Must hit by jackpot math question
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14 members have voted
November 21st, 2018 at 3:15:42 PM
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Let's say a "must hit by" jackpot starts at $2,500 and must hit by $5,000. The point at which a jackpot will hit is uniformly distributed in this range. If you survey a large number of such jackpots at any one time, what will be the average current jackpot size?
Free beer to the first correct answer and solution. A solution must be math-based, preferably geometry or calculus, as opposed to random simulation. Please put answers and solution in spoiler tags.
The question for the poll is what do you think the answer is?
p.s. Thanks ONM for correcting me on my poll options last time.
Free beer to the first correct answer and solution. A solution must be math-based, preferably geometry or calculus, as opposed to random simulation. Please put answers and solution in spoiler tags.
The question for the poll is what do you think the answer is?
p.s. Thanks ONM for correcting me on my poll options last time.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 21st, 2018 at 3:39:53 PM
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4723 I will post how I got to that shortly, probably not correct
November 21st, 2018 at 3:47:51 PM
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hmmmmm. It seems so easy, so it must not be.
Let's assume that players or potential players cannot see it for a moment.
Min value is 2,500
If it ever reached 5,000, it would instantly reset to 2,500, so it shouldn't be on 5,000 very long. so max value might be 4,999. And it's sometimes going to also reset at lower values.
Therefore it will be at say 4,000 less often than say 3,000......
So, the average value displayed will be less that the average (2,500 : 5,000) I'd expect considerably less.
Finger in the air.... about 2,500+(5,000-2,500)/e = $3,419
I've no real idea why, but I figured e has to come into it somewhere.
But then if the values are visible to potential players, then values close to 5,000 would be chased and would not spend long there. Thus further biasing the average displayed value lower?
Let's assume that players or potential players cannot see it for a moment.
Min value is 2,500
If it ever reached 5,000, it would instantly reset to 2,500, so it shouldn't be on 5,000 very long. so max value might be 4,999. And it's sometimes going to also reset at lower values.
Therefore it will be at say 4,000 less often than say 3,000......
So, the average value displayed will be less that the average (2,500 : 5,000) I'd expect considerably less.
Finger in the air.... about 2,500+(5,000-2,500)/e = $3,419
I've no real idea why, but I figured e has to come into it somewhere.
But then if the values are visible to potential players, then values close to 5,000 would be chased and would not spend long there. Thus further biasing the average displayed value lower?
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
November 21st, 2018 at 4:02:18 PM
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Keep those answers coming. I'll refrain from confirming or denying any for a while longer, to let everybody enjoy the problem.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 21st, 2018 at 4:05:57 PM
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OK,
I now have a calculated value of [redacted] give or take a dollar depending on granularity.
I didn't use e in that and ignored the effects of AP's
I now have a calculated value of [redacted] give or take a dollar depending on granularity.
I didn't use e in that and ignored the effects of AP's
Consider 2500 machines.
1st machine destined to pay out at 2501
At any instant in time it will show between 2500 and 2501 ( Let's take the average and call that a1 =2500.50)
2nd machine destined to pay at 2502
At any instant in time it will show between 2500 and 2502 ( Let's take the average and call that a2 =2501)
.
.
.
2500th machine destined to pay at 5000
At any instant in time it will show between 2500 and 5000 ( Let's take the average and call that a2500 =3725)
Take the average of those 2500 averages and get 3125.25
1st machine destined to pay out at 2501
At any instant in time it will show between 2500 and 2501 ( Let's take the average and call that a1 =2500.50)
2nd machine destined to pay at 2502
At any instant in time it will show between 2500 and 2502 ( Let's take the average and call that a2 =2501)
.
.
.
2500th machine destined to pay at 5000
At any instant in time it will show between 2500 and 5000 ( Let's take the average and call that a2500 =3725)
Take the average of those 2500 averages and get 3125.25
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
November 21st, 2018 at 5:10:46 PM
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Quote: WizardLet's say a "must hit by" jackpot starts at $2,500 and must hit by $5,000. The point at which a jackpot will hit is uniformly distributed in this range. If you survey a large number of such jackpots at any one time, what will be the average current jackpot size?
Free beer to the first correct answer and solution. A solution must be math-based, preferably geometry or calculus, as opposed to random simulation. Please put answers and solution in spoiler tags.
The question for the poll is what do you think the answer is?
p.s. Thanks ONM for correcting me on my poll options last time.
