Oh, dear. I wake up, and someone has posted a math problem. It's a cruel world. *sigh*
I disagree with your assumption that every starting place is equally likely. Anyone who watched Price is Right will know that the final distribution is that the most likely ending point is the one underneath where you started. So I'd drop the puck into the machine over top of the "$25" or the "BP" site.
(I need to convince myself that your symmetry arguments are right, but they have a plausible feel, at least? The thing I'd be worried about is the edge effects pushing me toward the edge with higher probability than planned for. You can really easily just compute the stationary distribution of the chain, of course.)
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. So just compute (E[X])^2 and E[X^2], subtract the former from the latter, and you have variance; then take its square root.
The argument in the last paragraph generalizes to what you really wanted very easily, at least for the expectation: the number of extra players is dependent on the number of times you've hit that site. 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). (You can also fairly easily extend to the case of the [higher] standard deviation.)
I guess you get the "professional me" icon, not the "economics" icon. *shrug*
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?
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.
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)
2008-06-28 01:43 pm (UTC)
Accounting assumption alerts
A couple of problems I found simulating this in Excel.
1. the chip reflecting off the wall is isomorphic to it wrapping around to the other side. Not quite true. In reality, no chip can move from one side to the other; all chips in an end slot will be moved to the inside. It's almost equivalent (in that .5 + .5 = 1) and would not make a difference, except for ...
2. we should assume that all starting positions are equally likely. As a social exercise, we can't assume that. As Dan pointed out, the average person doesn't make the assumption that all slots are equal, and therefore we can't take the easy route and calculate probabilities of outcomes. We have to simulate the expected payoff of each starting slot (including the difference caused by alert 1), then simulate various behaviours to discover upper and lower limits to the total winnings.
3. The distribution is affected by the number of rows of pegs on the plinko board. I've used the 9 rows plus the last virtual row from the slot edges for a total of 10. I'd have to simulate a range of number of rows to discover if that made a difference to the final distributions.
So, I built a spread sheet that calculated the probability of a chip ending up in any slot when it started in a particular slot. Odd numbered peg rows moved a chip L/R at .5 each. Even numbered rows did that for all slots except the end ones, which moved 100% toward the center. For example, raw distribution for slot 1 is:
0.451171875 0.322265625 0.161132813 0.053710938 0.010742188 0.000976563 0.00
For the BP slot, I then took the distribution for that slot, and spread it out over the other slots in the ratios like (slot 1/all slots except BP). Here's the final distribution for slot 1:
0.476780186 0.340557276 0.170278638 BP 0.011351909 0.001031992 0.00
Then I took the payoff for that slot and multiplied it by the probability of landing in that slot. The results:
10.34055728 12.56921373 15.79238329 17.44818653 15.79238329 12.56921373 10.34055728
Now we can run scenarios on 2100 drops with any assumptions of player's initial choice preferences that we want.
Maximum payout will be when all chips start in the BP slot: 36641.19
Minimum payout will be when all chips start in the $5 slot: 21715.17
If all slots are chosen equally, the payout would be 28455.75
So, the company would budget $36,000 conservatively and about $32000 on something approaching a bell curve distribution.
I'll send you my spreadsheet so you can check my math.
Edited at 2008-06-28 01:50 pm (UTC)
2008-06-28 03:02 pm (UTC)
Re: Accounting assumption alerts
I agree that people aren't likely to pick all starting slots with equal probability, but without data on what they do pick, I don't think you can legitimately assume anything else.
Given that the number of starting positions is small, though, you're right, it's easy enough to do the problem several times, once for each possible starting position, and at least get upper and lower bounds on the effects of behavior. That's a good point, and a very helpful elaboration of the analysis for planning purposes.
I haven't checked it in detail, but your results look right to me. If everything's starting in the same position, the random walk should give you a gaussian centered on the starting slot, and that appears to be what you've got.
Edited at 2008-06-28 03:03 pm (UTC)
2008-06-29 02:15 am (UTC)
Re: Accounting assumption alerts
Thank you. I remember doing economic problems like this.
Well, you're not the only one. Last Friday, the credit union in the building where I work had a big shindig outside. They were grilling hotdogs and giving away little items. But the big draw was a cash grab in one of those money blower things. You got to stand inside the thing for ten seconds, grab fake money, and then turn it in for real cash. So, I started to try and figure out how much they'd have to have budgeted for that.
The problem was, I know that they had mostly $1 bills in there, but, there was some proportion of $10s and $5s. (and I only learned this from talking to other people, and I'm not sure I believed that there were $10s in there). No idea if they had other denominations. I tried to figure out the average people were grabbing, assuming they were playing by the rules, which were that you couldn't grab any money off the ground, only airborne cash. I got a sample of about 30ish people from my office, and the average was about $10, (I got $6.) The highest grab I heard about (from the credit union folks), was $39. (anecdotally, I heard there was at least one rule breaker, who ended up with, again anecdotally, like $30 (there were conflicting tales, who knows if this was one guy or several). They were open for 2 hours.
I realized that I had not enough data to really figure it out, too many unknowns, so, I concluded that they probably basically figured out the amount they were willing to give away, broke it into denominations consisting mostly of ones, and then let it go at that.
But, I *wanted* it to be possible to figure out.
i <3 teh peoples who are doing teh maths.
Yes. (Mostly so I don't have to.)