## Poll

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**13 members have voted**

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. 😁

Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition. 🤗

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)

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?

It's not whether you win or lose; it's whether or not you had a good bet.

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 - y

^{2}, and the sum over the entire triangle is:

Integral (0, 1) (y - y

^{2}) dy

= (y

^{2}/ 2 - y

^{3}/ 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 - y^{2}, and the sum over the entire triangle is:

Integral (0, 1) (y - y^{2}) dy

= (y^{2}/ 2 - y^{3}/ 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.

It's not whether you win or lose; it's whether or not you had a good bet.

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 - y^{2}, and the sum over the entire triangle is:

Integral (0, 1) (y - y^{2}) dy

= (y^{2}/ 2 - y^{3}/ 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.

#FreeNATHAN
#Paytheslaves

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.