Interplay between Buy-It-Now price and last minute bidding on online bidding strategies

Abstract

Sellers and buyers on online auction sites like eBay have the option of setting and executing auction parameters such as auction length, Buy-It-Now price, starting price, reserve price, etc. Understanding why bidders choose to execute the Buy-It-Now price as opposed to submitting a bid at the last minute of an auction helps managers better choose auction parameters and generate more revenue. In this paper, we first study online bidder behavior pertaining to the execution of the Buy-It-Now option as opposed to the last minute bidding strategy on eBay. Consequently, we study the seller’s decision to set a Buy-It-Now option and the amount of it. Our main finding is that a dominant strategy for the seller is to offer a Buy-It-Now option at the beginning of the auction. An early bidder arriving at the auction site is indifferent in choosing threshold Buy-It-Now prices or moving onto the auction and under particular circumstances strategically waiting for the last minutes of the auction to submit a bid. We also provide conditions on the existence of a set of equilibria which predicts the outcome of the game of executing the Buy-It-Now option versus last-minute bidding on eBay.

Appendix

Lemma 1

For low value bidders, bidding {R} in the first stage and waiting for the last minute to snipe, (t ^L _i  = T) for i = 1, 2, is a dominant strategy.

Proof

For a low value bidder the choices and payoffs in the first stage of the game are as follows:

$$ \begin{aligned} \sigma_{1}^{L}(V_{1}^{L})&=\alpha (L-L)=0\\ \sigma_{1}^{L}(B)&=(L-B) < 0\\ \sigma_{1}^{L}(R)&=SSP\\ \end{aligned} $$

SSP stands for second stage payoffs. As will be shown, if the low type bidder continues to the second stage, there is a chance that she might win the auction and only pay the reserve price. Thus, expected second stage payoffs are strictly greater than zero (SSP > 0). Therefore, we can easily conclude that playing the {R} strategy strictly dominates other choices when the bidder is a low type.

Lemma 2

If

$$ p\in\left\{\begin{array}{ll} \left.\left({\frac{\alpha (H-L)}{(1-\alpha)H+\alpha L-R}},1 \right.\right]& \hbox{then}\left\{ (t_{1}^{H}=T,t_{1}^{L}=T), (t_{2}^{H}=T,t_{2}^{L}=T)\right\}\\ \left[\left.0,{\frac{\alpha(H-L)}{(1-\alpha)H+\alpha L-R}}\right)\right.& \hbox{then}\left\{ (t_{1}^{H}=T,t_{1}^{L}=T), (t_{2}^{H}=0,t_{2}^{L}=T)\right\} \end{array}\right. $$

is the Nash equilibrium.

Proof

(a) Since (H  − L) < (L − R) holds, $p<{\frac{L-R}{H-L}}$ has to hold, which means that Π ^C₁ (H) > Π ^A₁ (H). For Π ^B₁ (H) < Π ^D₁ (H) to hold, we need $p<{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ to be satisfied. Since ${\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}>{\frac{L-R}{H-L}}$ holds for all values of α ∈ [0, 1), existence of $p<{\frac{L-R}{H-L}}$ is enough to say that Player 1 will always submit her bid at the end of the auction. Player 2 knows that Player 1 will always play (t ^H₁  = T, t ^L₁  = T). If $p>{\frac{q(H-L)}{(1-q)H+qL-R}},$ then he prefers to wait until the last minute as well because Π ^D₂ (H) < Π ^C₂ (H). We know that

$$ {\frac{q(H-L)} {(1-q)H+qL-R}}=\left\{\begin{array}{ll} {\frac{H-L}{L-R}}& \hbox{if}\quad q=1\\ {\frac{\alpha(H-L)}{(1-\alpha)H+\alpha L-R}}& \hbox{if}\quad q=\alpha \end{array}\right. $$

is true. Since ${\frac{\alpha(H-L)}{(1-\alpha)H+\alpha L-R}}<{\frac{H-L}{L-R}}$ holds for all values of $\alpha\in[0,1), p>\frac{{\alpha(H-L)}}{(1-\alpha)H+\alpha L-R}$ is a sufficient condition for Player 2 to choose last minute bidding as his best response. As a result, both players wait until the end of the auction to submit their bids, and { (t ^H₁  = T, t ^L₁  = T), (t ^H₂  = T, t ^L₂  = T)} is the Nash equilibrium.

