As a refinement of Nash equilibrium, it is known that not all Nash equilibria are subgame perfect. But it seems to me in zero-sum games of perfect information, Nash equilibrium coincides with subgame perfect equilibrium. Is it true for all zero-sum games?
[Math] n example of zero-sum game that has a Nash equilibrium which is not subgame perfect
game theorynash-equilibrium
Related Solutions
Below the line is an answer to the original question, which was interesting because of subtle issues of indifference. The question has now been edited to include the prescription that indifferent players vote "no".
In that case there is no equilibrium in the continuous version of the game: At the last stage, player $7$ always votes "no", so the payoff is $(-1,100)$. At the penultimate stage, player $6$ always votes "yes", so the payoff is $(100,0,0)$. At the stage before that, there is no equilibrium, since offering something to player $6$ or $7$ is strictly dominated by offering half as much, and offering nothing to either of them is strictly dominated by offering something to one of them.
In the discrete case, the analysis is the same for the last two stages. Player $4$ must offer $1$ to either player $6$ or player $7$, with payoff $(99,0,1,0)$ or $(99,0,0,1)$. Player $3$ must offer $1$ to the two players who would otherwise get nothing, with payoff $(98,0,1,0,1)$ or $(98,0,1,1,0)$. Player $2$ must offer $1$ to the two players who would otherwise get nothing and $2$ to one of the players who would otherwise get $1$, with payoffs $(96,0,1,2,1,0)$, $(96,0,1,0,1,2)$, $(96,0,1,2,0,1)$ or $(96,0,1,0,2,1)$. Finally, player $1$ must offer $1$ to the players who would otherwise get $0$ and $2$ to one of the players who would otherwise get $1$, with payoffs $(96,0,1,2,0,0,1)$, $(96,0,1,0,0,2,1)$, $(96,0,1,2,1,0,0)$, $(96,0,1,0,1,2,0)$, $(96,0,1,2,0,1,0)$, $(96,0,1,0,0,1,2)$ or $(96,0,1,0,1,0,2)$. In all cases, the payoffs for players $1$ through $3$ are $96$, $0$ and $1$, respectively, and the remaining $3$ are divided in various ways among players $4$ through $7$.
In both versions, continuous and discrete, the question "Using the solution concept of subgame perfect Nash equilibrium, what happens?" doesn't make sense, since there is no unique subgame perfect Nash equilbrium in either.
This is a great question; it's interesting and quite relevant to understanding the underlying concepts, and you put considerable effort into describing what you've tried; I wonder why it hasn't received more upvotes.
crf's solution is the "right" solution in practical terms, since one would always offer otherwise indifferent players a small incentive to make sure they don't take an adverse decision; but I understood the question to be more formally about how to treat indifference in finding a (weak) subgame perfect Nash equilibrium. The answer to that is somewhat different, namely that the only subgame perfect Nash equilibrium outcome of the continuous version of this game is where player $1$ assigns herself all the money, without any infinitesimal bribes subtracted, whereas the discrete version (where the proposals have to be integer dollar amounts) has a plethora of subgame perfect Nash equilibria, in which various players are bribed with various small amounts of money if they threaten to vote "no" otherwise.
The reason lies in how the solution concept of a subgame perfect Nash equilibrium treats credible and non-credible threats and promises (let's call them announcements). An announcement is not credible if the player would be worse off by actually carrying it out. Whereas a plain Nash equilibrium allows strategies involving non-credible announcements, a subgame perfect Nash equilibrium doesn't. However, it still allows announcements about what you'll do when you're indifferent; that is, if in a subgame a player is indifferent between two options, either of the two options can be part of a strategy in a subgame perfect Nash equilibrium.
In the continuous version of the present game, none of the solutions found by crf are equilibria, since for any given $\epsilon\gt0$ player $1$ could always choose a smaller $\epsilon$ and get a higher payoff. So the question is whether there's a subgame perfect Nash equilibrium where player $1$ gets all the money. The crucial point here is that the condition for a (weak) Nash equilibrium (subgame perfect or not) is only that the players can't profit from switching strategies. Whether strategies are in equilibrium is not affected by whether one player switching their strategy would make it profitable for another player to also switch their strategy.
