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I and my friend have always had an interest in gambling. Every week, we each have $£100$ to spare, and we like to bet it on football games at the bookmaker’s. In the long term, we (unsurprisingly) don’t do well, and our losses over the years probably sum to a high amount.

Recently, my friend moved house, and now he lives far away from me, although we still keep in touch. This would’ve been the end to our weekly sessions had he not found another bookmaker in his new city.

There were a few interesting things we noted between the bookmaker’s in my city and his new city:

They are from different companies.

They run similar rules on how they bet.

Both of them use only machines for calculating the odds for the bets.

How the betting machines work:

All teams play against each other once a week. The outcome of each match is either a win or a loss (no ties - penalties are played until one team wins).

All of the teams are relatively even in skill — in a match, each team has an equally likely chance of winning.

The machine contains a list of teams. It will consider only matches between any pair of teams in that list. This list never changes.

Before every match, the machine tries to predict the outcome of the match (which team wins).

This prediction is known before you bet. You are only allowed to bet against this predicted outcome, or choose not to bet at all.

The success rates for the machines in betting the outcome of each match is constant.

The machines accept bets of $\small £100$ only, and charge an extra $£1$ per bet.

If you win the bet against the machine, it pays out $£\left(\frac{100}{1-p}\right)$ rounded down to the nearest pound, where $p$ is the success rate for that machine.

Example:

A particular machine has a success rate of $60\%$ of predicting the outcome of the match. To bet, I pay $£101$. If the machine predicts incorrectly, it pays out $£\left({100 \over {1-0.6}}\right) = £250$. If it predicts correctly, I lose all of my money. Therefore I can expect to lose around $£1$ per bet in the long run.

We then noted that although the majority of machines only listed local teams, there were a few teams which machines from both of the bookmakers’ shared:

Bookmaker’s in my city:

Machine 1:
Success rate: $60\%$
Teams: A, B, C

Bookmaker’s in his city:

Machine 2:
Success rate: $40\%$
Teams: B, C, F

This means that we can both bet on the weekly match of B vs C.

With all of this in mind, is it possible for me and my friend to work together to beat the system?

Each of us can only afford to make 1 bet of $£101$ each per week (and note that this $£101$ cannot be transferred between us). We have agreed to try and devise a strategy to minimise our losses (or maximise our profits), and to divide our winnings and any unused money equally between us each week. Note that a strategy to minimise our losses could be just to not bet at all, although the unused $£101$ or $£202$ from that week will not be carried over to the next.

With the best strategy, how much should I expect to lose/gain per week?

You can assume that we can fully trust each other (as we have known each other for many years) and you can assume that the way in which machines from different companies predict matches are completely independent of each other. Each prediction for a machine is independent of any of its previous predictions. In addition, assume that we have little knowledge of football, and don't watch the games to try and gauge which team is better. Note that the bookmakers’ are fair, and do not lie about their machines’ success rates. (For example, notice that machine 2 has a success rate of $40\%$ which is worse than flipping a coin to try to predict which team wins. The machine, however, will not try to ‘cheat’ to improve it’s success rate by flipping a coin instead.)


Now answer the same questions, but for these machines:

Bookmaker’s in my city:

Machine 3:
Success rate: $60\%$
Teams: C, D, E

Bookmaker’s in his city:

Machine 4:
Success rate: $70\%$
Teams: B, D, E


Final question:

Using all 4 machines together, can we improve our gain/losses per week? And if so, by how much?


Not part of the puzzle, but bounty will be awarded for good answers which also discuss these:

  • Let us consider a pair of machines which both share 2 teams (A and B, say). Let their success rates be $p_1$ and $p_2$. As the values of $p_1$ and $p_2$ change, how does our optimal strategy against the machines change? (consider the different cases for $p_1$ and $p_2$)

  • What if there were more machines which share 2 teams?

  • Why wouldn’t these strategies necessarily work in real life?