(b) Alternatively, $p<{\frac{\alpha(H-L)} {(1-\alpha)H+\alpha L-R}}$ is a sufficient condition for Player 2 to choose proxy bidding as his best response to Player 1’s action. Thus, Player 1 chooses to snipe and Player 2 decides to submit his proxy bid right away. As a result, {(t ^H₁  = T, t ^L₁  = T), (t ^H₂  = 0, t ^L₂  = T)} is the Nash equilibrium.

Lemma 3

If

$$ p\in\left\{\begin{array}{ll} \left[0,{\frac{L-R}{H-L}}\right)& \hbox{then}\Rightarrow\left\{ \begin{array}{l} \left\{(t_{1}^{H}=T,t_{1}^{L}=T), (t_{2}^{H}=0,t_{2}^{L}=T)\right\}\\ \hbox{or}\\ \left\{(t_{1}^{H}=T,t_{1}^{L}=T),(t_{2}^{H}=T,t_{2}^{L}=T)\right\} \end{array}\right.\\ \left({\frac{L-R}{H-L}},{\frac{(1-\alpha)H+\alpha L-R}{\alpha (H-L)}}\right)& \hbox{then}\Rightarrow\left\{ \begin{array}{l} \left\{(t_{1}^{H}=0,t_{1}^{L}=T),(t_{2}^{H}=0,t_{2}^{L}=T)\right\}\\ \hbox{or}\\ \left\{(t_{1}^{H}=T,t_{1}^{L}=T),(t_{2}^{H}=T,t_{2}^{L}=T)\right\} \end{array}\right.\\ \left.\left({\frac{\left(1-\alpha\right)H+\alpha L-R} {\alpha\left(H-L\right)}},1\right.\right]&\hbox{then}\Rightarrow\left\{ (t_{1}^{H}=0,t_{1}^{L}=T),(t_{2}^{H}=0,t_{2}^{L}=T)\right\} \end{array}\right. $$

is the Nash equilibrium.

Proof

(a) When $p<{\frac{L-R}{H-L}}$ holds we can easily show that Π ^A₁ (H) < Π ^C₁ (H) also holds. Similarly, we can show that Π ^D₁ (H) > Π ^B₁ (H) holds as long as $p<{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ is satisfied. Since ${\frac{L-R}{H-L}}<{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ is true for all α ∈ [0,1), we can conclude that as long as $p<{\frac{L-R}{H-L}}$ holds, sniping is the dominant strategy for Player 1. Since the value of p is common knowledge for both players, Player 2 knows that Player 1 will be sniping. Thus, for Player 2, the choice is between Π ^C₂ and Π ^D₂ . Since α ∈ [0, 1], we can identify a cut-off value of α such that for $\hat{\alpha}\in(0,1)$ we have:

$$ \left\{\begin{array}{ll} \Uppi_{2}^{D}(H)>\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha<\hat{\alpha}\\ \Uppi_{2}^{D}(H)=\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha=\hat{\alpha}\\ \Uppi_{2}^{D}(H)<\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha>\hat{\alpha} \end{array}\right. . $$

Therefore, depending on the level of α, Player 2 either chooses to snipe or to proxy bid. As a result, we have either {(t ^H₁  = T, t ^L₁  = T), (t ^H₂  = T, t ^L₂  = T)} or {(t ^H₁  = T, t ^L₁  = T), (t ^H₂  = 0, t ^L₂  = T)} as the Nash equilibrium.

