# Cheating at Bets with the Even Odds Algorithm

> You put 80% probability on an event and two different people offer you bets on it, Tom offers you 3:2 (you have to bet 3 parts to gain 2) and Ophelia offers you 3:1 (you have to bet 3 parts to gain 1). You win if the event takes place. The exact monetary amounts are determined by the other person (Tom or Ophelia) and are undisclosed. It's guaranteed you won't be put into financial trouble (e.g. your bet will never exceed $100). Which bet do you take? You can accept at most one bet.
> 
> a) Tom (offering 3:2)
> 
> b) Ophelia (offering 3:1)
> 
> c) I can't choose (missing info)
> 
> d) The answer seems obvious, there must be a trick

I encourage you to choose one answer before continuing reading.

If you are only interested in the solution to this puzzle you can jump to the "Tom and Ophelia - Solution" section. If you already have a clear picture what problem the *Even Odds Algorithm* is solving you can jump to the "Conclusion" section.

You can read about the *Even Odds Algorithm* in the ["Even Odds" post](https://www.lesswrong.com/posts/aiz4FCKTgFBtKiWsE/even-odds), but it's not a prerequisite.

Artem and Baani are in disagreement whether tomorrow it's going to rain. As good rationalists they immediately decide to put their money where their mouth is and bet on it. Artem thinks it's 80% likely it's going to rain and Baani thinks it's 60% likely. As if often done, they use the arithmetic mean of their beliefs $\frac{80\%+60\%}2=70\%$ meaning they will bet with $$$70:30=7:3$ odds. They agree that the maximum amount they want to bet is $10, so Artem bets $10 on the event and Baani bets $4.29 against the event ($\frac{10}{4.29}=\frac{7}{3}$). Should it rain tomorrow, Artem will get the $4.29 and should it not rain tomorrow Baani will get the $10.

Both get a positive EV out of this bet, namely for Artem:

$EV_A = 0.8 \times $4.29 - 0.2 \times $10 = $1.43$

and for Baani:

$EV_B = 0.4 \times $10 - 0.6 \times $4.29 = $1.43$

Now let's assume Artem likes having an edge where possible and he is OK with a bit of dishonesty. Let's again assume the same scenario as above except that Artem waits for Baani to state her 60% probability and then he claims he thinks it's only 65% likely to rain, even though he really think it's 80%. So they use the arithmetic mean of 65 and 60, which is 62.5, meaning $6.25:3.75$ odds. Artem bets $10 on the event and Baani bets $6 against the event ($\frac{10}{6}=\frac{6.25}{3.75}$). Should it rain tomorrow, Artem will get the $6 and should it not rain tomorrow Baani will get the $10. You may already see that Artem is getting a better deal than when he was being honest because now he stands to gain almost $2 more even though he still only loses at most $10. Note that for Artem we are using his true belief 80% for the EV calculation!

$EV_A = 0.8 \times $6 - 0.2 \times $10 = $2.80$

$EV_B = 0.4 \times $10 - 0.6 \times $6 = $0.40$

Baani is still getting a positive EV so she should still take the bet, but she is being cheated out of some of her EV, which would have been higher had Artem been honest.

Why does this problem occur? Because the algorithm we are using to decide the odds and the betting amounts is not a strictly proper scoring rule. A strictly proper scoring rule ensures that all participants get the highest score (EV in this case) if and only if they report their true beliefs.

Instead we can apply the *Even Odds Algorithm* as follows. Quoted from the original [Even Odds post](https://www.lesswrong.com/posts/aiz4FCKTgFBtKiWsE/even-odds), with better formatting.

> Alice and Bob are willing to bet up to $d$ dollars, Alice thinks S is true with probability $p$, and Bob thinks S is false with probability $q$. Assuming $p+q>1$, Alice should put in $d\cdot(p^2-(1-q)^2)$, while Bob should put in $d\cdot(q^2-(1-p)^2)$.

If Artem reports his belief as 80%:

$d=$10; p=0.8; q=0.4$

Artem should put in $$10\times(0.8^2-(1-0.4)^2)=$2.80$

Baani should put in $$10\times(0.4^2-(1-0.8)^2)=$1.20$

Should it rain tomorrow, Artem will get $1.20 and should it not rain, Baani will get $2.80.

$EV_A = 0.8 \times $1.20 - 0.2 \times $2.80 = $0.40$

$EV_B = 0.4 \times $2.80 - 0.6 \times $1.20 = $0.40$

If Artem reports his belief as 65%:

$d=$10; p=0.65; q=0.4$

Artem should put in $$10\times(0.65^2-(1-0.4)^2)=$0.63$

Baani should put in $$10\times(0.4^2-(1-0.65)^2)=$0.38$

Should it rain tomorrow, Artem will get $0.38 and should it not rain, Baani will get $0.63. Note that for Artem we are using his true belief 80% for the EV calculation.

$EV_A = 0.8 \times $0.38 - 0.2 \times $0.63 = $0.18$

$EV_B = 0.4 \times $0.63 - 0.6 \times $0.38 = $0.03$

Artem got an EV of $0.40 when being honest and and EV of $0.18 when cheating. He is therefore incentivized to be honest. This is what the *Even Odds Algorithm* is doing for us, ensuring dishonesty does not pay.

How could he *still* cheat?

