Mewtwo Joined: Jul 28, 2010
• Posts: 8
January 30th, 2019 at 10:47:29 PM permalink
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?
Ayecarumba Joined: Nov 17, 2009
• Posts: 6763
January 31st, 2019 at 10:25:57 AM permalink
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.
Simplicity is the ultimate sophistication - Leonardo da Vinci
7craps Joined: Jan 23, 2010
• Posts: 1977
January 31st, 2019 at 1:05:24 PM permalink
Quote: Mewtwo

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.

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?

1st Q
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
wayssum
64600
144550
252500
192475
435450
144435
432425
477400
216385
612375
669350
297335
897325
288310
867300
450285
846275
108270
432260
720250
297235
649225
81220
378210
633200
360185
552175
81170
450160
433150
108145
360135
252125
108120
216110
27105
108100
8195
8185
2775
13824total

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)
ChesterDog Joined: Jul 26, 2010
• Posts: 858
January 31st, 2019 at 5:47:24 PM permalink
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:

Probability100200300
at least 2500.465570.617910.61791
at least 4500.015480.078630.20038
at least 5000.003980.033280.12768
7craps Joined: Jan 23, 2010
• Posts: 1977
January 31st, 2019 at 9:24:56 PM permalink
Quote: ChesterDog

I got that, too, for when the 3 mystery prizes are 200. That gives me confidence in my other results:

Probability100200300
at least 2500.465570.617910.61791
at least 4500.015480.078630.20038
at least 5000.003980.033280.12768

Looks to me we agree
Probability100200300
at least 2500.465567130.617910880.61791088
at least 4500.0154803240.0786313660.200376157
at least 5000.0039785880.0332754630.127676505

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?"

have not looked at the other 2
looks to be simple expected value type math

enjoy

Mystery values = 100
ways/sums
WAYSSUM
1600
9550
45500
12475
147450
9435
81425
342400
54385
306375
624350
189335
717325
72310
912300
396285
1116275
27270
270260
1107250
540235
1180225
81220
594210
993200
576185
768175
162170
612160
514150
108145
360135
252125
108120
216110
27105
108100
8195
8185
2775
13824TOTAL

Mystery values = 300
ways/sums
wayssum
27900
27800
81750
90700
108675
162650
81635
81625
136600
72575
243550
54535
270525
333500
162485
390475
453450
171435
531425
216410
540400
270385
450375
81370
162360
390350
135335
447325
72310
507300
234285
630275
27270
270260
639250
297235
649225
81220
378210
633200
360185
552175
81170
450160
433150
108145
360135
252125
108120
216110
27105
108100
8195
8185
2775
13824TOTAL
winsome johnny (not Win some johnny)
tringlomane Joined: Aug 25, 2012
• Posts: 6255
January 31st, 2019 at 11:43:20 PM permalink
Quote: 7craps

Quote: ChesterDog

I got that, too, for when the 3 mystery prizes are 200. That gives me confidence in my other results:

Probability100200300
at least 2500.465570.617910.61791
at least 4500.015480.078630.20038
at least 5000.003980.033280.12768

Looks to me we agree
Probability100200300
at least 2500.465567130.617910880.61791088
at least 4500.0154803240.0786313660.200376157
at least 5000.0039785880.0332754630.127676505

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?"

have not looked at the other 2
looks to be simple expected value type math

enjoy

Mystery values = 100
ways/sums
WAYSSUM
1600
9550
45500
12475
147450
9435
81425
342400
54385
306375
624350
189335
717325
72310
912300
396285
1116275
27270
270260
1107250
540235
1180225
81220
594210
993200
576185
768175
162170
612160
514150
108145
360135
252125
108120
216110
27105
108100
8195
8185
2775
13824TOTAL

Mystery values = 300
ways/sums
wayssum
27900
27800
81750
90700
108675
162650
81635
81625
136600
72575
243550
54535
270525
333500
162485
390475
453450
171435
531425
216410
540400
270385
450375
81370
162360
390350
135335
447325
72310
507300
234285
630275
27270
270260
639250
297235
649225
81220
378210
633200
360185
552175
81170
450160
433150
108145
360135
252125
108120
216110
27105
108100
8195
8185
2775
13824TOTAL

It looks like you should never go for the \$100.
Mewtwo Joined: Jul 28, 2010
• Posts: 8
February 1st, 2019 at 8:57:19 PM permalink
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...)
tringlomane Joined: Aug 25, 2012
• Posts: 6255
Thanks for this post from: February 1st, 2019 at 9:14:25 PM permalink
Quote: Mewtwo

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...)

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.
7craps Joined: Jan 23, 2010
• Posts: 1977
February 2nd, 2019 at 8:03:54 AM permalink
Quote: Mewtwo

7craps - 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

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)
Mewtwo Joined: Jul 28, 2010