# Foolproof double betting

Just made up a problem. Have ideas, but no proof yet – came here for professional help.

Tonight an important match of Metagame (box/football/baseball/chess) takes place. Everyone is hyped. Everyone is making bets.

The result of the game will be represented with a real number between $0$ and $1$.

We have intel on $2$ (two) mathematicians. Mathematician $A$ believes that the probability density distribution of the game result $x$ goes as $F_a(x) = 2-2x$. At the same time, mathematician $B$ believes that game result is distributed as $F_b(x) = 2x$.

Both matematicians take bets rationally: if their expected winning is more than zero (read: equals zero, you stingy fox) they will take the bet. At the same time, the maximum that each of them will bet is 10 dollars.

You need to make two (different) bets with $A$ and $B$, so that no matter what game results are, you get profit. The money you can bet is technically non-limited, however, risking more than 10 dollars for a single bet is obviously a bad step – if you lose this bet, winning the other will barely compensate for your loss.

Make a strategy with maximal guaranteed profit, and a proof of its perfection.

• Can you explain what is meant by "distribution of the game result goes as Fa(x) = 2-2x". My first thought was that this is probability of the score being x is 2-2x but that doesn't work... May 14 '18 at 14:37
• @Chris Fa is what is known as a probability distribution function -- the probability that x lies between a and b is the area under the function curve between a and b. You can check that the whole area is a triangle with base 1 and height 2, giving a total probability of 1 as desired.
– ffao
May 14 '18 at 19:14
• Ah thanks. Despite being a mathematician stats was always a blindspot for me. :) May 14 '18 at 20:39
• Not at all clear to me how the game is played. What is the actual distribution? May 16 '18 at 18:58
• when the first bet is made is my stake in escrow, or can I use it to also make a hedging bet? May 19 '18 at 10:35

Main idea:

You can arbitrage the two mathematicians' mismatched priors; in particular, their priors most mismatch at the ends. So you want to always bet over-under bets with the two mathematicians; it doesn't make sense to bet any other union of intervals, since that would reduce the range of arbitrage.

Ugly overly formalized calculations that really should be intuitively obvious:

Moreover, it follows that you want to give the $$x < P_1$$ option to the mathematician whose priors are $$F_a(x) = 2-2x$$, and the $$x > P_2$$ option to the mathematician whose priors are $$F_b(x) = 2x$$, as that would give you favorable odds. Next we argue that $$P_1 = P_2$$, since if $$P_1 < P_2$$ there is the possibility that you lose both bets if $$P_{1} < x < P_2$$. If $$P_{2} < P_{1}$$, then you could earn more on either end by tightening one of the bounds, thus improving your guaranteed profit.
Now we compute what odds the mathematicians will be willing to take. This turns out to just be $$\int^P_0 F_a(x):\int^1_P F_a(x) = 1 - (1-P)^2 : (1-P)^2$$ for mathematician A and $$\int^1_P F_b(x):\int^P_0 F_b(x) =1 - P^2 : P^2$$. Assuming we choose a $$P$$ that gives us favorable odds to both mathematicians (and really, if you don't, it'll be hard to get much of a guaranteed prize), there is no downside to maximizing our bet to both mathematicians. Therefore, if A wins the bet, we earn $$10(1 - \frac{P^2}{1-P^2})$$ dollars, and if B wins the bet, we earn $$10(1 - \frac{(1-P)^2}{1-(1-P)^2})$$ dollars. Since we want to maximize the minimum of the two values, we can just use the fact that $$\frac{x^2}{1-x^2}$$ is monotonically increasing to show that the optimal $$P = 0.5$$. This nets us a guaranteed $$\boxed{\\\\frac{20}{3}}$$.

To summarize results:

Propose a bet with Mathematician A that $$x < \frac{1}{2}$$. Since he thinks this event will happen with $$\frac{3}{4}$$ probability, he will accept a bet at $$3:1$$ odds. If he wins the bet, he gets $$\\\\frac{10}{3}$$ and if he loses he loses $$\\\10$$.
Likewise, propose a bet with Mathematician B that $$x > \frac{1}{2}$$. He agrees to the same conditions. No matter whether $$x < \frac{1}{2}$$ or $$x > \frac{1}{2}$$, you will pay $$\\\\frac{10}{3}$$ to one party and get $$\\\10$$ from the other. Hence $$\\\\frac{20}{3}$$ guaranteed profit.

