Trying to figure out the math behind an arcade game
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January 30th, 2019 at 10:47:29 PM
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I tried looking for proability and odds calculators but wasn't able to find one that would do what I need. The closest thing I could come up with is a multi dice roller that showed the distribution of results but none allowed for custom value input let alone multiple events.
The game I'm trying to figure out the odds on is called "The Amazing Road Trip", and it works as follows: A wheel with 24 equally sized spaces is spun by the player 3 times. (While the game claims to be skill, it's generally assumed the outcome of any individual spin will be random and equally distributed between the possible outcomes.)
The possible values in order from low to high are: 25, 25, 25, 35, 35, 35, 50, 50, 50, 50, 75, 75, 75, 75, 100, 100, 100, 150, 150, 150, 200, and three ??? spaces, which will always be between 100 and 300, inclusive. (These are set to a payout control; in general, if the game has paid out more than the average recently, these values will be on the lower end, and if the game has not paid out a significant prize in some time, it will be on the higher end.)
In the first round, if the player totals less than 250 points, their game is over.
If the player meets or exceeds 250 points, they can quit and take a nominal prize (which they should never do, it's a bouncy ball in this case), or play the second level.
In the second level, the player starts over with 0 score and 3 spins, and must attempt to meet or exceed 450 points. If they fail, their game is over; if they succeed, they may either take a prize worth $25, or forfeit it to play the third level.
In the third level, they start again with 0 points and 3 spins, and must attempt to meet or exceed 500 points. If they fail, they get nothing and the game is over; if they succeed, they win a prize worth $100.
Due to promotional offers, the true cost of an individual game is either $0.70 or $0.80; without any promotional offers the cost of the game is $1.00.
What I'm trying to figure out and be able to plug in specific values into equations:
What are the chances of making at least 250 points in 3 spins, or 450 points in 3 spins, or 500 points in 3 spins, if the mystery values are 100 points? Or 200 points? Or 300 points?
Assuming a player will always attempt to get to the second level and win there, what are their odds of success on any individual game? Is there any scenario in which it is in the player's best interest to forfeit the $25 prize obtained on the second level to attempt to win the $100 prize on the third level, or does the math not support it even if the mystery values would somehow always award their maximum of 300?
The game I'm trying to figure out the odds on is called "The Amazing Road Trip", and it works as follows: A wheel with 24 equally sized spaces is spun by the player 3 times. (While the game claims to be skill, it's generally assumed the outcome of any individual spin will be random and equally distributed between the possible outcomes.)
The possible values in order from low to high are: 25, 25, 25, 35, 35, 35, 50, 50, 50, 50, 75, 75, 75, 75, 100, 100, 100, 150, 150, 150, 200, and three ??? spaces, which will always be between 100 and 300, inclusive. (These are set to a payout control; in general, if the game has paid out more than the average recently, these values will be on the lower end, and if the game has not paid out a significant prize in some time, it will be on the higher end.)
In the first round, if the player totals less than 250 points, their game is over.
If the player meets or exceeds 250 points, they can quit and take a nominal prize (which they should never do, it's a bouncy ball in this case), or play the second level.
In the second level, the player starts over with 0 score and 3 spins, and must attempt to meet or exceed 450 points. If they fail, their game is over; if they succeed, they may either take a prize worth $25, or forfeit it to play the third level.
In the third level, they start again with 0 points and 3 spins, and must attempt to meet or exceed 500 points. If they fail, they get nothing and the game is over; if they succeed, they win a prize worth $100.
Due to promotional offers, the true cost of an individual game is either $0.70 or $0.80; without any promotional offers the cost of the game is $1.00.
What I'm trying to figure out and be able to plug in specific values into equations:
What are the chances of making at least 250 points in 3 spins, or 450 points in 3 spins, or 500 points in 3 spins, if the mystery values are 100 points? Or 200 points? Or 300 points?
Assuming a player will always attempt to get to the second level and win there, what are their odds of success on any individual game? Is there any scenario in which it is in the player's best interest to forfeit the $25 prize obtained on the second level to attempt to win the $100 prize on the third level, or does the math not support it even if the mystery values would somehow always award their maximum of 300?
January 31st, 2019 at 10:25:57 AM
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There's a standard deviation calculator here.
I don't believe the values on the wheel are equally weighted, since it is supposed to be a game of skill. But you can do the calculation assuming they are.
With an average value of 85 - 95 per spin, but variance in the mid to upper 50's, it will be fairly common to hit the small prize total of 250, but 450 - 500 will be very rare.