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  • $\begingroup$ Is the outcome a binary thing? Is the bet about who wins? Or exact scores? what about ties? In other words, can two machines have different outcomes and both lose? $\endgroup$ – Ivo Beckers Apr 18 '16 at 11:50
  • $\begingroup$ It's a bet about who wins. I should've made that clear. And there are no such things as ties. $\endgroup$ – Shuri2060 Apr 18 '16 at 11:51
  • $\begingroup$ Should the answer assume that you have no info on the actual probability distribution of the outcome of the matches? $\endgroup$ – JiK Apr 18 '16 at 12:21
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    $\begingroup$ the machine might in order to predict the outcomes (or try to), but you don't. You only have those success rates of the machines to go by. $\endgroup$ – Shuri2060 Apr 18 '16 at 12:26
  • $\begingroup$ "[Y]ou can assume that the way in which machines from different companies predict matches are completely independent of each other. Each prediction for a machine is independent of any of its previous predictions." This still doesn't imply that the events "Machine 1 predicts correctly" and "Machine 2 predicts correctly" are independent, which the answers seem to assume. $\endgroup$ – JiK Apr 19 '16 at 12:16
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Yes, it is possible for you and your friend to beat the system. The key to this is sharing with each other what predictions the machines are making, and changing your bets based on whether the predictions are the same or are different.

For the first two machines, we can construct the following probability table:

$\begin{matrix} & 1W & 1L \\ 2W & 0.24 & 0.16 \\ 2L & 0.36 & 0.24 \end{matrix}$

These are the four possible outcomes, and their probabilities were calculated by multiplying each machine's probability. For example, the 1L 2W cell, which holds the probability that machine 1 loses and machine 2 wins, is the probability that machine 1 loses times the probability that machine 2 wins.

$0.4(1-0.6) = 0.16$

At this point, we need to combine the knowledge of the two machines' predictions to form two groups of results. Either the machines agree, or they disagree. If both machines predict the same team, then either both must win, or both must lose. This puts us in the upper-left or lower-right quadrants of the table. Conversely, if the machines predict different teams, then we are in either the upper-right or lower-left quadrants. Once we have restricted the possible positions, we can recalculate the probabilities.

The probability of both machines winning was equal to the probability of both machines losing, so if the machines chose the same result, each now has a

$50\%$ chance of winning. $\frac{0.24}{0.24+0.24}$

If the machines have different predictions, the probability that the first machine will win (and the second will lose) is

$\frac{0.36}{0.36+0.16} = 0.69230769230769230769230769230769 = {9\over13}$

Similiarily, the probability that the second machine will win given that the machines have different predictions is

$\frac{0.16}{0.36+0.16} = 0.30769230769230769230769230769231 = {4\over13}$

Each machine's payout is rounded down to the nearest pound, so our results will be slightly less than one might initially calculate.

$P_1 = \left\lfloor \frac{£100}{1-0.6} \right\rfloor = £250$

$P_2 = \left\lfloor \frac{£100}{1-0.4} \right\rfloor = £166$

Now we can multiply each outcome's probability with its payout to find its expected value. Any bets placed on a machine that wins pays $£0$, so we don't need to bother including them. $E(M_n|S)$ means the expected value of betting on machine $n$ given that the predictions were the same. Likewise, $E(M_n|D)$ means the expected value of betting on machine $n$ given that the predictions were different. Each expected value is lowered by $£101$ to account for the initial bet and the betting fee.

$E(M_1|S) = 0.5 \times £250-101 = £24$
$E(M_2|S) = 0.5 \times £166-101 = -£18$
$E(M_1|D) = {4\over13} \times £250-101 = -£24.076923076923076923076923076923$
$E(M_2|D) = {9\over13} \times £166-101 = £13.923076923076923076923076923077$

From these results, we can see that

Bets should be placed on machine 1 when the machines predict the same team, and on machine 2 when the machines predict different teams.

The estimated return following this strategy is the sum of (the probability of each action taken times the expected value of that action). The probability of taking the first action is

$0.24 + 0.24 = 0.48$

The probability of taking the second action is

$0.36 + 0.16 = 0.52$

So the strategy's expected return is

$(0.48 \times £24) + (0.52 \times £13.923076923076923076923076923077) = £18.76$

Splitting the betting profits with your friend, each person can expect to make

$\frac{£18.76}{2} = £9.38$ per week.

