One of the most fundamental concepts in poker is "The Power of Position." The player who acts last generally should beat the player who acts first due to the imbalance of knowledge. Any data-driven poker player could tell you that their win rate is highest on the button (the last position on the table), lowest from the blinds (the first positions), and generally cascades down the positions in between. It is not rocket science -- they win more when they have the advantage of seeing how their opponent acts and lose more when it's the other way around.
This also applies to negotiations, as Chris Voss wrote about in his book Never Split The Difference. Voss, a former hostage negotiator for the FBI, advocated for trying to withhold from making the first offer. The person throwing out the first offer can only meet or exceed the expectation of the person receiving it, but the person receiving does not have a take an offer they deem to be unfavorable. If you offered more money for the car than what the salesmen would have sold it for, then the salesmen will just accept the higher offer. If you offered less, then he will just counter with what he wanted for it anyway. It is what's known a reverse freeroll, a contest you could only lose or tie but never win.
Suppose I told you The Power of Position also applies to one of America's favorite game shows as well, The Price is Right. Recently, The Riddler on 538 asked the following riddle:
In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.
Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price without going over. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.
In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, someone has to win, right?
If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?
The answer was that each contestant would have an exactly even 1/3 chance of winning an optimal game. However, the riddle overlooks one important element of the actual show: players cannot guess fractional dollars. In other words, the first player cannot wager the item to be worth $66.66666666666 because you can't split pennies like that. If all three players evenly bid $33.33 as The Riddler suggests, then there would still be one additional penny that only one player could win with. Which player is it though?
To demonstrate the game theory solution, I am just going to round all wagers to the whole dollar (in actuality you cannot bet any change on The Price is Right). The solution works whether you use change or not, but I just think the whole dollar makes an easier demonstration. If we changed the rules of the riddle to only allow whole dollars between $1 - $100, then The Riddler solution would assume to just round up the $66.67 ($100 - 1/3) up to $67 to get the same optimal answer. Except that would be wrong -- it's actually $68.
There's a critical game theory element the original answer overlooked, which is the idea of indifference. A Nash Equilibrium occurs in game theory when a player is indifferent between two equal decisions. The goal of Player 1 on The Price is Right is not to create a Nash Equilibrium for P2 and P3 but to create one optimal answer for them to always choose. This is because of the threat of what I will term as "sniping," which is when a player goes $1 over another player's wager to take their bet's coverage area for themselves. When a player gets sniped, the only chance they would have to win is guess the actual price perfectly.
Don't worry, it is actually possible!
Sidebar: I love the guy who heard the first guy confidently say, "It's $2148. I saw it on the show last week," thinks it over for several seconds, and casually comes up with a bid of...$1800!?!?
Back to our riddle, let's say P1 had bid $67, which would lock in the 34 spots from $67 - $100 and leave 66 spots for P2 and P3 to split beneath him. P2 would now have three realistic options: bid $1, split the difference under P1, or snipe P1.
While bidding $1 temporarily gives 66 spots of coverage, we know P3 would immediately snipe at $2 to lock in 65 spots for himself. So the only two realistic options are split the difference or snipe P1, which are both worth 33 spots. This is a Nash Equilibrium for P2, which means he should randomly pick either one. That poses a problem for P1 because if P2 is going to snipe half the time, then the expected value wagering $67 is not 34 spots but actually 17.5. P1 can only avoid this by moving up $1 to $68, which would leave 32 spots above and thus make P2 never want to snipe.
EV (P1 wagers $67) = [50% * 34 spots] + [50% * 1 spot] = 17.5 spots
EV (P1 wagers $68) = [100% * 33 spots] + [0% * 1 spot] = 33 spots
So we know P1 optimally bids $68 and leaves 67 spots below, which changes P2's options to this:
P2 could also try to wager $34 instead of $35 on the splitting option, which would get 34 spots instead of 33. However, he would now face the same dilemma P1 had. If he chooses $34 (34 spots up to $67), then P3 would now be indifferent between $1 and sniping P2 at $35 because they would both be worth 33 spots. So P3 would now choose the two options at random.
P2 knows he cannot leave P3 indifferent between the two options in order to maximize his own EV. Therefore, P2 must go $1 higher to lock in the guaranteed 33 spots between $35 and $67 and ensure he is never sniped. This finally leaves P3's options as follows:
P3 obviously will choose $1 since it has more coverage than sniping either P1 or P2. The game has finally reached its optimal solution with P3 winning 34% and P2 and P1 winning 33% each, with both P1 and P2 having willingly conceded the extra $1 to P3.
So the next time you're watching The Price is Right and the crowd is yelling at the last contestant to wager $1, remember they're not just trolling. They are actually reminding them to use The Power of Position!
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