Who’s the weakest link?

In the Weakest Link game show, eight contestants answer a series of trivia questions to accumulate a “bank” of money, with one contestant (the “weakest link”) voted off at each round. When there are 2 contestants remaining, they face off for a series of 5 questions each, with the winner receiving the entire amount that was banked. In theory the champion could win “up to a million dollars,” but in practice the total bank ends up being in the $40k-$80k range.

For the first several rounds, it makes a lot of sense to vote for players who are actually the “weakest,” since they will be less likely to answer questions correctly and contribute to increasing the amount in the bank throughout the game. But in the final voting round (when 3 contestants remain), it becomes pretty clear that you should actually vote off the “strongest” player so that you can go up against a weaker opponent in the final round.

However, this analysis for the final voting round makes several assumptions, which may not hold in practice. First, it assumes that it is clear to you, and to the other players, who the strongest player remaining actually is (and if it is you, then you are screwed). This may be difficult to assess over the relatively small sample of questions each player is given during the game. For example, player A may have correctly answered more questions than player B, but A might have also received easier questions. Furthermore, while it may be evident to you that player A is the strongest contestant, it may not be obvious to player B. If B incorrectly perceives you to be stronger than A and votes for you, while A votes for B, then it is clearly optimal for you to vote for B despite the fact that B is weaker than A.

A second issue is that, regardless of whether the opponents’ perceptions of abilities are correct, they may not understand that they actually want to eliminate the strongest player as opposed to the weakest player in the final voting round. While it seems obvious that one wants to go head-to-head against the weaker remaining opponent, often I see players voting for the clearly weaker remaining opponent. And in the interview of the final contestant eliminated, often their explanation makes it clear that they are not aware of the fact that players would prefer to vote off the strongest player in the last round.

Considering these additional factors, it may actually be optimal under certain circumstances to vote for the weaker remaining contestant, as opposed to the “obviously optimal strategy” of voting for the strongest one (I already gave one example above).

To construct our model, suppose that the total amount of money in the bank to be awarded to the winner is W>0 (the loser gets $0). Suppose that if you are head-to-head against opponent 1 you will win with probability p₁, and against opponent 2 you will win with probability p₂. Assume that player 2 is stronger than player 1, so that p₁ > p₂. Finally, assume that player 1 will vote for you with probability y₁ (and therefore will vote for player 2 with probability 1-y₁), and that player 2 will vote for you with probability y₂. We will assume that obviously no player will vote for themselves.

If there is a three-way tie, we will assume that each player is voted out with probability 1/3. (In reality, the “statistically strongest link” from the previous round gets to cast a tie-breaking vote, which is obviously very relevant, but for simplicity I will ignore this aspect of the problem to simplify the analysis). So under this assumption, our expected payoff in the case of a tie equals:

1/3 (W * p₁) + 1/3 (W * p₂) + 1/3 * 0 = W(p₁ + p₂)/3

Given this model and assumptions, we now compute your optimal voting strategy. Observe that if both players vote for you, then your vote is irrelevant, since you will be eliminated regardless. So the only relevant cases to consider are when P1 votes for you and P2 votes for P1, and when P2 votes for you and P1 votes for P2.

Case 1 (P1 votes for you, P2 votes for P1):

If you vote for P1, you will go head-to-head against P2 and obtain expected payoff (W * p₂).

If you vote for P2, it will be a three-way tie and you will obtain expected payoff W(p₁ + p₂)/3, which was calculated above.

Case 2 (P2 votes for you, P1 votes for P2):

If you vote for P2, you will go head-to-head against P1 and obtain expected payoff (W * p₁).

If you vote for P1, it will be a tie and you obtain W(p₁ + p₂)/3.