You don't need calculus. Once a JP hits, most people don't play them. They play the biggest one they can find that day, so therefore, the average will be below the halfway point, since most are untouched.
I am a robot.
November 21st, 2018 at 5:29:09 PM
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ONM makes a good point and it is safe to read his spoiler.
Let's assume that player behavior is unaffected by the jackpot amount. In other words, they will pick a machine at random to play, regardless of the jackpot.
Let's assume that player behavior is unaffected by the jackpot amount. In other words, they will pick a machine at random to play, regardless of the jackpot.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 21st, 2018 at 7:59:08 PM
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If it's truly random, $3750? ;-)
Playing it correctly means you've already won.
November 21st, 2018 at 8:17:30 PM
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Consider one random machine. This is the same as playing many random machines. This machine has the probability of being in a state with must hit jackpot at X with probability density (X-2500)/3125000, since it is proportional to the size of the interval of possible values the machine can hit (2500 to X). Each machine will be on average at half of its hit jackpot (X+2500)/2. So, we get an Expectation of Integrate[(X+2500)/2*(X-2500)/3125000, x, 2500, 5000] = $3333.33.
November 21st, 2018 at 9:26:23 PM
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Quote: djtehch34t
Consider one random machine. This is the same as playing many random machines. This machine has the probability of being in a state with must hit jackpot at X with probability density (X-2500)/3125000, since it is proportional to the size of the interval of possible values the machine can hit (2500 to X). Each machine will be on average at half of its hit jackpot (X+2500)/2. So, we get an Expectation of Integrate[(X+2500)/2*(X-2500)/3125000, x, 2500, 5000] = $3333.33.
Congratulations! That is correct. Good solution too. I owe you beer.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 22nd, 2018 at 6:20:51 AM
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Player behavior is a factor, even if they pick a machine at random.Quote: WizardIn other words, they will pick a machine at random to play, regardless of the jackpot.
Being Devil's Advocate here.
(1) What if all players play until they hit the jackpot, then immediately leave? Then the jackpots will always read $2,500.
(2) In reality, people don't play a random amount, and then randomly leave.
------
When non-random machine picking is allowed:
(3) In one geographic area, people believed that jackpots which hit high would hit again soon, so low jackpots (less than $100 movement; e.g. $2,500-2,599) were hard to find.
November 22nd, 2018 at 11:51:05 AM
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Quote: WizardQuote: djtehch34t
Consider one random machine. This is the same as playing many random machines. This machine has the probability of being in a state with must hit jackpot at X with probability density (X-2500)/3125000, since it is proportional to the size of the interval of possible values the machine can hit (2500 to X). Each machine will be on average at half of its hit jackpot (X+2500)/2. So, we get an Expectation of Integrate[(X+2500)/2*(X-2500)/3125000, x, 2500, 5000] = $3333.33.
Congratulations! That is correct. Good solution too. I owe you beer.
Great! I hope to make it out to Vegas sometime in 2019. Fun problem!
November 22nd, 2018 at 7:08:10 PM
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Note that the beer prize states neither one beer nor lifetime supply nor any quantity in between.Quote: Wizard...
Congratulations! That is correct. Good solution too. I owe you beer.
I sense an upcoming advantage play for beer. 😁
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
November 22nd, 2018 at 7:13:42 PM
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Quote: DJTeddyBearI sense an upcoming advantage play for beer. 😁
That summarizes most of my Vegas vacations. I think Wiz owes me a beer too, tbh. I think for helping verify Hot Roll slot math. Lol
November 23rd, 2018 at 7:37:18 PM
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I don’t believe calculus can provide an exact answer to this question since it assumes an infinite amount of jackpot values, which is not possible in the real world.
Calculus will tell you that the average value will be the low value plus exactly 1/3 of any range specified. That average value is a good estimate for larger ranges but a poor estimate for low ranges.
Fortunately this problem can be easily calculated exactly with basic math and a couple series summation tricks.
Since the range specified here is $2500 to $5000, there are 2,501 possible jackpot values (high but not infinite)
The average value of the sample will be :
2,500 * (1 + (2 * 2,500 + 1)/(3! * 2,501))
= 8,336,250 / 2,501 =~ 3,333.166733
The calculus estimate of 10,000/3 =~ 3333.333333 is very close since the range of 2500 is effectively approaching infinity in this scenario. But the average value of every finite value set will be always be less than that. If you run the calculation for 250,001 values (less plausible but that would be maximum possible number of values( $2500.00 to $5000.00) it’s even closer to 10,000/3, but still less.
