# Pig Wheel question

A friend of mine was playing the bar game Pig Wheel recently and posed some interesting questions to me. He was playing with others as a group of four and, acting collectively, they came out about even after two hours, which surprised him. That got me thinking about the game.

So the game:

There are 45 numbers on a wheel, you place a bet on a number, the wheel is spun, and if you win, the payout is 40-to-1.

Let x = number of spots bet on.

Let y = amount place on each spot (assuming evenly distributed - which doesn't change any of the math below)

$$\frac{x}{45}\left(40-x\right)y - \frac{45-x}{45}xy = -\frac{xy}{9}$$

Here $xy$ is the total amount bet on the spin. So you're losing on average roughly 11 cents on dollar you put in (per spin).

On to what has stumped me:

They were betting \$10 on 8 numbers every spin and had a bank of \$400. Let's say the group saw a spin every two minutes, for a total of 60 spins. What is probability they come out even at the end of those 60 spins i.e. what is the probability that they 'succeed' on 12 spins?

Thoughts:

$$\binom{60}{12}\left(\frac{8}{45}\right)^{12}\left(\frac{37}{45}\right)^{48}\approx .116$$

But this is an overestimation, since it includes junk sequences such as 48 failures followed by 12 successes, which clearly would not be possible in this real-life example. It seems like quite a significant overestimation since the reverse of most 'good' sequences are 'bad' sequences but not vice versa.

I've thought about this more and have more I could say but I'll stop here for now and toss it out for others to think over.

• "since it includes junk sequences such as 48 failures followed by 12 successes, which clearly would not be possible in this real-life example." --- I was about to take you to task for falling for a gambler's fallacy here, but then I realized that you're actually right: you'd run out of bank after a mere 5 bad spins! Sep 17 '14 at 2:49

I had a go of it in Excel. If you manage to get through 60 spins and still have money -- 84.6% of days you won't -- coming out even is the third most common possibility: 2.56% of all days, and 16.7% of all successful days.

Here's my workings. It's quite number-crunchy, which is unfortunate, but it felt the most straightforward. https://docs.google.com/spreadsheets/d/1GZGzHHbSSQFSpk_yRerQY1gnjVSWr0DXfkRJozmoEH4/edit?usp=sharing

• Your results add up to more than $100\%$. A brief simulation confirms $84.6\%$ and $2.6\%$ for loss and break-even, with the remaining $12.8\%$ having net gains. Jan 19 '19 at 13:15
• Take care: the 16.7% isn't supposed to add up with the other two, it's 0.0256/(1-0.846), the proportion of the time you break even given that you finish the run of 60 spins. Jan 19 '19 at 13:22
• Ah, I see now. That is break-even on $16.7\%$ of the days you don't lose. Jan 19 '19 at 13:42

Update: Thanks to Dan Uznanski in a comment below for raising the possibility that the OP intended for winning bets to be paid out at odds of $$\mathit{39}$$ to $$\mathit{1}$$, rather than $$40$$ to $$1$$. This would simplify the problem a great deal, because the players' bank would then always remain a multiple of $$\80$$. It could be treated as a much simpler Markov chain—whose state is simply the multiple of $$\80$$ which it contains—, and the problem raised in the first paragraph of my original answer (included below) disappears.

If we assume that's what was intended, and we let $$\ 80S_n\$$ be the size of the players' bank, in dollars, after $$\ n\$$ spins, then $$\ \left\{S_n\right\}_{n=0}^{60}\$$ is a Markov chain, whose state distributions can be calculated from the recursion: $$\begin{eqnarray} \mathrm{Pr}\left(S_0 =s\right) &=& \delta_{s5}\\ \mathrm{Pr}\left(S_n=s\right)&=& \frac{37}{45}\mathrm{Pr}\left(S_{n-1}=s+1\right)\\ &&+\, \frac{8}{45}\mathrm{Pr}\left(S_{n-1}=s-4\right)\ \mbox{for }s\ge 5\\ \mathrm{Pr}\left(S_n=s\right)&=&\frac{37}{45}\mathrm{Pr}\left(S_{n-1}=s+1\right)\ \mbox{for }\ 1\le s\le 4\\ \mathrm{Pr}\left(S_n=0\right)&=&\mathrm{Pr}\left(S_{n-1}=0\right)+\frac{37}{45}\mathrm{Pr}\left(S_{n-1}=1\right)\ . \end{eqnarray}$$ Under these assumptions, according to my script for performing the above calculations, there's an $$84.6\%$$ chance that the players will go broke (i.e. $$\mathrm{Pr}\left(S_{60}=0\right)\approx 0.846\$$), and a $$12.8\%$$ chance they will come out ahead (i.e. $$\mathrm{Pr}\left(S_{60}\ge 6\right)\approx 0.128\$$). There's only a small chance of about $$2.6\%$$ of the players coming out "about even", which is actually $$\ \mathrm{Pr}\left(S_{60}=5\right)\$$, the probability that the players win exactly $$12$$ times, and therefore come out exactly even. It turns out that $$\ \mathrm{Pr}\left(S_{60}=s\right)=0\$$ for $$\ 1\le s\le 4$$, and $$\ 6\le s\le 9\$$.

