# Mixed strategies and wagering theory

Welcome back to The Final Wager’s Guide to Game Theory. In this installment, we’re finally going to resolve a cliffhanger by looking at mixed-strategy Nash equilibria.

We return to the concept of minimax, which we discussed in a previous installment. Minimax is used to find your optimal approach in a zero-sum game by minimizing your opponent’s maximum return.

We’ve already looked at sequential games like chess; here, we’re going to look at simultaneous games, like Final Jeopardy! wagering. We solve for the optimal approach by using a probability distribution.

There are two main types of Nash equilibria. Think of what we saw in the Prisoner’s Dilemma: each player will always confess. This is called “pure strategy” – you do the same thing every time.

The second is called “mixed strategy”, where you randomly choose from among your options with a predetermined probability. Between pure and mixed strategies, every game has a Nash equilibrium.

We’ll come back to our favorite wagering scenario, where the leader has 20,000 and the trailer has 14,000. Each player has two wagering ranges: small and large.

Calculating the mixed-strategy Nash equilibrium is a three step process. First, we’ll assign a variable to each of our opponent’s decisions.

We return to the payoff matrix. On the left, in the green boxes, we’ll come up with some variables. Let’s call the probability that the leader wagers “small” “y”. Since the leader must pick one or the other, the probability of these two must sum to 1. Therefore, we represent the probability that he wagers “large” as “1 minus y”.

We do the same with “x” for the trailer.

Step two: determine your payoffs for each of your opponent’s decisions. Let’s look at it from the trailer’s perspective to see how he should wager. We start with when the trailer goes “small” – zero. If the leader wagers zero, the trailer’s payoff will be -4. We multiply that by the probability that the leader wagers zero – “y”.

If the leader instead wagers large, the trailer’s expected payoff will be zero. We multiply that by the probability of the leader’s large wager – “1 minus y”.

When the trailer wagers zero, therefore, he expects his payoff to be “negative 4 y” plus zero times the quantity “1 minus y”.

We do the same thing with the trailer’s large wager. His expected payoff will be 0 times y, for when the leader wagers small, plus negative 2 times the quantity “1 minus y”, for when the leader wagers big.

Still with me? OK, good.

Our final step is to set the two formulas equal to each other.

We’ll walk through the algebra in the video. The equation reduces to 1/3. So y – the probability that the leader will wager small in the optimal strategy – is 1/3.

You can do the math on your own, but take my word for it that because this is a symmetrical game, the probability that the trailer will wager small is also 1/3.

We now return to our matrix, with our probabilities put in. Remember: the probabilities for a given player’s choices must add to 1, so if there’s a 1/3 probability a player will wager small, there must be a 2/3 probability he will wager large.

Now, to solve this, we need to calculate expected value, or EV. If you’re unfamiliar with this concept, have a look at this video.

When both players play the optimal mixed strategy, the leader will expect to win 2/3 of the time, and the trailer, 1/3 of the time.

Now let’s see what happens if one player deviates from the optimal strategy. If the trailer maintains his mixed strategy and the leader only wagers large, we calculate the probability of each happening, and sum the totals to find the expected value for each player. Again, it turns out to be 2/3 for the leader, 1/3 for the trailer.

The same is true when the leader always wagers large, when the trailer always wagers small, and when the trailer always wagers large.

Let’s return one last time to our matrix, and add in a new column and row for our mixed strategy. In any case when either player uses the mixed strategy, the leader can expect to win 67% of the time, and the trailer can expect to win 33% of the time.

Let’s draw in the arrows to show how this works.

Same as before: if we follow the blue arrows around the outside of the matrix, we enter an endless pattern.

But see what happens when we highlight the middle cell, where both players choose the mixed strategy: neither player can do better by unilaterally changing his choice. Therefore, both players playing a mixed strategy is a Nash equilibrium – which we call, fittingly, a mixed-strategy Nash equilibrium.

How does one choose randomly? I like to use the second hand on my watch: if it’s a multiple of 3, I might pick the “small” wager; otherwise, I’ll pick the “large” wager. Poker players do this sometimes when they want to randomize their play. It’s a neat little trick.

And we’ll learn about more neat tricks in future installments of The Final Wager.

One thing is unclear to me. You state: “Our final step is to set the two formulas equal to each other.” but you don’t explain why you are logically entitled to do this. I know that this step is crucial to being able to determine the probability values of 2/3 for large and 1/3 for small, but why does it logically follow that the two formulae SHOULD or MUST equal eachother?

