Wagering Question - Four-Fifths Rule

[opusthepenguin]

I'm wondering if the J! Archive's entry on Four-Fifths should be revised or expanded or something. But I'm not sure what it should say. Currently, it says:

The four-fifths scenario is a special case of the two-thirds scenario in which a trailing player with a score at least four-fifths the score of the leading player going into Final Jeopardy! safely can, and therefore should, wager to beat the leader's maximum safe bet of the difference between scores with a bet of twice the difference between the two scores, without losing the double stumper against a shut-out bet by the leader. According to the Four-fifths Rule, the trailing player should wager between (2*(leader's score - own score) + $1) (venusian) and 3*(own score) - 2*(leader's score) (martian).

Suppose we have two players and an empty podium. The scores going into FJ are:

Player 1: $500
Player 2: $400

First, the rule refers to "the leader's maximum safe bet of the difference between scores"--i.e. $100 in this case. Does "maximum safe bet" just mean that's the maximum Player 1 can wager without dipping below Player 2's current score? Because I don't think it's a safe bet, let alone the largest safe bet. The wagering calculator says Player 1 should bet $300 in this scenario. That's larger and, as far as I can tell, no less safe than the "maximum safe bet" of $100.

Furthermore, isn't there a distinction between a scenario where Player 2 has exactly four-fifths vs more than four-fifths? At exactly four-fifths, I think Player 1 should bet to tie Player 2's doubled score--i.e. $300 in the case above. The shut-out bet of $301 is too risky. If Player 2 bets $200 (as the theory recommends), then Player 1 loses rather than ties on a double stumper. And if Player 2 bets all $400 (as such players often will), Player 1 ties rather than wins outright on the double get. So betting $301 means giving up advancing on a double stumper in order to advance by a win rather than a tie on a double get. Not a good trade.

But if Player 2 has more than four-fifths of Player 1, then a shut-out bet doesn't cost Player 1 any opportunities. Right?

[Bob78164]

I'm wondering if the J! Archive's entry on Four-Fifths should be revised or expanded or something. But I'm not sure what it should say. Currently, it says:

The four-fifths scenario is a special case of the two-thirds scenario in which a trailing player with a score at least four-fifths the score of the leading player going into Final Jeopardy! safely can, and therefore should, wager to beat the leader's maximum safe bet of the difference between scores with a bet of twice the difference between the two scores, without losing the double stumper against a shut-out bet by the leader. According to the Four-fifths Rule, the trailing player should wager between (2*(leader's score - own score) + $1) (venusian) and 3*(own score) - 2*(leader's score) (martian).

Suppose we have two players and an empty podium. The scores going into FJ are:

Player 1: $500
Player 2: $400

First, the rule refers to "the leader's maximum safe bet of the difference between scores"--i.e. $100 in this case. Does "maximum safe bet" just mean that's the maximum Player 1 can wager without dipping below Player 2's current score? Because I don't think it's a safe bet, let alone the largest safe bet. The wagering calculator says Player 1 should bet $300 in this scenario. That's larger and, as far as I can tell, no less safe than the "maximum safe bet" of $100.

Furthermore, isn't there a distinction between a scenario where Player 2 has exactly four-fifths vs more than four-fifths? At exactly four-fifths, I think Player 1 should bet to tie Player 2's doubled score--i.e. $300 in the case above. The shut-out bet of $301 is too risky. If Player 2 bets $200 (as the theory recommends), then Player 1 loses rather than ties on a double stumper. And if Player 2 bets all $400 (as such players often will), Player 1 ties rather than wins outright on the double get. So betting $301 means giving up advancing on a double stumper in order to advance by a win rather than a tie on a double get. Not a good trade.

But if Player 2 has more than four-fifths of Player 1, then a shut-out bet doesn't cost Player 1 any opportunities. Right?

I was the one who originally identified the four-fifths rule, so I'll respond. You're correct that there is a distinction between exactly four fifths and more than four fifths. The "maximum safe bet" is the largest wager by the leader that guarantees a win if the trailer does not respond correctly to Final Jeopardy!.

Either slam or Gneq (I think) later pointed me to a research paper from UCLA (if memory serves) demonstrating the both the two-thirds and the four-fifth rules generalize. --Bob

[opusthepenguin]

I was the one who originally identified the four-fifths rule, so I'll respond. You're correct that there is a distinction between exactly four fifths and more than four fifths. The "maximum safe bet" is the largest wager by the leader that guarantees a win if the trailer does not respond correctly to Final Jeopardy!.

