We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips; in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.
A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence
- $p_{k,d}=0$ if $2k<d$
- $p_{k,1}=1$ if $k\ge 150$
- $p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$
This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction
$$ p_{100,1}\approx0.56097114279613511032732301110367086534$$
should be good enough for us.
Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):
Start=100;Target=150
A=vector(Target-Start+1,n,-1);
getP(k,d)={local(pkd);
if(2*k<d, 0, if(k>=Target, 1,
pkd=if(d==1&&A[k-Start+1]>=0,
A[k-Start+1],
12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
);
if(d==1,A[k-Start+1]=pkd);
pkd))
}
getP(Start,1) +.0
The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.
Best Answer
This is an interesting problem. I thought about it for some time and I believe I have a complete solution.
First, let's see how we would answer the question if we had only 2 people. Say, Alice believes that an event will happen with probability $90\%$, while Bob believes that it will happen with probability $80\%$. How do you make a bet out of this event? It is not obvious.
Intuitively we can think that the two people must take opposite positions, otherwise they will both be winners or losers at the same time. Moreover, the one with more confidence in the event (Alice) should take the
for
position and the other person should take theagainst
position. We will prove this later, but there is also another major question: on what odds should they bet? Let's explore the problem:Assume that Alice bets $\$1$ on the
for
position (i.e. the event actually happening) How much should she make back to consider this a favourable bet for her? Let's name the return (i.e. the amount won) as $x$. Just to be clear, $x$ is the amount returned above the $\$1$ she's already put in. In other words, if Alice is betting $\$1$ and wins, she is getting back $\$1+\$x$. To find $x$ of a favourable bet, we just make the expected value of the bet positive: $$0.9\cdot x - 0.1\cdot 1 > 0 \iff x > \frac{1}{9}$$So if Alice bets $\$1$, she would like her opponent to bet at least $\$1/9$. If we write this in the typical form of odds as opponent_bet:your_bet then Alice is looking for a $1/9:1 \iff 1:9$ bet or better.
What if Alice wants to bet on the
against
position (i.e., the event not happening)? A favourable bet of $\$1$ would be one with return $x$ where $$0.1\cdot x - 0.9\cdot 1 > 0 \iff x > 9$$The odds now are $9:1$. It is the reciprocal as we expected.
Doing the analysis for Bob we find that
for
position wants odds $1:4$ or betteragainst
position wants odds $4:1$ or betterHow would we create a fair bet between the two? Let's make a table with all the possible position combinations and conditions for each combination
$$ \begin{array}{l|c|c} & \text{Alice for} & \text{Alice against} \\ \hline \text{Bob for} & \text{not a bet} & \begin{array}{c} & \text{Alice wants}\space x:1, \space \color{red}{x>9}\\ & \text{Bob wants}\space 1:x, \space \color{red}{x<4} \\ \end{array}\\ \hline \text{Bob against} & \begin{array}{c} & \text{Alice wants}\space 1:x, \space \color{green}{x<9}\\ & \text{Bob wants}\space x:1, \space \color{green}{x>4} \\ \end{array} & \text{not a bet} \\ \end{array} $$
As we see from the table, the only possible bet is when Bob bets
against
and Alice betsfor
. This is inline with our intuition that the person most confident should take thefor
position. We also know that an $x$ such that $4<x<9$ is acceptable for both Alice and Bob (in the sense that they both expect a positive gain). What is the most fair value however? Should we just take the middle value ($x=6.5$)? This might seem reasonable, but remember that odds are "multiplicative" entities, and taking "additive" averages might not work. Indeed the most fair selection of $x$ is the one that yields the same expected values for both parties: $$0.2\cdot x -0.8 \cdot 1 = 0.9 \cdot 1 - 0.1 \cdot x\iff 0.3 \cdot x = 1.7 \iff x = \frac{17}{3} \approx 5.666$$ So Alice should make thefor
bet with odds $3:17$ and Bob make (the opposite) bet with odds $17:3$. And if we wanted for both of them to bet $\$100$ in total, the bets would be $100\cdot 17/(17+3) = \$85$ for Alice and $100\cdot 3/(17+3) = \$15$ for Bob. Notice how Alice's bet is the average of their perceived likelihoods.So if we know the perceived likelihoods for each of the two players, we can construct a fair bet. It's interesting to note that we can always construct a fair bet for two people, and that the closer the likelihoods are to each other, the smaller the gain both players expect. So if both Alice and Bob believe that the event happens with probability $80\%$, then the only fair bet is $1:4 \space\space(4:1)$ and the expected gain for both of them is $0$. If, on the other hand, Alice believes the event will happen $90\%$ and Bob believes it will happen $10\%$ of the time, then the fair bet is $1:1$ and the expected gain of both players is $\$0.8$ for each $\$1$ played. Just to be clear, this is the expected gain based on a player's perceived likelihood, it does not have to agree with the actual expected gain (in other words: people can have mistaken beliefs)
The case for 3 players
Let's examine what happens with 3 players. Let's take the particular example given in the question.