It seems to me that the probability of the leading playing betting large is more reliably derived from past history. In the last 3 seasons the leader has bet to cover second place 84.16%, 80.17% and 90.0% … all significantly more than the 66.7% that your theory would suggest. (Although I recognize most of these games are 3 player games, not 2 player games.)

Your analysis does, however, suggest a possible alternative wagering strategy by the leader based primarily on the leaders assessment of the difficulty of the final jeopardy catecory.

It would go something like this.

In general, Final Jeopardy is only guessed correctly about 50% of the time. If, as the leader, after you see the final jeopardy category, you believe the answer/question will more difficult than average (less than 50% chance of getting it right), you should strongly consider betting small (say 3999 in this example) particularly if doing so will lock out the 3rd place player (or he is already locked out).

If the second place player bets small enough to win if you both get it wrong and you bet large (which, in general is the recommended strategy for both the leader and the follower) you will win or tie 100% of the time with a small wager by the follower, while still winning 50% of the time if he bets large to counter the possibility of you betting small. By betting small you actually improve the probability of moving on in this situation. You give up some potential for maximum $ won today, and you no longer “control your own destiny” but if you really think the category is difficult, betting small is best strategy probability-wise, based solely on the perceived difficulty of the answer/question and not on “mind games.”

One final caveat…if you’re wrong about the difficulty assessment…and player B gets it right AND bets large, you will lose even if you get it right. Would you feel really bad if this happened? Would you be perceived as a “chicken” by your friends, family and Jeopardy fans? If you don’t want to risk the possibility of getting it right and losing, then you have to go with the lock-out bet, no matter what your assessment of the odds tells you.

We need to figure out the variable; to do that, we set the known formulas equal to each other and solve for the variable.

The example assumes you are both flipping coins to determine whether you gain or lose points. Any additional information — the probability you or your opponent will respond correctly, your opponent’s strategy — will affect the mixed strategy.

Keith:

Just to make sure you don’t misunderstand my previous post, I didn’t mean to imply that you are NOT logically entitled to assume that the two payoffs are equal. I’m assuming that you probably have a good reason to assume this. I’m just looking for a little explanation regarding why you can assume this. I’m guessing that the explanation might be something like this.

It has been proven mathematically that all zero sum games such as this have a mixed strategy Nash equilibrium and that the equilibrium occurs where the payoffs from the two strategies are equal.

The other question still remains, however. Let’s say as player B (the follower) that you notice by repeatedly playing this game against one player A that he always bets large (the lock-out bet). Isn’t your optimal strategy to always bet small in this case? That way it would seem you would improve your odds to 50%, winning in the WR and WW cases while losing in the RR and RW cases. (Winning whenever the A player gets it wrong, which is the basis of your strategy #2.)

I made a spreadsheet to calculate the optimum betting strategy for the follower if he knows the leader’s betting “track record”. As you indicated, if the leader bets large 2/3rds of the time, it doesn’t matter from a “probability of going on” perspective what he wagers. He will win only 1/3 of the time no matter what he bets. In this case I think he is justified in betting it all since 1/3 of a doubled score is worth more than 1/3 of a score of current score +/- small bet.

Interestingly (and as you might have suspected) if he knows the leader deviates from a 2/3 large strategy, his optimum strategy is to always bet small if the probability of the leader betting large is more than 2/3 and to always bet large if the probability of the leader betting large is less than 2/3.

Here are the results. If you want me to send you the full spreadsheet with all the calculations let me know.

Follower Strategy:

Always 2/3 Always

Large Large Small

If leader bets:

Probability of Follower Winning

100.00% Large 25.00% 33.33% 50.00%

90.00% Large 27.50% 33.33% 45.00%

80.00% Large 30.00% 33.33% 40.00%

70.00% Large 32.50% 33.33% 35.00%

66.67% Large 33.33% 33.33% 33.33%

65.00% Large 33.75% 33.33% 32.50%

60.00% Large 35.00% 33.33% 30.00%

50.00% Large 37.50% 33.33% 25.00%

40.00% Large 40.00% 33.33% 20.00%

30.00% Large 42.50% 33.33% 15.00%

20.00% Large 45.00% 33.33% 10.00%

10.00% Large 47.50% 33.33% 5.00%

0.00% Large 50.00% 33.33% 0.00%

Your analysis is correct: if the trailer knows the leader will randomly select a “large” wager more than he should, the trailer should always select the “small” wager.

Keith:

The table in my previous post got garbled since leading spaces were ignored. The first 3 and 5th line should be centered over the last 3 columns, while the 4th line should be to the far left. Is there any way of fixing this on your end? And for future reference, is there anything I could do differently on a future post to avoid this problem?

Thanks,

Ken