(b) Since $p>{\frac{L-R}{H-L}}$ holds we can easily show that Π ^A₁ (H) > Π ^C₁ (H) also holds. Thus, if Player 2 decides to proxy bid right away in the third stage, i.e., t ^H₂  = 0, then best response of Player 1 is to do the same and proxy bid, i.e., t ^H₁  = 0. Similarly, since $p<{\frac{H-L} {L-R}}$ has to hold due to Case 2, we know that Π ^A₂ (H) > Π ^B₂ (H) is always true. Thus, if Player 1 decides to proxy bid right away in the third stage, i.e., t ^H₁  = 0, then best response of Player 2 is to do the same and proxy bid, i.e., t ^H₂  = 0. Thus, { (t ^H₁  = 0, t ^L₁  = T), (t ^H₂  = 0, t ^L₂  = T)}, the case where both players submit proxy bids, is one of the two possible Nash equilibria of this auction. Alternatively, since $p<{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ holds, if Player 2 decides to snipe, i.e., t ^H₂  = T, then best response of Player 1 is to do the same, i.e., t ^H₁  = T because as we noted earlier it is always true that Π ^D₁ (H) > Π ^B₁ (H). If Player 1 chooses to snipe, then for Player 2, the choice is between Π ^C₂ (H) and Π ^D₂ (H). Similar to part (a) of this proof, we can identify a cut-off value of α such that for $\hat{\alpha}\in(0,1)$ we have:

$$ \left\{\begin{array}{ll} \Uppi_{2}^{D}(H)>\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha<\hat{\alpha}\\ \Uppi_{2}^{D}(H)=\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha=\hat{\alpha}\\ \Uppi_{2}^{D}(H)<\Uppi_{2}^{C}(H)& \hbox{when}\quad\alpha>\hat{\alpha} \end{array}\right. . $$

If $\alpha>\hat{\alpha}$ holds, then Π ^C₂ (H) > Π ^D₂ (H) is true and Player 2 chooses to submit a proxy bid. Then, the abovementioned Nash equilibrium is the single one of this auction. However, if $\alpha<\hat{\alpha}$ holds, then Π ^C₂ (H) < Π ^D₂ (H) is true and Player 2 also chooses to snipe. Then, we have {(t ^H₁  = T, t ^L₁  = T), (t ^H₂  = T, t ^L₂  = T)} as the second Nash equilibrium of this auction.

(c) Lastly, when $p>{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ holds we can easily show that Π ^D₁ (H) < Π ^B₁ (H) also holds. Similarly, we can show that Π ^A₁ (H) > Π ^C₁ (H) holds as long as $p>{\frac{L-R}{H-L}}$ is satisfied. Since ${\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}>{\frac{L-R}{H-L}}$ is true for all α ∈ [0,1), we can conclude that as long as $p>{\frac{(1-\alpha)H+\alpha L-R}{\alpha(H-L)}}$ holds, proxy bidding is the dominant strategy for Player 1. Since $p < {\frac{H-L} {L-R}}$ has to hold due to Case 2, we know that Π ^A₂ (H) > Π ^B₂ (H) is always true. Thus, {(t ^H₁  = 0, t ^L₁  = T), (t ^H₂  = 0, t ^L₂  = T)}, the case where both players submit proxy bids, is the only Nash equilibrium of this auction.

Proposition 1

Result of the second stage of the game depends on the BIN price that the seller assigns in the first stage. The high type first player becomes indifferent between executing the BIN option and moving to the next stage at the following threshold BIN prices:

$$ \begin{aligned} B_{1}&=R+p(H-R)-\alpha p^{2}(H-L)\\ B_{2}&=H-\alpha(1-p)(H-R)-\alpha p^{2}(H-L)\\ B_{3}&=H(1-\alpha)+\alpha L \end{aligned} $$

Proof

The BIN price that seller sets depends on the expected Nash equilibrium of the third stage, which is dependant on the auction parameters p and α. Following from proofs of Lemmas 2 and 3, it is easy to show that there are three main outcomes of the third stage game:

$$ \left\{\begin{array}{l} \Uppi_{1}^{D}(H)=(1-p)(H-R)+\alpha p^{2}(H-L)\\ \Uppi_{1}^{C}(H)=\alpha(1-p)(H-R)+\alpha p^{2}(H-L)\\ \Uppi_{1}^{A}(H)=\alpha(H-L) \end{array}\right. . $$

If we set each one of them equal to Π(BIN) = H−B, and then solve for the BIN value that leaves Player 1 indifferent regarding her choices of {B} or moving onto the third stage of the auction. As a result, the seller would set one of the following BIN prices depending on the auction parameters.

$$ \begin{aligned} B_{1}&=R+p(H-R)-\alpha p^{2}(H-L)\\ B_{2}&=H-\alpha(1-p)(H-R)-\alpha p^{2}(H-L)\\ B_{3}&=H(1-\alpha)+\alpha L \end{aligned} $$

Proposition 2

It is a dominant strategy for the seller to offer a Buy-It-Now option at the beginning of the auction.