This gives us the following favourable odds for each person:
for
with $13:7$ (or better), oragainst
with $7:13$ (or better)for
oragainst
with $1:1$ (or better)for
with $3:2$ (or better), oragainst
with $2:3$ (or better)Let's build the betting table again. With $3$ people we have $2^3=8$ combinations. Similarly to the 2-people case, combinations where everyone bets
for
or everyone betsagainst
are not valid bets. So we are left with 6 betting combinations. What should the conditions be for each combination? Things are more complicated now that we have 3 people. First let's revisit the conditions for the 2-person case. In the analysis above, I stated without much explanation that conditions for the 2-person case look like this: $x:1$ for Alice and $1:x$ for Bob. It seems intuitive, but it would be helpful to derive it so we can understand the restrictions we are taking and the assumptions we are making. The initial conditions for the 2-person example should have been: $$ \begin{array}{rc} \text{Alice wants odds } \space x:a, & \frac{x}{a}>\frac{1}{9} \\ \text{Bob wants odds } \space y:b, & \frac{y}{b}>\frac{1}{4} \\ \end{array} $$ But now we have 4 variables instead of one! This is because these two conditions describe only what the favourable bets are for each player. So if Alice and Bob were betting any amount against a bookmaker, these are the conditions they would use. But if we want to make them bet against each other then we need to impose more restrictions. We need to restrict these quantities so that whatever one player bets is the (potential) winning of the other player. This means that $a=y$ and $b=x$. With these restrictions we are down to two variables. How do we eliminate another variable? Since odds are essentially ratios, we can simply normalise one of them by setting $a=1$. This is what we have implicitly done when analysing the 2-person case above. Another (perhaps better) option is to treat the $x,y,a,b$ variables as the actual quantities being betted and won. In this case, if we know the total amount we want our players to bet (say $\$100$) we simply set $a+b=100 \iff a+x = 100$ which together with the equation of making the expected gains equal, yields a $2\times 2$ system. Its solution for the 2-person example we gave earlier results in $a=85$ and $x=15$ as we have found before.How does all these translate to the 3-person case? Let's take one particular combination where Alice bets
against
, while Bob and Charlie betfor
. Also assume, as the question states, that all together are betting $\$100$. Here are the initial three conditions for favourable bets, along with the three extra restrictions that make this a valid bet among Alice, Bob, and Charlie : $$ \begin{array}{rl} \text{Alice wants odds } \space x:a, & \frac{x}{a}>\frac{7}{13} \\ \text{Bob wants odds } \space y:b, & \frac{y}{b}>\frac{1}{1} \\ \text{Charlie wants odds } \space z:c, & \frac{z}{c}>\frac{3}{2} \\ a &= y+z\\ x &= b +c\\ a+b+c &= 100 \end{array} $$If we make the expected gains to be equal to each other in order to ensure the fairness criterion, we will get $2$ equations which along with the $3$ restriction make an underdetermined system with $5$ equations and $6$ variables. We could bring another restriction to force the system to have a unique solution (unique for a given betting combination out of the possible $6$), but there is no strong restriction I can think of. There are desirable properties such as maximising the product $a\cdot b\cdot c$ which means that we are trying to make the 3 bets as equal as possible (in other words, make it unlikely to have one person betting $\$98$ and the other two betting $\$1$). Or we can maximise the (equalised) expected gain, but note that this does not mean that everyone will get their best odds. In any case, I do not see these as necessary restrictions. We can have one or not. If not, we will have a infinite number of solutions (under the remaining restrictions) for each betting combination. The solutions ensure that everyone has a perceived positive gain, and that all the perceived gains are equal, so we have a fair solution.
Here are the 5 equations and the parametrised solutions for the case where Alice bets
against
, while Bob and Charlie betfor
, and all together are betting $\$100$:$$ \left. \begin{array}{rl} 0.65\cdot x - 0.35\cdot a &= 0.5\cdot y - 0.5\cdot b \\ 0.5\cdot y - 0.5\cdot b &= 0.4\cdot z - 0.6\cdot c \\ a &= y+z\\ x &= b +c\\ a+b+c &= 100 \end{array} \right\} \iff \begin{array}{rl} x &= 39.62 - 0.077 \cdot t\\ y &= 48.85 - 1.231 \cdot t\\ z &= 11.54 + 1.310 \cdot t\\ a &= 60.38 + 0.077 \cdot t\\ b &= 39.62 - 1.077 \cdot t\\ c &= t \end{array} $$ (Solution provided by this online tool)
Our parameter (free variable) is $t$, and we can notice that in this particular case it mostly affects $b,c,y,z$ and minimally affects $a,x$. By changing the parameter we are essentially changing the relative "weight" between Bob and Charlie as they both take on Alice. For example, try $t=19$ for a fairly balanced "weight" between Bob and Charlie (where Bob and Charlie bet almost equal amounts). This results in the following bet:
against
. If she wins she'll earn $\$38.15$ (so she'll receive back $\$38.15 + \$61.85 = \$100$)for
. If he wins he'll earn $\$25.46$ (so he'll receive back $\$25.46 + \$19.15 = \$44.61$)for
. If he wins he'll earn $\$36.39$ (so he'll receive back $\$36.39 + \$19.00 = \$55.39$)Notice how the sum of players' bets is $\$100$, and what winners get back also totals $\$100$ (as it should).