In the present case, if player $1$ allocates all the money to herself and everyone else's strategy is to always vote "yes", then no player can profit from switching strategies, so this is a (weak) subgame perfect Nash equilibrium. The fact that players threatening to vote "no" would make it profitable for player $1$ to offer them something doesn't matter; it's not profitable, given the given strategies, for player $1$ to switch strategies.
Here's the analysis in detail. At the last stage, with two players left, the equilibrium strategy profiles are: player $6$ makes any allocation and player $7$ votes "no" no matter what player $6$ does; or player $6$ allocates all the money to player $7$ and player $7$ votes yes if and only if player $6$ allocates all the money to her. The payoffs are $(-1,100)$ in the first case and $(0,100)$ in the second.
At the penultimate stage, with three players left, subgame perfection eliminates non-credible threats by player $6$ to vote "no" to any proposal allocating her any money (since she'd get at most $0$ if the proposal fails), so there can't be an equilibrium allocating any money to player $6$, since player $5$ would be better off allocating half as much to player $6$. Thus in equilibrium player $5$ must allocate all the money to herself. Of course player $7$ votes "no", but player $6$ can't profit from voting "no", so player $6$ voting "yes" yields an equilibrium. (This is independent of which of the two equilibria for the last stage we select; for $(0,100)$ player $6$ is indifferent and for $(-1,100)$ she even profits from voting "yes".) The strategy of player $6$ in case player $5$ offers her something can be anything since it doesn't affect the outcome, so there's a continuum of equilibrium strategy profiles, but they all have the same payoff, $(100,0,0)$.
The analysis proceeds analogously at all stages. With player $4$ proposing, player $5$ will always vote "no", and there can't be an equilibrium where player $4$ offers anything to player $6$ or $7$ because she could offer half as much instead (non-credible threats to vote "no" in that case being eliminated by subgame perfection); so the only equilibrium is that player $4$ allocates all to herself, player $5$ votes "no" and players $6$ and $7$ vote "yes"; again the strategies of players $6$ and $7$ if player $4$ offers them something are arbitrary and don't affect the outcome, which is always $(100,0,0,0)$.
At the next stage, with player $3$ proposing, player $4$ votes "no", and any of the four subsets of players $5$ through $7$ with at least two players voting "yes" yields an equilibrium (as always, the strategies in case player $3$ offers them something being arbitrary). The payoff is $(100,0,0,0,0)$, and the analysis continues, with more different strategy profiles in equilibrium at each stage since there are more majority subsets at each stage, but the payoff is always the same: $(100,0,0,0,0,0)$ at the second stage and $(100,0,0,0,0,0,0)$ at the first.
In the discrete version, this is still a subgame perfect Nash equilibrium, but now there are also ones similar to crf's result.
Again, at the last stage, there are two equilibrium payoffs, $(0,100)$ and $(-1,100)$. At the penultimate stage, in the $(-1,100)$ case player $6$ will always vote "yes" to any proposal, so the only equilibrium payoff is $(100,0,0)$, whereas in the $(0,100)$ case player $1$ is indifferent between $(100,0,0)$ and $(0,100)$ and can therefore threaten to vote "no" when offered $0$, but can't threaten to vote "no" when offered $1$, so in this case there's an additional equilibrium with payoffs $(99,1,0)$.
With player $4$ proposing, in the $(100,0,0)$ case there are additional equilibria with payoffs $(99,0,1,0)$, $(99,0,0,1)$ and $(98,0,1,1)$, where in each case the players offered $1$ threaten to vote "no" if offered $0$. In the $(99,1,0)$ case, there are additional equilibria with payoffs $(98,0,1,1)$, $(98,0,2,0)$ and $(97,0,2,1)$. In summary, the possible equilibrium payoffs at this stage are $(100-x-y,x,y)$, with $0\le x\le2$ and $0\le y\le1$.