Assuming we are in either case 1 or case 2 (since the other cases are irrelevant, as I showed already), the probability that we are in case 1 is

y₁(1-y₂), and the probability we are in case 2 is y₂(1-y₁). We need to normalize these so they sum to 1, so the real probability we are in case 1 is:

y₁(1-y₂) / (y₁(1-y₂) + y₂(1-y₁))

And the probability we are in case 2 is:

y₂(1-y₁) / (y₁(1-y₂) + y₂(1-y₁))

Putting this all together, our expected payoff of voting for player 1 is:

[y₁(1-y₂) / (y₁(1-y₂) + y₂(1-y₁))] * (W * p₂)

+[y₂(1-y₁) / (y₁(1-y₂) + y₂(1-y₁))] * W(p₁ + p₂)/3

= [y₁(1-y₂) * (W * p₂) + y₂(1-y₁) * W(p₁ + p₂)/3]

/ [(y₁(1-y₂) + y₂(1-y₁))]

Similarly, our expected payoff of voting for player 2 is:

[y₁(1-y₂) / (y₁(1-y₂) + y₂(1-y₁))] * W(p₁ + p₂)/3

+[y₂(1-y₁) / (y₁(1-y₂) + y₂(1-y₁))] * (W * p₁)

= [y₁(1-y₂) * W(p₁ + p₂)/3 + y₂(1-y₁) * (W * p₁)]

/ [(y₁(1-y₂) + y₂(1-y₁))]

So we should vote for player 1 if

[y₁(1-y₂) * (W * p₂) + y₂(1-y₁) * W(p₁ + p₂)/3]

/ [(y₁(1-y₂) + y₂(1-y₁))]

≥ [y₁(1-y₂) * W(p₁ + p₂)/3 + y₂(1-y₁) * (W * p₁)]

/ [(y₁(1-y₂) + y₂(1-y₁))]

We can multiply both sides by the denominator to eliminate it and obtain an equivalent condition:

[y₁(1-y₂) * (W * p₂) + y₂(1-y₁) * W(p₁ + p₂)/3]

≥ [y₁(1-y₂) * W(p₁ + p₂)/3 + y₂(1-y₁) * (W * p₁)]

If we multiply through and expand both sides, we obtain:

y₁Wp₂-y₁y₂Wp₂ + y₂Wp₁/3 + y₂Wp₂/3-y₁y₂Wp₁/3-y₁y₂Wp₂/3

≥y₁Wp₁/3+y₁Wp₂/3-y₁y₂Wp₁/3-y₁y₂Wp₂/3+y₂Wp₁-y₁y₂Wp₁

We can simplify this to obtain:

y₁Wp₂ - y₁y₂Wp₂ + y₂Wp₁/3 + y₂Wp₂/3

≥ y₁Wp₁/3 + y₁Wp₂/3 + y₂Wp₁ - y₁y₂Wp₁

Multiplying both sides by 3:

3y₁Wp₂ - 3y₁y₂Wp₂ + y₂Wp₁ + y₂Wp₂

≥ y₁Wp₁ + y₁Wp₂ + 3y₂Wp₁ -3y₁y₂Wp₁

Simplifying further:

2y₁Wp₂ + y₂Wp₂ + 3y₁y₂Wp₁ ≥ 2y₂Wp₁ + y₁Wp₁ + 3y₁y₂Wp₂

2y₁p₂ + y₂p₂ + 3y₁y₂p₁ ≥ 2y₂p₁ + y₁p₁ + 3y₁y₂p₂

One immediate observation is that the optimal strategy does not depend on W. Regardless of whether the payoff is $10k or $1M, your optimal strategy should be the same, and should only depend on p₁, p₂, y₁, and y₂.

As an example, consider the case where p₁ = 1 and p₂ = 0 (you will always win head-to-head against player 1, and will lose head-to-head against player 2). The optimal equation becomes:

3y₁y₂ ≥ 2y₂ + y₁, or equivalently 3y₁y₂ - 2y₂ - y₁ ≥ 0

One can easily verify that 3y₁y₂ - 2y₂ - y₁ is always ≤ 0. So in this case, you always obtain at least as large a payoff by voting for player 2 as for voting for player 1.