As a caveat, my calculus is weak as I just started using it a few months ago. So forgive me if what I wrote is BS (but I don’t think so)
Calculus will tell you that the average value will be the low value plus exactly 1/3 of any range specified. That average value is a good estimate for larger ranges but a poor estimate for low ranges.
Fortunately this problem can be easily calculated exactly with basic math and a couple series summation tricks.
Since the range specified here is $2500 to $5000, there are 2,501 possible jackpot values (high but not infinite)
The average value of the sample will be :
2,500 * (1 + (2 * 2,500 + 1)/(3! * 2,501))
= 8,336,250 / 2,501 =~ 3,333.166733
The calculus estimate of 10,000/3 =~ 3333.333333 is very close since the range of 2500 is effectively approaching infinity in this scenario. But the average value of every finite value set will be always be less than that. If you run the calculation for 250,001 values (less plausible but that would be maximum possible number of values( $2500.00 to $5000.00) it’s even closer to 10,000/3, but still less.
As a caveat, my calculus is weak as I just started using it a few months ago. So forgive me if what I wrote is BS (but I don’t think so)
It’s all about making that GTA
November 25th, 2018 at 5:47:14 PM
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Quote: Ace2I don’t believe calculus can provide an exact answer to this question since it assumes an infinite amount of jackpot values, which is not possible in the real world.
I know you can't divide a penny, but in comparison to thousands of dollars, it is a negligible issue.
For those who must have a pure calculus problem, here you go.
Quote: I love integral calculusConsider the triangle bounded by (0,0), (1,0) and (1,1). What is the mean distance within the triangle from the nearest point on the x-axis?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 25th, 2018 at 7:01:15 PM
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Quote: WizardFor those who must have a pure calculus problem, here you go.
For any particular value y, the distance to the nearest point on the X-axis is y.
The length of the line segment in the triangle for that value y is 1 - y.
Thus, the sum of the values for a particular y is y (1 - y) = y - y2, and the sum over the entire triangle is:
Integral (0, 1) (y - y2) dy
= (y2 / 2 - y3 / 3) | {0, 1}
= (1/2 - 1/3) - (0 - 0) = 1/6
The area of the triangle is 1/2, so the mean is (1/6) / (1/2) = 1/3.
November 25th, 2018 at 7:04:03 PM
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Quote: ThatDonGuy
For any particular value y, the distance to the nearest point on the X-axis is y.
The length of the line segment in the triangle for that value y is 1 - y.
Thus, the sum of the values for a particular y is y (1 - y) = y - y2, and the sum over the entire triangle is:
Integral (0, 1) (y - y2) dy
= (y2 / 2 - y3 / 3) | {0, 1}
= (1/2 - 1/3) - (0 - 0) = 1/6
The area of the triangle is 1/2, so the mean is (1/6) / (1/2) = 1/3.
I agree! That is worthy of yet another Wizard beer.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 25th, 2018 at 7:12:49 PM
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I'm assuming since you're giving these beers out so freely, I don't want one. Must be piss in them.Quote: WizardQuote: ThatDonGuy
For any particular value y, the distance to the nearest point on the X-axis is y.
The length of the line segment in the triangle for that value y is 1 - y.
Thus, the sum of the values for a particular y is y (1 - y) = y - y2, and the sum over the entire triangle is:
Integral (0, 1) (y - y2) dy
= (y2 / 2 - y3 / 3) | {0, 1}
= (1/2 - 1/3) - (0 - 0) = 1/6
The area of the triangle is 1/2, so the mean is (1/6) / (1/2) = 1/3.
I agree! That is worthy of yet another Wizard beer.
I am a robot.
November 26th, 2018 at 4:54:38 PM
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There's actually a non-calculus way to calculate this.
Graph this in 3 dimensions, with z = y; the mean is the volume divided by the area at the base.
As it turns out, the graph is a cone; since the volume of a cone is 1/3 x its base area x its height, and, in this case, the height is 1, the mean = 1/3 x 1 = 1/3.
Any idea when the Spring Fling will be this year? If I can make it out there, I'm holding you to that "one round for the house" you promised.
Graph this in 3 dimensions, with z = y; the mean is the volume divided by the area at the base.
As it turns out, the graph is a cone; since the volume of a cone is 1/3 x its base area x its height, and, in this case, the height is 1, the mean = 1/3 x 1 = 1/3.
Any idea when the Spring Fling will be this year? If I can make it out there, I'm holding you to that "one round for the house" you promised.