At the next stage, with player $3$ proposing, there is for the first time a choice of majorities, and player $3$ can pick the cheapest majority to bribe. The equilbrium payoffs are:
$$ \begin{array}{c|c} \text{stage }4&\text{stage }3\\\hline (100,0,0,0)&(100,0,0,0,0),(99,0,0,0,1),(99,0,1,0),(99,1,0,0),\\&(98,0,1,1),(98,1,0,1),(98,1,1,0)\\\hline (99,0,0,1)&(99,0,0,0,1),(98,0,0,1,1),(98,0,1,0,1),(97,0,1,1,1)\\\hline (99,0,1,0)&(99,0,0,1,0),(98,0,0,1,1),(98,0,1,1,0),(97,0,1,1,1)\\\hline (98,0,1,1)&(98,0,0,1,1),(97,0,0,1,2),(97,0,0,2,1),(97,0,1,1,1),\\&(96,0,1,2,1),(96,0,1,1,2)\\\hline (98,0,2,0)&(98,0,0,2,0),(97,0,1,2,0),(97,0,0,2,1),(97,0,1,2,1)\\\hline (97,0,2,1)&(97,0,0,2,1),(96,0,1,2,1),(96,0,0,2,2),(96,0,1,2,2)\\ \end{array} $$
I'll leave the boring details of the first two stages to you :-)
I think the point of this problem is to illustrate the difference between a Nash equilibrium and a subgame-perfect Nash equilibrium.
For subgame-perfect equilibria, the situation is clear. Player $2$ cannot make an empty threat to play $(0,0)$ in the left-hand branch; she plays $(5,5)$, and thus Player $1$ plays $A$; there is no other equilibrium.
If the Nash equilibrium doesn't have to be subgame-perfect, the situation is quite a bit more complicated. Player $2$ can threaten to play $(0,0)$ with non-zero probability $1-p$ in the left branch, thus making this branch worth an expected payoff of $5p$ for Player $1$. In the right-hand branch, as you wrote, Player $1$ will always play "No", and Player $2$ can play L with probability $q$ and NL with probability $1-q$, giving an expected payoff of $3q+4(1-q)=4-q$ to Player $1$.
There are two types of equilibria, depending on whether Player $1$ prefers A or not.
Player $1$ prefers A: Here Player $2$ must choose $p=1$, since the left-hand branch actually gets played. $q$ can be anything, since the right-hand branch doesn't get played and any value of $q$ lets Player $1$ prefer $A$ if $p=1$.
Player $1$ doesn't prefer A: Here Player $2$ can have mixed strategies in both branches, subject to the condition $4-q\ge5p$ so that Player $1$ doesn't prefer A. In these equilibria, Player $1$ always plays B. Note that this includes (weak) equilibria with $4-q=5p$ in which Player $1$ is indifferent between A and B. Nevertheless, for an equilibrium Player $1$ must play B and cannot mix strategies, since Player $2$'s empty threat in the left branch wouldn't be in equilibrium otherwise.
Best Answer
It is true that the set of Nash equilibrium and subgame-perfect equilibrium outcomes (distributions over terminal nodes) are equivalent for zero-sum games. However, the sets of equilibrium strategies need not be equivalent. Under subgame-perfect equilibrium, any off-equilibrium play is permitted.
Consider an extensive game $\Gamma$ and some subgame-pefect equilibrium strategies in which some subgame is reached with probability $0$. A Nash equilibrium strategy which induces the same outcome can be achieved by modifying some player's strategy to be non-credible in the subgame which is never reached along the equilibrium path.
Here is an example. Let player 1 move first. If 1 moves left, the game ends and payoffs are (5,-5). If player 1 moves right, player 2 moves. If player 2 moves left, payoffs are (3,-3) and if player moves right, payoffs are (2,-2).
Moving right is strictly dominated by left for player 1. The only NE and Subgame outcome is (5,-5) with probability 1. However, while (Left, Right) is the only subgame perfect strategy profile, (Left, Right) and (Left, Left) are Nash equilibrium strategies.