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I am a gigantic nerd [Jun. 27th, 2008|09:34 pm]
Lifto forwarded an email from his mom to a mailing list I'm on asking for help figuring out how much money her company (or whatever) will spend on a promotion that involves a Plinko board. And I just spent about an hour and a half figuring it out. Because I can, dammit. And because I'm sort of compulsive about that kind of challenge.

Here's the basic setup (I'll put the answer behind the cut, in case people want to think about it themselves).

Here are some pictures of Plinko boards. You drop a chip in at the top, it bounces around and falls into a slot at the bottom. They want to do one with 7 slots: $5 $10 $25 BP $25 $10 $5. BP means "Birthday Present", and if you get that, you get to go again and double your money. (But only one doubling -- after that, BP is just a re-roll.)

If they have 2100 people play this game, how much will it probably cost?

So the first thing I'll note is that the pictures show 8 slots at the bottom, not 7. I'm ignoring that. The technique is the same.

It turns out that there are a couple simplifying assumptions about this problem that make it a whole lot simpler than it might look. The first is that the edges of the board don't actually matter. If the layout of the prize slots is symmetric, then the chip reflecting off the wall is isomorphic to it wrapping around to the other side. So we can pretend it's a cylinder.

Second, the players can drop the chips in from any point at the top, and absent any actual data about where people like to drop chips for psychological reasons, we should assume that all starting positions are equally likely.

So, unless the pegs are seriously different from one part of the board to another, the actual dynamics of the drop are irrelevant, because whatever kind distribution they generate, it's cloned across all the possible starting positions, making all the ending positions equally likely as well. Which means it's basically just rolling a 7-sided die.

Thus, p = 1/7 for each slot, and with the doubling on BP that gives us the following outcomes:

 x     p(x)
$05  @ 6/21
$10  @ 7/21
$20  @ 1/21
$25  @ 6/21
$50  @ 1/21

The average payout per play is $(30+70+20+150+50)/21 = $15.24. Multiply that by 2100 players and you get $32,000 for the total.

But in terms of budgeting, they'll be happier if they look at the distribution a little bit, too. The plays are IID, so by the Central Limit Theorem, it'll converge to a gaussian. I ought to have been able to figure out what the variance is from pure theory, but math was hating me tonight, so I just simulated 10,000 sets of 2100 plays and determined that sigma is pretty close to 500. (Anybody know what the right number is, and why?)

So it'll probably cost them around $32,000, give or take $500, but there's a 1 in 50 chance it could cost as much as $33,000. Depending on how their budgets work, it might be important to budget $33k instead of $32k. This is something I've been doing a lot of at work: trying to communicate to people that the right way to respond to uncertainty is find out what the range of likely outcomes is and come up with a plan that will work wherever reality actually ends up.

Of course, they also need to figure out how much padding they need to cope with the uncertainty in the number of players, but I figured that was something they already knew about, being that this is a promotion to get people to come into their store or whatever.

They were originally thinking about letting players double indefinitely with the BP, but didn't know how much that would cost. The fun bit is that I realized you can actually figure that out really easily if you modify it just a little. If you change BP to "get an extra chip", then it's equivalent to playing with a six-slot board and adding an extra player with probability 1/7 every time someone plays. And you can figure out how many extra players you'll get because it's a geometric series: sum_k (1/n^k) = 1/(1-n) = 7/6 for n = 7. So your 2100 players become 2450 playing for $5, $10, or $25 (p = 1/3 for each), and you get a mean of $32,666 and a sigma of about $425.

Whee! Math is fun!

[User Picture]From: dr_tectonic
2008-06-28 02:39 pm (UTC)
The reason for the standard deviation being what it is is that it's sqrt(n) times whatever the standard deviation is for one toss.

Ah! I never massaged the math around properly to get to that conclusion. Thanks!

You want something like sum_k=0...infty (2^(k-1)/7^k), not what you have (I think I'm right, but, um, I just woke up).

Hmm... I think I basically forgot to give the extra players the opportunity to win extra chips. Is there as simple explanation of how that converts to 0.5*(2/7)^k instead of 1/7^k?
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[User Picture]From: melted_snowball
2008-06-28 03:59 pm (UTC)
Well, you're basically saying that there's a 6/7 probability that they toss one chip, a 6/49 probability that they toss two chips, a 6/343 probability that they toss 4 chips, and so on.

So the expected number of chips they toss is then
(6/7) sum (k=1..infty) 2^(k-1) * (1/7)^k

Also, the claim below to detailbear that the distribution will look like a Gaussian centred (it mildly amuses me that he would spell that word the same, given that he lives 100 km from me) at the drop site is not actually true; the chain converges to a uniform distribution because of the walls. (To see this, you have to recall that the steps of the random walk are not all iid: once you hit a wall, the walk can only go straight down or to the one side, not to left or right. If the walk had no boundaries, then, yes, the walk steps would all be iid, and then the CLT would apply.) In fact, once there's a reasonably good probability that you've hit both walls, you might as well start from any random site.
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[User Picture]From: dr_tectonic
2008-06-28 07:34 pm (UTC)
Ah, I see the confusion. I was trying to say that you don't double your chips when you hit BP, you get one extra. So in that case it would be, hmm, 6/7*sum(k=1..inf)[(k+1)*(1/7)^k], no?

The way they have the walls set up, a chip that hits the wall can only reflect. There's no dropping straight down. So I think it really does behave like a cylinder. So it becomes more Gaussian the taller it is, but the Gaussian also widens and wraps around more times so that it approximates a uniform distribution. Had some lovely mental pictures thinking about it in the shower this morning...

Edited at 2008-06-28 07:35 pm (UTC)
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