Pig Wheel question
$begingroup$
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)
$$frac{x}{45}left(40-xright)y - frac{45-x}{45}xy = -frac{xy}{9}$$
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.
probability recreational-mathematics
asked Sep 17 '14 at 2:34
camagacamaga
111
$endgroup$
add a comment |
$begingroup$
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)
$$frac{x}{45}left(40-xright)y - frac{45-x}{45}xy = -frac{xy}{9}$$
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.
probability recreational-mathematics
asked Sep 17 '14 at 2:34
camagacamaga
111
$endgroup$
1
$begingroup$
"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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49
add a comment |
$begingroup$
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)
$$frac{x}{45}left(40-xright)y - frac{45-x}{45}xy = -frac{xy}{9}$$
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.
probability recreational-mathematics
asked Sep 17 '14 at 2:34
camagacamaga
111
$endgroup$
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)
$$frac{x}{45}left(40-xright)y - frac{45-x}{45}xy = -frac{xy}{9}$$
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.
probability recreational-mathematics
probability recreational-mathematics
asked Sep 17 '14 at 2:34
camagacamaga
111
asked Sep 17 '14 at 2:34
camagacamaga
111
asked Sep 17 '14 at 2:34
camagacamaga
111
asked Sep 17 '14 at 2:34
camagacamaga
111
asked Sep 17 '14 at 2:34
camagacamaga
111
111
1
$begingroup$
"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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49
add a comment |
1
$begingroup$
"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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49
1
1
$begingroup$
"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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49
$begingroup$
"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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49
add a comment |
1 Answer
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votes
$begingroup$
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
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
$endgroup$
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
add a comment |
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$begingroup$
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
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
$endgroup$
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
add a comment |
$begingroup$
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
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
$endgroup$
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
add a comment |
$begingroup$
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
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
$endgroup$
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
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
answered Sep 17 '14 at 3:50
Dan UznanskiDan Uznanski
6,65521427
6,65521427
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
add a comment |
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 19 at 13:15
$begingroup$
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.
$endgroup$
– Daniel Mathias
Jan 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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
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.
$endgroup$
– Dan Uznanski
Jan 19 at 13:22
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
$begingroup$
Ah, I see now. That is break-even on $16.7%$ of the days you don't lose.
$endgroup$
– Daniel Mathias
Jan 19 at 13:42
add a comment |
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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!
$endgroup$
– Dan Uznanski
Sep 17 '14 at 2:49