## Poll

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**13 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.

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

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?

If you are enjoying the game, you're already winning.

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.

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

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

If you are enjoying the game, you're already winning.

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.

They tried to kill us, Jimmy. They did. They're dirty f**king cops! *photo is not of an AP, just a morbidly obese hill billy. Jar Jar Binks was supposed to be the Sith Lord . Nathan is going to run and own this place

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.

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

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.

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