As another example, suppose that p₁ = 3/4 and p₂ = 1/4 (you win 75% of the time head-to-head against player 1, and win 25% of the time head-to-head against player 2). The optimal equation now becomes:

y₁/2 + y₂/4 + 9y₁y₂/4 ≥ 3y₂/2 + 3y₁/4 + 3y₁y₂/4

Multiplying both sides by 4:

2y₁ + y₂ + 9y₁y₂ ≥ 6y₂ + 3y₁ + 3y₁y₂

Simplifying:

6y₁y₂ - y₁- 5y₂ ≥ 0

Again one can show that the payoff of voting for player 2 is at least as big as the payoff of voting for player 1(6y₁y₂ - y₁- 5y₂ is always ≤ 0).

Now consider the example p₁ = 0.6, p₂ = 0.4.

The optimal equation now becomes:

2y₁*0.4 + y₂*0.4 + 3y₁y₂*0.6 ≥ 2y₂*0.6 + y₁*0.6 + 3y₁y₂*0.4

Multiplying both sides by 5:

4y₁ +2y₂ + 9y₁y₂ ≥ 6y₂ + 3 y₁ + 6y₁y₂

Simplifying:

3y₁y₂ + y₁- 4y₂ ≥ 0

Or equivalently:

y₁ ≥ 4y₂ / (3y₂ + 1)

In this case, for every y₂, there exists region of values for y₁ for which voting for player 1 is optimal. For example, if y₂ = 0.5, then voting for player 1 is optimal if:

y₁ ≥ 4 * (0.5)/(3*0.5+1) = 0.8.

So for the situation p₁ = 0.6, p₂ = 0.4, the optimal strategy depends on the values of y₁ and y₂.

We can next consider the more general question: what are the values of (p₁,p₂) for which it is optimal to never vs. sometimes vote for player 1 (we showed cases of each situation above). We can see from the optimal equation that it is strictly optimal to vote for player 1 if there exist values of y₁ and y₂ for which:

f(y₁,y₂) = 2y₁p₂ + y₂p₂ + 3y₁y₂p₁ - 2y₂p₁ - y₁p₁ - 3y₁y₂p₂ > 0

So in other words, for fixed (p₁,p₂), it is never strictly optimal to vote for player 1 if (and only if) the maximum value of f(y₁,y₂) is less than or equal to zero.

The maximum of f is attained where y₁ = 0, 1, or df(y₁,y₂)/dy₁ = 0, and similarly y₂ = 0, 1, or df(y₁,y₂)/dy₂ = 0.

If y₁ = 0, then f(y₁,y₂) = f(0,y₂) = y₂p₂ - 2y₂p₁ = y₂(p₂-2p₁)

So f(0,y₂) > 0 if y₂ > 0 and p₂ > 2p₁. But since we are assuming that p₁ > p₂, this condition will never be satisfied.

If y₁ = 1, then

f(y₁,y₂) = f(1,y₂) = 2p₂ + y₂p₂ + 3y₂p₁ - 2y₂p₁ - p₁ - 3y₂p₂

= 2p₂ - 2y₂p₂ + y₂p₁ - p₁

If y₂ = 1 then f(1,1) = 0.

If y₂ <1, then f(1,y₂) > 0 iff p₂ > p₁/2.

So this shows that one set of conditions under which it is possible for us to strictly prefer to vote for player 1 is p₂ > p₁/2.

If y₂ = 0, then f(y₁,y₂) = f(y₁,0) = 2p₂ - p₁

So f(y₁,0) > 0 if p₂ > p₁/2, which is the same condition we just saw.

If y₂ = 1, then

f(y₁,y₂) = f(y₁,1) = 2p₂ + p₂ + 3p₁ - 2p₁ - p₁ - 3p₂ = 0.