Proof

First, we need to calculate the seller’s expected revenue from not offering a BIN option.

$$ \Uppi(Auction)=(1-p^{2})R+p^{2}\left[(1-\alpha)^{2}H+ \left[1-(1-\alpha)^{2}\right]L\right] $$

The first term above represents the case where either both bids do not get recorded or only one of them gets recorded. In this case, the revenue for the seller is only the reserve price R. The second term represents the case where both bids get recorded. In this case, both bidders can be high type, which means the seller earns H. If either one is a low type, then the seller’s revenue is only the bid of the low type, that is L. Thus, we obtain the above expected revenue for the seller for not offering a BIN price. Next, we need to show that all of the three possible BIN prices raise more revenue for the seller than the expected revenue of no-BIN case.

$$ \begin{aligned} \Uppi(Auction)&=(1-p^{2})R+p^{2}\left[(1-\alpha)^{2}H+\left[1-(1-\alpha)^{2}\right]L\right]\\ \Rightarrow\Uppi(Auction)&=R-p^{2}R+p^{2}\left[L+(1-\alpha)^{2}(H-L)\right]\\ \Rightarrow\Uppi(Auction)&=R+p^{2}(L-R)+p^{2}(H-L)-2p^{2}\alpha(H-L)+p^{2}\alpha^{2}(H-L)\\ \Rightarrow\Uppi(Auction)&=R+p^{2}(H-R)-p^{2}\alpha(H-L)+p^{2}\alpha^{2}(H-L)-p^{2}\alpha(H-L) \end{aligned} $$

Now, we can compare the above expected revenue with B ₁ = R + p(H − R) − αp ²(H − L). Since p ²(H − R) < p(H − R) and p ²α²(H − L) − p ²α(H − L) < 0, it is easy to see that B ₁ > Π(Auction) for all values of p and α. Now, we compare the above expected revenue with B ₂ = H − α(1 − p)(H − R) − αp ²(H − L). Let us first show that B ₂ > B ₁.

$$ \begin{aligned} &B_{2}-B_{1}=(H-R)-\alpha(1-p)(H-R)-p(H-R)\\ &\Rightarrow B_{2}-B_{1}=(1-p)(H-R)-\alpha(1-p)(H-R)\\ &\Rightarrow B_{2}-B_{1}=(1-p)(1-\alpha)(H-R)>0 \end{aligned} $$

Thus, we can easily conclude that B ₂ > Π(Auction) for all values of p and α. Lastly, we compare the above expected revenue with B ₃ = H(1 − α) + α L.

$$ \begin{aligned} B_{3}&=H(1-\alpha)+\alpha L\\ \Rightarrow B_{3}&=H-\alpha(H-L)\\ \Rightarrow B_{3}&=R+(H-R)-\alpha(H-L)\\ \Rightarrow B_{3}-\Uppi(Auction)&=(H-R)-\alpha(H-L)-p^{2}(H-R)+2\alpha p^{2}(H-L)-\alpha^{2}p^{2}(H-L)\\ \Rightarrow B_{3}-\Uppi(Auction)&=(1-p^{2})(H-R)-\alpha(H-L)\left[1-2p^{2}+\alpha p^{2}\right]\\ \Rightarrow B_{3}-\Uppi(Auction)&=(1-p^{2})(H-R)-\alpha(H-L)\left[1-(2-\alpha)p^{2}\right] \end{aligned} $$

since (1 − p ²) > [1 − (2 − α)p ²] and (H − R) > α(H − L)

$$ \Rightarrow B_{3}-\Uppi(Auction)=(1-p^{2})(H-R)-\alpha(H-L) \left[1-(2-\alpha)p^{2}\right]>0 $$

Thus, we can easily conclude that B ₃ > Π(Auction) for all values of p and α. As a result, the seller always prefers to offer a Buy-It-Now option to the bidders since his or her expected payoffs are always greater than not offering it.