I don't believe the values on the wheel are equally weighted, since it is supposed to be a game of skill. But you can do the calculation assuming they are.
With an average value of 85 - 95 per spin, but variance in the mid to upper 50's, it will be fairly common to hit the small prize total of 250, but 450 - 500 will be very rare.
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 31st, 2019 at 1:05:24 PM
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1st QQuote: MewtwoThe possible values in order from low to high are: 25, 25, 25, 35, 35, 35, 50, 50, 50, 50, 75, 75, 75, 75, 100, 100, 100, 150, 150, 150, 200, and three ??? spaces, which will always be between 100 and 300, inclusive.
What are the chances of making at least 250 points in 3 spins, or 450 points in 3 spins, or 500 points in 3 spins, if the mystery values are 100 points? Or 200 points? Or 300 points?
what ARE the mystery values when the game 1st starts before any games are played?
there must be some values given to each one.
and is it assumed that all mystery values are the same?
case 1
M1:200
M2:200
M3:200
or could it be this
case 2
M1:160
M2:210
M3:250
for case 1
here is the distribution of the 1st 3 spins
just expand this polynomial
(3*x^25+3*x^35+4*x^50+4*x^75+3*x^100+3*x^150+x^200+3*x^200)^3
using pari/gp
64*x^600 + 144*x^550 + 252*x^500 + 192*x^475 + 435*x^450 + 144*x^435 + 432*x^425 + 477*x^400 + 216*x^385 + 612*x^375 + 669*x^350 + 297*x^335 + 897*x^325 + 288*x^310 + 867*x^300 + 450*x^285 + 846*x^275 + 108*x^270 + 432*x^260 + 720*x^250 + 297*x^235 + 649*x^225 + 81*x^220 + 378*x^210 + 633*x^200 + 360*x^185 + 552*x^175 + 81*x^170 + 450*x^160 + 433*x^150 + 108*x^145 + 360*x^135 + 252*x^125 + 108*x^120 + 216*x^110 + 27*x^105 + 108*x^100 + 81*x^95 + 81*x^85 + 27*x^75
you should be able to get the probabilities from this
Mvalues all = 200
27*x^75 means there are 27 ways to get a sum of 75
like so
a = total sums = 13,824
b = less than 250 = 5,282
probability less than 250 = b/a = 2641/6912
close to 0.38208912
| ways | sum |
|---|---|
| 64 | 600 |
| 144 | 550 |
| 252 | 500 |
| 192 | 475 |
| 435 | 450 |
| 144 | 435 |
| 432 | 425 |
| 477 | 400 |
| 216 | 385 |
| 612 | 375 |
| 669 | 350 |
| 297 | 335 |
| 897 | 325 |
| 288 | 310 |
| 867 | 300 |
| 450 | 285 |
| 846 | 275 |
| 108 | 270 |
| 432 | 260 |
| 720 | 250 |
| 297 | 235 |
| 649 | 225 |
| 81 | 220 |
| 378 | 210 |
| 633 | 200 |
| 360 | 185 |
| 552 | 175 |
| 81 | 170 |
| 450 | 160 |
| 433 | 150 |
| 108 | 145 |
| 360 | 135 |
| 252 | 125 |
| 108 | 120 |
| 216 | 110 |
| 27 | 105 |
| 108 | 100 |
| 81 | 95 |
| 81 | 85 |
| 27 | 75 |
| 13824 | total |
time is short
hope this helps some
those mystery values mess things up
knowing the default values of each before any game is played
would be of great help
winsome johnny (not Win some johnny)
January 31st, 2019 at 5:47:24 PM
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Quote: 7craps...probability less than 250 = b/a = 2641/6912...close to 0.38208912...
I got that, too, for when the 3 mystery prizes are 200. That gives me confidence in my other results:
| Probability | 100 | 200 | 300 |
|---|---|---|---|
| at least 250 | 0.46557 | 0.61791 | 0.61791 |
| at least 450 | 0.01548 | 0.07863 | 0.20038 |
| at least 500 | 0.00398 | 0.03328 | 0.12768 |
January 31st, 2019 at 9:24:56 PM
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Looks to me we agreeQuote: ChesterDogI got that, too, for when the 3 mystery prizes are 200. That gives me confidence in my other results:
Probability 100 200 300 at least 250 0.46557 0.61791 0.61791 at least 450 0.01548 0.07863 0.20038 at least 500 0.00398 0.03328 0.12768
| Probability | 100 | 200 | 300 |
|---|---|---|---|
| at least 250 | 0.46556713 | 0.61791088 | 0.61791088 |
| at least 450 | 0.015480324 | 0.078631366 | 0.200376157 |
| at least 500 | 0.003978588 | 0.033275463 | 0.127676505 |
as to the second OP set of Qs
"1)Assuming a player will always attempt to get to the second level and win there, what are their odds of success on any individual game?