The only remaining situation is when df(y₁,y₂)/dy₁ = df(y₁,y₂)/dy₂ = 0.

df(y₁,y₂)/dy₁ = 2p₂ + 3y₂p₁ - p₁ - 3y₂p₂

Setting this equal to zero gives us y₂* = (p₁ - 2p₂)/(3(p₁ - p₂))

Note that we must have 0 ≤ y₂ ≤ 1. y₂ ≥ 0 implies that p₁ ≥ 2p₂, and y₂ ≤ 1 implies that p₂ ≤ 2p₁ (which always holds since p₂ < p₁).

df(y₁,y₂)/dy₂ = p₂ + 3y₁p₁ - 2p₁ - 3y₁p₂

Setting this equal to zero gives us y₁* = (2p₁ - p₂)/(3(p₁ - p₂))

Note that we must have 0 ≤ y₁ ≤ 1. y₁ ≥ 0 implies that 2p₁ ≥ p₂ (which always holds since p₁ >p₂), and y₂ ≤ 1 implies that 2p₂ ≤ p₁.

Substituting these values into f, and multiplying through by [3(p₁ - p₂)]², gives us that f(y₁*,y₂*) > 0 if:

2(2p₁- p₂)p₂*[3(p₁ - p₂)] + (p₁ - 2p₂)p₂*[3(p₁ - p₂)] + 3(2p₁- p₂)(p₁ - 2p₂)p₁ - 2(p₁ - 2p₂)p₁*[3(p₁ - p₂)] - (2p₁ - p₂)p₁*[3(p₁ - p₂)]- 3(2p₁ - p₂)(p₁ - 2p₂)p₂ > 0

Multiplying through, we have:

12p₁²p₂ - 18p₁p₂² + 6p₂³ + 3p₁²p₂ - 9p₁p₂² + 6p₂³ + 6p₁³ - 15p₁²p₂ + 6p₁p₂² - 6p₁³ + 18p₁²p₂ - 12p₁p₂² - 12p₁³ + 18p₁²p₂ - 6p₁p₂² - 6p₁²p₂ + 15p₁p₂² - 6p₂³ > 0

Simplifying:

30p₁²p₂ - 24p₁p₂² + 6p₂³ - 12 p₁³ > 0

g(p₁, p₂) = 5p₁²p₂ - 4p₁p₂² + p₂³ - 2p₁³ > 0

Let us find the values (p₁, p₂) that maximize (p₁, p₂). This can occur when p₁=0, 1, or dg(p₁, p₂)/dp₁ = 0, and similarly for p₂ = 0, 1, or dg(p₁, p₂)/dp2 = 0.

We cannot have p₁ = 0, since our model assumes that p₁ > p₂.

If p₁ = 1, then g(p₁, p₂) = g(1, p₂) = 5p₂ - 4p₂² + p₂³ - 2.

This is always ≤ 0 for 0 ≤ p₂ ≤ 1.

If p₂ = 0, then g(p₁, p₂) = g(p₁, 0) = -2p₁³, which is clearly always ≤ 0.

We cannot have p₂ = 1, since p₁ > p₂.

Next we consider the case dg(p₁, p₂)/dp₁ = dg(p₁, p₂)/dp2 = 0.

dg(p₁, p₂)/dp₁ = 10p₁p₂ - 4p₂² - 6p₁²

Setting this equal to zero gives us:

3p₁² - 5p₁p₂ + 2p₂² = 0

(3p₁ - 2p₂)(p₁ - p₂) = 0

p₁ = p₂ or p₁ = 2p₂/3

Neither of these is possible since they contradict p₁ > p₂.

dg(p₁, p₂)/dp₂ = 5p₁² - 8p₁p₂ + 3p₂²

Setting this equal to zero gives us:

3p₂² - 8p₁p₂ + 5p₁² = 0

(3p₂ - 5p₁)(p₂ - p₁) = 0

p₂ = p₁ or p₂ = 5p₁/3.

Again neither of these is possible since they contradict p₁ > p₂.

So we have shown that the only set of conditions for which there exist values of (y₁,y₂) where voting for player 1 is strictly optimal is p₂ > p₁/2 (assuming that p₁ > p₂). If p₂ ≤p₁/2, then it is always optimal to vote for player 2 (the stronger player) regardless of your beliefs of the strategies taken by the other players.