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Final Jeopardy Wagering Strategy
Are contestants throwing away a chance at winning?
If you watched IBM's Watson supercomputer take on Ken Jennings and another former champion on Jeopardy, you probably noticed that when given the chance, Watson made odd wagers on Daily Doubles and in Final Jeopardy. Watson would wagen something like $5,389 instead of just a nice round number like $5,000. The reason for this is that IBM programmers applied basic game theory to Watson's wagering algorithm to give him the best chance of winning.
That shouldn't be surprising. What is surprising to me is that more contestants don't try applying some simple game theory of their own to improve their odds of winning. That doesn't mean they need to memorize slick formulas and make esoteric wagers, because there is some very simple math that thousands of contestants could have used throughout the years to win instead of going home with a consolation prize.
Why second place contestants are wagering poorly in Final Jeopardy
Here's a typical Final Jeopardy scenario: the person in first has $16000, the person in 2nd has $12000, and the person in third is far behind with $2000. You've probably seen this played out so many times you already know what everyone will wager. Third place will wager $1999, so even if they get the final clue wrong they still have $1 and maybe get 2nd place (if someone wagers it all and gets it wrong and ends up with $0) and take home the $2000 consolation prize instead of the $1000 prize for 3rd place. The person in 2nd place wagers everything ($12000). The person in 1st wagers $8001. That way, if they get it right and the 2nd place person gets it right, the person in 1st wins by $1 ($24001 to $24000). It's happened so many times it's boring to watch.
And best of all, this strategy is wrong. Here's why:
With this wagering strategy (which assumes the chances of getting the final Jeopardy question right are a coin flip and are the same for all contestants), the person in 2nd place is only giving themselves a 1 in 4 chance of winning. That is, they win if they get it right and 1st place gets it wrong. All other outcomes result in 2nd place not winning. Believe it or not, they can do better than this. They can make a wager here that would have them winning in 2 out of 4 outcomes, assuming the 1st place contestant wagers just enough to win by $1 -- which is what they typically do, because nobody wants to go into final Jeopardy with the lead, get the question correct, and then not win because they didn't wager enough.
So what is the right wager in Final Jeopardy?
If the person in 2nd place going into Final Jeopardy has at least 2/3rds of the total of the 1st place contestant and the 3rd place contestant either didn't make Final Jeopardy or is far out of the running, the 2nd place contestant should use the following formula for their wager:
3y - 2x
x = 1st place total
y = 2nd place total
Using this formula with the example above, where 1st place had $16000 and 2nd place had $12000, the second place contestant will wager the following:
3y - 2x
3(12000) - 2(16000)
36000 - 32000 = 4000
There are 4 scenarios here:
1st gets the question correct, 2nd gets the question correct
1st gets the question correct, 2nd gets the question wrong
1st gets the question wrong, 2nd gets the question correct
1st gets the question wrong, 2nd gets the question wrong
Assuming the person in 1st wagers $8001, they will only win if they get the question correct. 2nd place will win if 1st place gets the question wrong, even if 2nd place also gets the question wrong. This is an improvement over typical strategy. Plug this into other scenarios where 2nd place has at least 2/3 of the total of 1st place, and you'll find the strategy works.
Summary: if you're ever on Jeopardy and find yourself in 2nd place but close to 1st place, just remember this formula for wagering in Final Jeopardy:
3 times the second place total, minus 2 times the 1st place total