If Artem knows all of the above he could still report his belief as 65% and when the final bet is agreed upon he says to Baani: "Hey look, the amounts we are betting are ridiculously low, what do you think we just scale it up 10x?" No harm done, right? A bet with a positive EV still has a positive EV when you scale it up. Artem now has to put in $10\times$0.63=$6.30$ and Baani $10\times$0.38=$3.80$.

$EV_A=10\times$0.18=$1.80$

$EV_B=10\times$0.03=$0.30$

And voilà, suddenly Artem has a higher EV than he would have gotten had he been honest about his belief (since $$1.80 > $0.40$). He does have to put up a bit more money upfront (namely $6.30 instead of the $2.80 from the honest case) but if instead of proposing to scale up the bet 10x he proposes scaling it up by 7.37x (i.e. the proportion of the two amounts $$$\frac{$2.80}{$0.38}=7.37$) he will achieve a higher EV than when being honest without having to invest a single cent more.

Scaling bets up or down is not harmless and can in fact be used to cheat.

If you are using the *Even Odds Algorithm* you should always choose a maximum monetary amount first and then stick to the amounts calculated at the end.

More generally, the value of a bet is not only determined by the odds of the bet but also by the amounts being bet. This may seem like a trivial insight, but it was surprisingly hard to intuitively "get it" for myself and some other people I asked.

To highlight that last point, let's go back to question at the start of the post.

Tom and Ophelia - Solution
==========================

> If you put 80% probability on an event and two different people offer you bets on it, Tom offers you 3:2 (you have to bet 3 parts to gain 2) and Ophelia offers you 3:1 (you have to bet 3 parts to gain 1). You win if the event takes place. The exact monetary amounts are determined by the other person (Tom or Ophelia) and are undisclosed. It's guaranteed you won't be put into financial trouble (e.g. your bet will never exceed $100). Which bet do you take? You can accept at most one bet.

I asked this question in two rationality chat groups and got the following responses. Of course the sample is very small (34 people), not statistically significant etc.

a) Tom (offering 3:2) - 18 people (53%)

b) Ophelia (offering 3:1) - 0 people (0%)

c) I can't choose (missing info) - 5 people (15%)

d) The answer seems obvious, there must be a trick - 11 people (32%)

I would in fact count answers d) the same as answer a), meaning that 85% of people thought Tom's bet was better.

Which bet has the higher EV? Depends on the amounts that are bet. We can solve this. Let's say in the Tom case he bets amount $2x$ and therefore asks you to bet amount $3x$ . In the Ophelia case she bets amount $1y$ and asks you to bet amount $3y$ .

Can Ophelia's bet have a higher EV?

$0.8\times y - 0.2\times3y>0.8\times2x-0.2\times3x$

Which simplifies to: $y>5x$

So if Ophelia is asking you to put exactly 5 times as much money as Tom is asking you to put, then both bets have the same EV. If she is asking you to bet more than that, then her bet is preferable. Otherwise Tom's bet is preferable.

For example if Tom offers the bet $$6:$4$ (which is equivalent to 3:2) meaning you have to put in $6 and he will put in $4 and Ophelia offers the bet $$36:$12$ (which is equivalent to 3:1) meaning you have to put in $36 and she will put in $12 then your EV for the bet Tom is proposing is $0.8\times$4-0.2\times$6=$2$ and your EV for the bet Ophelia is proposing is $0.8\times$12-0.2\times$36=$2.40$ . In this specific case Ophelia's bet would yield a higher EV for you.

Therefore the correct answer to the quiz is "c) I can't choose (missing info)". Without knowing the exact monetary amounts Tom and Ophelia are asking you to bet it's not possible to compare the bets because odds alone are not enough to determine which bet has the higher EV.

Conclusion
==========

To tie it all together, the *Even Odds Algorithm* can be exploited when you don't stick to the algorithm and convince the other player to reveal their probability before agreeing on the max bet. You then make your own probability closer to the probability they named and propose a higher max bet than you normally would. Why does this work? Because naming a probability closer to the probability of the other player gives you better odds. The *Even Odds Algorithm* counteracts this trick by reducing the monetary amounts, therefore being dishonest does not pay. However, if you don't stick to the monetary amounts the algorithm outputs and scale them up, you work around that protection. The EV of a bet does not only depend on the odds but also on the specific monetary amounts being bet.

Interestingly enough the same exploit will also work if the other player does not tell you their probability but you can approximately guess what it is. In that case you can still propose a higher bet and make your own probability closer to theirs.

How do you protect yourself against this exploit?

*   Strictly stick to the algorithm by defining the max bet before any probabilities are exchanged.
*   Avoid leaking information about what your probability is before agreeing on the exact terms of the bet.
*   Generally precommit to always using the same max bet amount for all bets. For example a good rule of thumb could be: Consider what is the amount of money you *generally* want to win or lose in *most* bets. Then multiply that by 4. That should be your general max bet. Why? Because often the final amounts being exchanged are about 1/4 of the max bet since the probabilities of the participants are closish to each other. Of course under rare circumstances the whole max bet will be in play, namely if one player thinks the event is 100% likely to occur and the other 0% likely. The higher the gap between the player's probabilities, the more we will approximate max bet.

Thanks to Hauke, Nawid and Valentin for reviewing a draft of this post and proposing improvements which I have incorporated.