In fact, $$\ \mathrm{Pr}\left(S_{60}=s\right)\ne0\$$ only if $$\ s\$$ is a multiple of $$5$$, because if the players win $$\ w\$$ times, and therefore lose $$\ 60-w\$$ times if they complete $$60$$ spins, they will win $$4w \times \80\$$ and lose $$\ \left(60-w\right)\times \80\$$ for a total nett gain of $$\ 5\left(w-12\right) \times \80\$$.

Original answer: As stated, the problem is incompletely specified, because we're not told what the players will do if their bank of $$\ \10b\$$, say, drops below $$\80$$. I will presume they simply bet their whole remaining bank, in lots of $$\10$$, on $$\ b\$$ numbers. While they will still need only a total of at most $$12$$ successes to at least break even, their probability of success on this spin is no longer $$\ \frac{8}{45}\$$, but only $$\ \frac{b}{45}\$$, and their winnings will be $$\ \10\left(41-b\right)\$$ on that spin, instead of only $$\330$$.

I doubt if there's any simple expression for the probability that the players come out "about" even, but if $$\ 10B_n\$$ is the size of their bank after $$\ n\$$ spins, it's easy enough to calculate the distribution of $$\ B_n\$$ recursively from the following equations: $$\begin{eqnarray} \mathrm{Pr}\left(B_0 =j\right) &=& \delta_{j\,40}\\ \mathrm{Pr}\left(B_n=j\right)&=& \frac{37}{45}\mathrm{Pr}\left(B_{n-1}=j+8\right)\\ &&+\, \frac{8}{45}\mathrm{Pr}\left(B_{n-1}=j-33\right)\ \mbox{for }j\ge 42\\ \mathrm{Pr}\left(B_n=41\right)&=& \frac{37}{45}\mathrm{Pr}\left(B_{n-1}=49\right)\\ &&+\,\sum_\limits{d=1}^{7} \frac{d}{45}\mathrm{Pr}\left(B_{n-1}=d\right)\\ \mathrm{Pr}\left(B_n=j\right)&=&\frac{37}{45}\mathrm{Pr}\left(B_{n-1}=j+8\right)\ \mbox{for }\ 1\le j\le 40\ . \end{eqnarray}$$ According to my script to carry out this computation, there's an $$\ 82.3\%$$ chance that the players will go broke, and a $$\ 16.3\%\$$ chance that they will come out ahead. The probability that they will come out "about even" is minuscule. The probability that they end up winning or losing no more than $$\100$$ is $$2.1\times10^{-7}\$$. The second most likely outcome (after them going broke), with a probability of $$0.034$$ is that the players come out $$\530$$ ahead.

The OP is correct in his or her conclusion that the binomial estimate, $$\ B\left(60,\frac{8}{45}\right)\left(12\right)\$$, significantly overstates the probability of obtaining $$12$$ successes in $$60$$ spins. The probability of winning $$\330$$ exactly $$12$$ times, for a total gain of $$\120$$, is $$\ 0.0277\$$, about a quarter of the binomial estimate. There are a few extra, but much smaller, chances of winning exactly $$12$$ times (e.g. a probability of $$\ 3.8\times10^{-4}\$$ of winning $$\330$$ exactly $$11$$ times, and $$\340$$ exactly once). However, their contribution to the total probability of exactly $$12$$ successes is negligible.

• Recheck your assumption about what dollar values are available. The only possible outcomes of a spin are -$80 and +$320, so since we start on a multiple of $80 we will always remain on such a number Jun 23 '19 at 11:29 • I don't beleive that is correct. Odds of$40$to$1$means a winning bet of$\$10$ wins $\$400$, so the players get their$\$10$ back plus the $\$400$winnings, for a total payout of$\$410$. Thus, since their bank goes down by $\$80$when they place their bets, and then up by$\$410$ if one of their bets wins, their bank will increase by $\$330$whenever any of their numbers wins. An alternative way of looking at it is that whenever one of their numbers wins, it wins$\$400$, while the other $7$ bets, totalling $\$70$, are lost, for a total nett gain of$\$400-\$70 = \$330\$. Jun 23 '19 at 12:48
• On looking more closely at the OP's calculation, $$\frac{x}{45}\left(40-x\right)y - \frac{45-x}{45}xy = -\frac{xy}{9}\ ,$$ it would appear that he or she was assuming that a winning bet would only pay out at odds of $\mathit{39}$ to $\mathit{1}$, not $40$ to $1$. If that is what was actually meant, then your observation would be correct, and it would simplify the problem a lot. Jun 23 '19 at 13:15