Either slam or Gneq (I think) later pointed me to a research paper from UCLA (if memory serves) demonstrating the both the two-thirds and the four-fifth rules generalize. --Bob

Interesting. Thanks. As I play with the numbers, I'm wondering if this generalization works: At all Break Points over one-half--2/3, 3/4, 4/5, 5/6, etc.--the first place player should wager to tie a doubled score by second place. And the second place player should calculate what that wager would be for the first place player and wager to tie on a double stumper if that bet is made. In other cases, the first place player doesn't lose any opportunities by adding $1 to make the shut-out bet.

This would suggest the following wagers at Break Points.

Two-Thirds
Player 1 $300 --> $100 wager to tie Player 2's double
Player 2 $200 --> $0 wager to tie on double stumper

Three-Fourths
Player 1 $400 --> $200 wager to tie Player 2's double
Player 2 $300 --> $100 wager to tie on double stumper

Four-Fifths
Player 1 $500 --> $300 wager to tie Player 2's double
Player 2 $400 --> $200 wager to tie on double stumper

Five-Sixths
Player 1 $600 --> $400 wager to tie Player 2's double
Player 2 $500 --> $300 wager to tie on double stumper

It seems to me that this generalization helps each player guard most effectively against players who don't use the rule, without sacrificing any opportunities against those who do. The wagering calculator agrees with me except in two cases. First, it suggests an all-in bet by Player 2 in the two-thirds scenario is also good. I disagree. That loses 6 out of 8 times against the shut-out bet. Then on Five-Sixths, the wagering calculator suggests the shut-out bet for Player 1. But that throws away a chance to win on a double stumper (if Player 2 bets $300) in order to gain a win rather than a tie on a double get (if Player 2 bets $500). Just as we argued with Four-Fifths, that's not a good trade.

If I'm right, do I get a rule named after me? If so, I'd like to point out that "Opus's Rule" sounds a little clumsy. I'd recommend Regula Operis.

[jeff6286]

I may be missing something obvious here, but in the 4/5ths and 5/6ths cases you mention, why would second place bet to tie on a double stumper, rather than to win? At $500 to $400, a total of 600 vs 599 would not seem to make much difference, so couldn't the trailer bet 199 so that they would win 201 to 200 on a double stumper? Any bet from 0 to 199 would seem to have the same effect, so why would the leader be expecting a bet of 200 from second place to begin with? Is the theory simply that the leader is anticipating a savvy wager by second place, so if second plans to win by a dollar if first makes the shutout bet, first counters this by taking off the extra dollar and thus winding up in a tie? While that is a nice idea in theory, I don't foresee it being successful in a very high percentage of cases, due mainly to the wild unpredictability of second place wagers.

[GMB]

I believe the wagering calculator suggest the 4/5ths wagering at the breakpoints, so that second will tie with first on the triple stumper if first place bets rationally, but also win on a double get if first "underwagers" with the maximum safe bet. However, I agree, from second place the chances that the first place person will bet the "maximum safe bet" is so small, that second does not really have anything to lose by betting small.

As for the "two thirds breakpoint" the wagering calculator suggests all or nothing, because if 2nd bets all and 1st offers the tie, B wins/ties for the win in 4 out of 8 scenarios, the same as if second bets 0. However, if first doesn't goes for the tie and adds the extra dollar, than B, as you said loses in 6 out of 8 scenarios. B should only chose to bet "all" in the two-thirds breakpoint if they are confident that first will wager correctly (and that they will likely get final right.)

On a similar issue I really don't understand about the wagering calculator is that it suggests that in the 3/4ths scenario, it suggests that the second place person should try top or tie a 0 wager on the first place contestant even if it risks the lockout.

For example with these scores:
A - $20,000
B - $15,000
C - $7,5000

The wagering calculator suggests that B bets $5,000 (to tie with A's 0 wager, and tie with A on the triple stumper.) Why wouldn't second place person bet $0 in this scenario. The weird thing is that in the four-fifths scenarios (where the second place person has more than or equal to twice the third place contestant) the wagering calculator does suggest that the second place contestant should keep second locked out. Why not in 3/4s too.