangle _{\text{savvy}}= \int_0^{V_m} \frac{1}{V_m}dV\int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB=\left[n-\left(1+\frac{f}{2}\right)\right]\frac{V_m}{4}.$$With $V_m=1000$, the theoretical predictions are $$\langle \text{Profit}\rangle _{\pm~30\%}=250.0~n-287.5~;~~\langle \text{Profit}\rangle _{\pm 5\%}=250.0~n-256.25.$$These compare favorably with the simulation fits.

Note that the breakeven value is $n=1+\dfrac{f}{2}$ and decreases as $f$ decreases. The more accurately one is capable of assessing the value of the property, the more likely one is to turn a profit on average. This is in sharp contrast with the clueless bidder who needs $n > 2$ in order to turn a profit.

Moral: A pig in a poke is worth about as much as two birds in the bush.

On edit: I forgot to say that the "experimental" results shown in the figure come from a Monte Carlo simulation on an Excel sheet. Values and bids were randomly selected after which the filter was applied and the profits calculated. There were 40,000 bids of which half (as expected) were successful.

 

[etotheipi]

@kuruman I'm a little confused as to where this formula comes from, I wondered if you could explain?$$\langle \text{Profit}\rangle _{\text{savvy}}= \int_0^{V_m} \frac{1}{V_m}dV\int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB=\left[n-\left(1+\frac{f}{2}\right)\right]\frac{V_m}{4}$$Also when I do the integrals I get$$\int_0^{V_m} \frac{1}{V_m}dV = V_m$$ $$\int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB=\frac{V}{4}(2n - 2 - f)$$Which would appear to give$$\langle \text{Profit}\rangle _{\text{savvy}} = \frac{V V_m}{4}(2n - 2 - f)$$which can't be right, because for instance we still have a dummy $V$.

 

[kuruman]

@kuruman I'm a little confused as to where this formula comes from, I wondered if you could explain?$$\langle \text{Profit}\rangle _{\text{savvy}}= \int_0^{V_m} \frac{1}{V_m}dV\int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB=\left[n-\left(1+\frac{f}{2}\right)\right]\frac{V_m}{4}$$

It's a double integral, not the product of two integrals. The inner integral is the expectation value of profit when the value is $V$ and you have to add over all acceptable bids for that value. It is a function of $V$, $$\text{Profit(V)} =\int_{V}^{V+fV}\frac{(n V-B)}{2fV}dB=\frac{V}{4}[2n-(f+2)]$$ To get the final answer you do the outer integral and sum over all possible values of $V$ $$\int_0^{V_m} \frac{1}{V_m} \text{Profit(V)} dV=\int_0^{V_m} \frac{1}{V_m} \frac{V}{4}[2n-(f+2)] dV.$$

 

[etotheipi]

Ah, thanks for clarifying! That makes sense. But shouldn't we write the integral as$$\langle \text{Profit}\rangle _{\text{savvy}}= \int_0^{V_m} \frac{1}{V_m} \left( \int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB \right) dV$$because the inner integral is not a constant (depends on $V$) so we can't pull it out of the outer integral?

 

[kuruman]

Ah, thanks for clarifying! That makes sense. But shouldn't we write the integral as$$\langle \text{Profit}\rangle _{\text{savvy}}= \int_0^{V_m} \frac{1}{V_m} \left( \int_{V}^{V+fV}\dfrac{(n V-B)}{2fV}dB \right) dV$$because the inner integral is not a constant (depends on $V$) so we can't pull it out of the outer integral?

Yes we should. I got lazy.

 

[etotheipi]

I can hardly complain; I think one tiny bit of laziness is quite excused, given the rest of your excellent analysis and work on Excel . Thanks for taking the time to clarify!