We can find similar solutions (families of solutions) for the five remaining betting combinations. It turns out that with this example (these likelihood beliefs) all six combinations can yield fair solutions, albeit some combinations have a smaller range of solutions than others. You can explore the other families of solutions yourself, doing the same analysis and solving equivalent $5\times6$ systems.
I believe this is a quite interesting result and it fully solves this betting problem.
Addendum
I include here my earlier work on the 3-people case in order to show some of the progress of my ideas. I created a table with all the betting combinations and tried to provide the conditions for each combination. I arbitrarily restricted 3 of the variables to take specific values (but still abiding by all restrictions), and this resulted in a $3\times3$ equation system. For example, for the case we have been analysing above I took $x = 7$, $b = 3$, $c=4$, satisfying the condition $x=b+c$. I do not assume that $a+b+c=100$. This simplified some expressions and I could check the conditions of all the combinations manually. For three of the combinations there are no solutions with this arbitrary choice that I made (impossible conditions are written in red at the table below). Here's the table (note: I changed the original variable names used, so that they agreed with the names of the rest of the analysis above.)
$$ \begin{array}{l|c|c} & \text{Alice for} & \text{Alice against} \\ \hline \text{Bob for, Charlie for} & \text{not a bet} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space y:3, \space \color{green}{y>3} \\ & \text{Charlie wants}\space z:4, \space \color{green}{z>6} \\ & \color{green}{a = y + z} \end{array}\\ \hline \text{Bob for, Charlie against} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space y:1, \space \color{red}{y>1} \\ & \text{Charlie wants}\space 8:c, \space \color{red}{c<12} \\ & \color{red}{x+y = c} \end{array} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space y:9, \space \color{green}{y>9} \\ & \text{Charlie wants}\space 2:c, \space \color{green}{c<3} \\ & \color{green}{a+c=y} \end{array}\\ \hline \text{Bob against, Charlie for} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space 9:b, \space \color{red}{b<9} \\ & \text{Charlie wants}\space z:2, \space \color{red}{z>3} \\ & \color{red}{x+z=b} \end{array} & \begin{array}{c} & \text{Alice wants}\space 7:a, \space \color{green}{a<13}\\ & \text{Bob wants}\space 1:b, \space \color{green}{b<1} \\ & \text{Charlie wants}\space z:8, \space \color{green}{z>12} \\ & \color{green}{a+b=z} \end{array}\\ \hline \text{Bob against, Charlie against} & \begin{array}{c} & \text{Alice wants}\space x:7, \space \color{red}{x>13}\\ & \text{Bob wants}\space 3:b, \space \color{red}{b<3} \\ & \text{Charlie wants}\space 4:c, \space \color{red}{c<6} \\ & \color{red}{x=b+c} \end{array} & \text{not a bet} \\ \end{array} $$
Applying these arbitrary restrictions we can see that only three combinations (the ones with conditions in green) offer viable solutions. A minor side note: these restrictions were not so arbitrary. They were chosen so that we get integer values in the conditions, so we can check easily and quickly if the conditions hold.
If we take the first of these valid betting combinations: Alice
against
, Bobfor
, and Charliefor
, and apply the rule of equal expected gains we get:$$ \left. \begin{array}{rl} 0.65\cdot 7 - 0.35\cdot a &= 0.5\cdot y - 0.5\cdot 3\\ 0.5\cdot y - 0.5\cdot 4 &= 0.4 \cdot z - 0.6 \cdot 4\\ a = y + z \end{array} \right\} \iff \begin{array}{rl} a &= \frac{1179}{103} \approx 11.45\\ y &= \frac{421}{103} \approx 4.09\\ z &= \frac{758}{103} \approx 7.36 \end{array} $$
(Solution to the above $3\times3$ system provided by this online tool)
Which means Alice is betting
against
with odds $7:11.45$, Bob bettingfor
with odds $4.09:3$,and Charlie bettingfor
with odds $7.36:4$. How much should each person put in, if the total initial bet is $\$100$?For Alice it should be $\frac{100 \times 11.45}{11.45+3+4} = \$62.06 $, for Bob $\frac{100\times 3}{11.45+3+4} = \$16.26 $, and for Charlie $\frac{100\times 4}{11.45+3+4} = \$21.68$.
This solution corresponds to our general/parametric solution where $t = 21.68$