The third and fourth machines can be solved in the same manner.

$\begin{matrix} & 3W & 3L \\ 4W & 0.42 & 0.28 \\ 4L & 0.18 & 0.12 \end{matrix}$

$E(M_3|S) = {12\over54} \times \left\lfloor {£100\over0.4} \right\rfloor-101 = -£45.444444444444444444444444444444$
$E(M_4|S) = {12\over54} \times \left\lfloor {£100\over0.3} \right\rfloor-101 = -£26.925925925925925925925925925926$
$E(M_3|D) = {28\over46} \times \left\lfloor {£100\over0.4} \right\rfloor-101 = £51.173913043478260869565217391304$
$E(M_4|D) = {18\over46} \times \left\lfloor {£100\over0.3} \right\rfloor-101 = £29.304347826086956521739130434783$

The strategy we should use here is to

Bet on both machines when they differ, and neither when they agree.

The expected return for just using machines 3 and 4 is

$(0.28 \times £51.173913043478260869565217391304) + (0.18 \times £29.304347826086956521739130434783) = £19.603478260869565217391304347826$

or

$£9.801739130434782608695652173913$ per person per week

If we consider all four machines, then we should look at the actions we have so far. We have four actions that all gain money, but not all can be taken at the same time.

If machines 1 and 2 agree, and machines 3 and 4 also agree, then our strategy tells us to bet on machines 2, 3, and 4, but the friend (who has access to machines 2 and 4) only has enough money to bet on one machine. In this case, the friend should take the action that has the higher value, which is to bet on machine 4. This is the only situation in this scenario that has a conflict.

This scenario has only three different sets of actions to take.

If machines 3 and 4 differ, bet on both of them. Otherwise, follow the original strategy for machines 1 and 2, which itself has two possible actions.

The expected return for following our strategy using all four machines is

The expected value of machines 3 and 4 plus (the expected value of machines 1 and 2 times the probability of machines 3 and 4 being the same (causing us to consider 1 and 2))

That is,

$£19.603478260869565217391304347826 + (0.54 \times £18.76) = £29.733878260869565217391304347826$

or

$£14.866939130434782608695652173913$ per person per week

In the general case, two machines having probabilities $p_1$ and $p_2$, respectively, have the following probability table:

$\begin{matrix} & 1W & 1L \\ 2W & p_1p_2 & (1-p_1)p_2 \\ 2L & p_1(1-p_2) & (1-p_1)(1-p_2) \end{matrix}$

These machines then have the following expected values:

$E(M_1|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)}\times\frac{100}{1-p_1}-101$

$E(M_1|S) = \frac{1-p_2}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E(M_2|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)} \times \frac{100}{1-p_2}-101$

$E(M_2|S) = \frac{1-p_1}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E(M_1|D) = \frac{(1-p_1)p_2}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_1}-101$

$E(M_1|D) = \frac{p_2}{p_1+p_2-2p_1p_2} \times 100-101$

$E(M_2|D) = \frac{p_1(1-p-2)}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_2}-101$

$E(M_2|D) = \frac{p_1}{p_1+p_2-2p_1p_2} \times 100-101$

With two machines,

Simply bet on whichever outcomes ($S_1$, $S_2$, $D_1$, $D_2$) that occurred are positive.

With multiple sets of machines,

Bet on the highest (positive) outcomes that occurred. For example, if you have the following outcomes:

$S_1=5$
$S_2=-5$
$D_1=10$
$D_2=5$
$S_3=20$
$S_4=-20$
$D_3=-10$
$D_4=15$

And machines 1 and 2 are the same, and 3 and 4 are different, then bet on machines 1 and 4. This is because the valid cases are:

$S_1=5$
$S_2=-5$
$D_3=-10$
$D_4=15$

And of these, $S_1$ is greater than $D_3$ and $D_4$ is greater than $S_2$.

With more than two (say, $n$) machines for a single pair of teams,

Build a bigger ($n \times n$) probability table, calculate the expected values the same way, and bet on the outcomes with the highest (positive) expected values.