2)Is there any scenario in which it is in the player's best interest to forfeit the $25 prize obtained on the second level to attempt to win the $100 prize on the third level,
3)or does the math not support it even if the mystery values would somehow always award their maximum of 300?"
1) is now answered
have not looked at the other 2
looks to be simple expected value type math
super bowl weekend ready time
enjoy
Mystery values = 100
ways/sums
| WAYS | SUM |
|---|---|
| 1 | 600 |
| 9 | 550 |
| 45 | 500 |
| 12 | 475 |
| 147 | 450 |
| 9 | 435 |
| 81 | 425 |
| 342 | 400 |
| 54 | 385 |
| 306 | 375 |
| 624 | 350 |
| 189 | 335 |
| 717 | 325 |
| 72 | 310 |
| 912 | 300 |
| 396 | 285 |
| 1116 | 275 |
| 27 | 270 |
| 270 | 260 |
| 1107 | 250 |
| 540 | 235 |
| 1180 | 225 |
| 81 | 220 |
| 594 | 210 |
| 993 | 200 |
| 576 | 185 |
| 768 | 175 |
| 162 | 170 |
| 612 | 160 |
| 514 | 150 |
| 108 | 145 |
| 360 | 135 |
| 252 | 125 |
| 108 | 120 |
| 216 | 110 |
| 27 | 105 |
| 108 | 100 |
| 81 | 95 |
| 81 | 85 |
| 27 | 75 |
| 13824 | TOTAL |
Mystery values = 300
ways/sums
| ways | sum |
|---|---|
| 27 | 900 |
| 27 | 800 |
| 81 | 750 |
| 90 | 700 |
| 108 | 675 |
| 162 | 650 |
| 81 | 635 |
| 81 | 625 |
| 136 | 600 |
| 72 | 575 |
| 243 | 550 |
| 54 | 535 |
| 270 | 525 |
| 333 | 500 |
| 162 | 485 |
| 390 | 475 |
| 453 | 450 |
| 171 | 435 |
| 531 | 425 |
| 216 | 410 |
| 540 | 400 |
| 270 | 385 |
| 450 | 375 |
| 81 | 370 |
| 162 | 360 |
| 390 | 350 |
| 135 | 335 |
| 447 | 325 |
| 72 | 310 |
| 507 | 300 |
| 234 | 285 |
| 630 | 275 |
| 27 | 270 |
| 270 | 260 |
| 639 | 250 |
| 297 | 235 |
| 649 | 225 |
| 81 | 220 |
| 378 | 210 |
| 633 | 200 |
| 360 | 185 |
| 552 | 175 |
| 81 | 170 |
| 450 | 160 |
| 433 | 150 |
| 108 | 145 |
| 360 | 135 |
| 252 | 125 |
| 108 | 120 |
| 216 | 110 |
| 27 | 105 |
| 108 | 100 |
| 81 | 95 |
| 81 | 85 |
| 27 | 75 |
| 13824 | TOTAL |
winsome johnny (not Win some johnny)
January 31st, 2019 at 11:43:20 PM
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Quote: 7crapsLooks to me we agreeQuote: ChesterDogI got that, too, for when the 3 mystery prizes are 200. That gives me confidence in my other results:
Probability 100 200 300 at least 250 0.46557 0.61791 0.61791 at least 450 0.01548 0.07863 0.20038 at least 500 0.00398 0.03328 0.12768
Probability 100 200 300 at least 250 0.46556713 0.61791088 0.61791088 at least 450 0.015480324 0.078631366 0.200376157 at least 500 0.003978588 0.033275463 0.127676505
as to the second OP set of Qs
"1)Assuming a player will always attempt to get to the second level and win there, what are their odds of success on any individual game?
2)Is there any scenario in which it is in the player's best interest to forfeit the $25 prize obtained on the second level to attempt to win the $100 prize on the third level,
3)or does the math not support it even if the mystery values would somehow always award their maximum of 300?"