• Yeah, I suppose, it's formalized enough. Thanks. May 14 '18 at 11:42
• Any chance of getting an English description of the actual bet that we should be proposing? :-)
– Bass
May 14 '18 at 17:26
• So, you've bet > and you've bet < - doesn't this mean that you lose both if it's =? May 17 '18 at 14:51
• If it's =, then you win both bets. But if it puts your mind to ease, you can replace one of the < with <=. (Neither mathematician would care, since x=0.5 to them is a 0 probability event) May 17 '18 at 18:39
• that's not 3:1 odds. May 19 '18 at 10:41

For a payout of $9.99 A probability distribution is an expression of how probable an event is, the integral of a probability dist (called the cumulative probabilitiy distribution) tells you the probability that the result will be less than the chosen number subject A expecting PD$ 2-2x $has CPD$1-(1-x)^2$subject B expecting PD$ 2x $has CPD$x^2$so subject A expects a result higher than 0.5 at 75% probability and subject B expects lower than 0.5 at 75% probability you've got$10 to stake.

So offer just over 1:3 odds for either side of 0.5 result ( the best you can do with normal currency is: punter bets $15.99 and receives 19.99 on a win) then You'll have one taker for "higher" and one taker for "lower" and you'll you incoming will be 29:98 outgoing 19:99 and your profit$9.99.(I'm not using whole dollars because the bad has to be seen as favourable)

a proof is required that this strategy guarantees the most profit. to so that we need to look at how much we can be sure to win.

I feel that his should be at the sccore where the difference between the expectations of the mathematicians is greatest.

at 0.5

but I'm nit yet sure how to prove that.

Using poker like EV.

Question is not clear. I am assuming the real distribution is random.

Fb(x) = 2x
integral = xx
x
x = 1/2 = 1/sqrt(2) = 0.7071067812
take the under on x <= 0.7071067812
b will take it as they even money if x <= 0.7071067812
if you bet 10 your EV is 10*(1/(sqrt(2)) - (1 - 1/(sqrt(2))
= 10* (2/sqrt(2) - 1) = 4.142135624 EV per hand

for Fa(x) = 2-2x just take even money on x <= 1 - 1/sqrt(2) = 0.2928932188
and you have the same EV of 4.142135624 per hand
take odds at x >= 1/2 x <= 1/2 has better EV = 10(-1/2 + 3/2) = 10 per hand
actually take the odds on a very close to 0 or 1
say <= 1/10 against b and get 99:1 for an EV of 890 on a $10 bet you still have positive EV against a I ran the EV and actual distribution matters - assume distribution is x The EV is -b+b/x + -b + b*(1-x)*((2x - x^2) / (1 - 2x + x^2)) + 1) -2b + b/x + b(1-x)((2x - x^2) / (1 - 2x + x^2)) + 1) • The question asks you to maximixe "guaranteed profit", meaning the actual distribution is immaterial. May 17 '18 at 11:27 • I was going to ask this question for EV with even-distributed real result of the game, but considered it too hard of a puzzle. The bet you advice is valid, but not proven to be best. May 17 '18 at 11:30 • The question posits a single event (a "Metagame" match) and seeks a maximal guaranteed profit. I'm not sure how much more clear this can be. If you feel the need to continue to argue about the meaning of the question, which has been explained more than once now, please take it to chat. These comments are being culled. – Rubio May 29 '18 at 20:41 • @Rubio Argue? I said I read it wrong. May 29 '18 at 20:46 • @Rubio If you consider this argue then sorry. Even with 1 match if you assume a normal distribution and bet 1 / sqrt(2) you have a chance of making 0 but you also have an EV of$28.28 versus. \$20. As a poker player I looked at EV but it does say max guaranteed. May 29 '18 at 21:04