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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.

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    $\begingroup$ "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! $\endgroup$ Sep 17 '14 at 2:49
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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

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  • $\begingroup$ 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. $\endgroup$ Jan 19 '19 at 13:15
  • $\begingroup$ 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. $\endgroup$ Jan 19 '19 at 13:22
  • $\begingroup$ Ah, I see now. That is break-even on $16.7\%$ of the days you don't lose. $\endgroup$ Jan 19 '19 at 13:42
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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.

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    $\begingroup$ 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 $\endgroup$ Jun 23 '19 at 11:29
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    $\begingroup$ 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\ $. $\endgroup$ Jun 23 '19 at 12:48
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    $\begingroup$ 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. $\endgroup$ Jun 23 '19 at 13:15

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