1) is now answered
have not looked at the other 2
looks to be simple expected value type math
super bowl weekend ready time
enjoy
Mystery values = 100
ways/sums
WAYS SUM 1 600 9 550 45 500 12 475 147 450 9 435 81 425 342 400 54 385 306 375 624 350 189 335 717 325 72 310 912 300 396 285 1116 275 27 270 270 260 1107 250 540 235 1180 225 81 220 594 210 993 200 576 185 768 175 162 170 612 160 514 150 108 145 360 135 252 125 108 120 216 110 27 105 108 100 81 95 81 85 27 75 13824 TOTAL
Mystery values = 300
ways/sums
ways sum 27 900 27 800 81 750 90 700 108 675 162 650 81 635 81 625 136 600 72 575 243 550 54 535 270 525 333 500 162 485 390 475 453 450 171 435 531 425 216 410 540 400 270 385 450 375 81 370 162 360 390 350 135 335 447 325 72 310 507 300 234 285 630 275 27 270 270 260 639 250 297 235 649 225 81 220 378 210 633 200 360 185 552 175 81 170 450 160 433 150 108 145 360 135 252 125 108 120 216 110 27 105 108 100 81 95 81 85 27 75 13824 TOTAL
It looks like you should never go for the $100.
February 1st, 2019 at 8:57:19 PM
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I want to thank all of you so much for your help!
7craps - thank you for letting me know about Pari/GP as this will work for any possible combination I could throw at the problem!
And honestly everyone here went above and beyond going through and confirming the math on this.
Looks like the $25 gift card is +EV at even middle of the road Mystery Values (especially if playing under the promotion combination that allows for games to cost 70 cents, it looks like you need better than 1 win out of 35 to break even assuming 200 mystery, and odds of winning show to be 1/20.58 at that level...)
7craps - thank you for letting me know about Pari/GP as this will work for any possible combination I could throw at the problem!
And honestly everyone here went above and beyond going through and confirming the math on this.
Looks like the $25 gift card is +EV at even middle of the road Mystery Values (especially if playing under the promotion combination that allows for games to cost 70 cents, it looks like you need better than 1 win out of 35 to break even assuming 200 mystery, and odds of winning show to be 1/20.58 at that level...)
February 1st, 2019 at 9:14:25 PM
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Quote: MewtwoI want to thank all of you so much for your help!
7craps - thank you for letting me know about Pari/GP as this will work for any possible combination I could throw at the problem!
And honestly everyone here went above and beyond going through and confirming the math on this.
Looks like the $25 gift card is +EV at even middle of the road Mystery Values (especially if playing under the promotion combination that allows for games to cost 70 cents, it looks like you need better than 1 win out of 35 to break even assuming 200 mystery, and odds of winning show to be 1/20.58 at that level...)
One thing i would be concerned with is the equal weighting of the wheel assumption. Most casino games that use wheels to award bonuses are weighted heavily toward the lower prize values.
February 2nd, 2019 at 8:03:54 AM
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glad it helped out.Quote: Mewtwo7craps - thank you for letting me know about Pari/GP as this will work for any possible combination I could throw at the problem!
Pari/GP is a handy calculator tool. (not the only one of it's type)
(3*x^25+3*x^35+4*x^50+4*x^75+3*x^100+3*x^150+x^200+3*x^200)^3
is a representation of
one 24-sided die rolled 3 times
or
three 24-sided dice rolled 1 time (for dice lovers)
a standard 6-sided die (1d6) is this (x+x^2+x^3+x^4+x^5+x^6)
two standard 6-sided dice (2d6) is this (x+x^2+x^3+x^4+x^5+x^6)^2
and so on
(it is called a 'generating function')
this is also 'drawing with replacement' in math lingo
'drawing without replacement'
(Keno draw sums for example)
requires some different math (and formula)
can be found here
https://wizardofvegas.com/forum/questions-and-answers/gambling/10526-what-is-keno-exact-scores-house-edge/3/#post652407
the pari/gp formula there actually prints out a very large result
it could be shortened to return a result for drawing just 20 items
winsome johnny (not Win some johnny)
February 5th, 2019 at 5:37:03 PM
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It is a physical wheel, not a virtual one, where each space is the same width. There's no weighting to it.
If it were on a computer screen I absolutely understand it would be weighted against high values.
If it were on a computer screen I absolutely understand it would be weighted against high values.
February 5th, 2019 at 5:59:43 PM
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Quote: MewtwoIt is a physical wheel, not a virtual one, where each space is the same width. There's no weighting to it.
If it were on a computer screen I absolutely understand it would be weighted against high values.
How do you know for sure???
The Wheel of Fortune slot machine uses a physical wheel, but each equally sized slice have much different weightings. How do you KNOW for sure it's not set up this way? Do you have a bunch of data implying that it's unweighted?