These strategies wouldn't work in real life, because

They operate under the assumption that the machines generate predictions independently. In reality, the machines probably use a lot of the same input data, which makes their predictions partially dependent.

This is my first post on Stack Exchange (long time lurker), so forgive the poor formatting. I'm trying to figure out how it works, but would appreciate any tips.

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  • 2
    $\begingroup$ Welcome to Puzzling.SE! This is a fantastic answer - I hope to see you around here a lot more! c: $\endgroup$ – Deusovi Apr 18 '16 at 21:42
  • $\begingroup$ Thanks @Deusovi! I have wanted to interact more, but never cared enough to actually make an account until now. $\endgroup$ – Jake The Snake Apr 18 '16 at 22:04
  • $\begingroup$ Comment pinging with @ only works on people who have commented or edited the post you're commenting on. If you want to ask a question, I recommend commenting on their post. $\endgroup$ – Deusovi Apr 18 '16 at 22:59
  • $\begingroup$ Excellent answer - just what I was looking for! My apologies with the question, as there was a slight mistake when I said &*they share their profits*. I had actually meant them to split any money they decide not to bet with, too. (Essentially, I had intended me and my friend to gain an equal share of money from this). Nevertheless, another interesting followup question would be: How much more/less than my friend would I get per week if we used the perfect strategy but didn't share the money? $\endgroup$ – Shuri2060 Apr 19 '16 at 9:00
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    $\begingroup$ @JakeTheSnake Before I accept this answer, I'll have to go through some of the numbers to make sure they agree with mine - which I have written somewhere. In addition, I've suggested some improvements to the formatting (I hope I haven't changed the meaning of anything). $\endgroup$ – Shuri2060 Apr 19 '16 at 13:41
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The different combinations of predictions are as follows

$\frac{3}{5}\times\frac{2}{5}=\frac{6}{25}$ for both being right
$\frac{2}{5}\times\frac{3}{5}=\frac{6}{25}$ for both being wrong
$\frac{3}{5}\times\frac{3}{5}=\frac{9}{25}$ for 1 being right and 2 being wrong
$\frac{2}{5}\times\frac{2}{5}=\frac{4}{25}$ for 1 being wrong and 2 being right

If both machines agree on the outcome, there is a 50/50 chance that they are right. In this case you should

Bet on machine 1, since you would win £250 half the times for an average payout of
$250\times0.5=£125$
And an average profit of
$125-100-1=£24$

If the machines disagree, machine 1 i right $\frac{9}{13}$ of the time, and you should

Bet on machine 2, which gives £166 in 9 of 13 cases for an average payout of
$166\times\frac{9}{13}=£114.92$
And an average profit of
$114.92-100-1=£13.92$

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    $\begingroup$ Right idea. However, this doesn't fully answer the question. In addition, explain how you got 50/50 and $\frac{9}{13}$. (Also, in England, we write $£XX.XX$ rather than $XX.XX£$) $\endgroup$ – Shuri2060 Apr 18 '16 at 11:42
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I'm probably missing something but I can't see how having more machines makes any difference, especially when they're are independent from each other.

Therefore I say

whatever strategy you follow you can expect to lose £1 per bet in the long run

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    $\begingroup$ I'm afraid that this puzzle would be a lot of wasted text if that really were the case :) $\endgroup$ – Shuri2060 Apr 18 '16 at 11:07
  • $\begingroup$ @QuestionAsker yeah I guessed it would be a lot of wasted text if this was the answer. But there is always a possibility for it being a trick question and I honestly don't know how it would work otherwise, so I just thought I give it a try :) $\endgroup$ – Ivo Beckers Apr 18 '16 at 11:09
  • $\begingroup$ That's not a problem - at least we now know we can't do worse than that in the long run :) $\endgroup$ – Shuri2060 Apr 18 '16 at 11:10
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    $\begingroup$ As an extreme example, consider one machine that predicts the outcome correctly 100 % of the time and another that predicts the outcome correctly 50 % of the time. If you had only the second machine, you wouldn't have a winning strategy, but with the first machine you'll make a lot of money by betting against the second machine every time the machines give different predictions. $\endgroup$ – JiK Apr 18 '16 at 13:16

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