Guess how much 56 pounds of change is worth and win 2% (coinstar)

[ptmusic]

$89

-ptmusic

p.s. when I win, will you pm me? I can't keep up with the wacky pace of OOT sometimes. Thanks.

[Falker11]

479.11

Falker

[Ogre]

456

[AndysDaddy]

Assumptions:
- These coins are leftovers from purchases
- The cents-part of total purchase prices are random

Then any random collection of coins should be apportioned thusly:
Pennies: 42.55% (Rp)
Nickles: 8.51% (Rn)
Dimes: 17.21% (Rd)
Quarters: 31.91% (Rq)

------------------

Given that 56 pounds is 25,401.17 grams and the weights of coins are:
Penny: 2.5g (Wp)
Nickle: 5g (Wn)
Dime: 2.268g (Wd)
Quarter: 5.6g (Wq)

Then the total number of coins should be (T):
P + N + D + Q = T
P = T(Rp)
N = T(Rn)
D = T(Rd)
Q = T(Rq)

T(Rp)(Wp) + T(Rn)(Wn) + T(Rd)(Wd) + T(Rq)(Wq) = 25,401.17
Solve for T = (about)6,935 coins

P = 2951 pennies
N = 590 nickles
D = 1180 dimes
Q = 2213 quarters

Total value: $730.41
Total after 8.9% rake (no maximum): $665.30

I didn't check to see if anyone else already guessed this amount.
--
Scott

[jakethebake]

[ QUOTE ]
Assumptions:
- These coins are leftovers from purchases
- The cents-part of total purchase prices are random

Then any random collection of coins should be apportioned thusly:
Pennies: 42.55% (Rp)
Nickles: 8.51% (Rn)
Dimes: 17.21% (Rd)
Quarters: 31.91% (Rq)

------------------

Given that 56 pounds is 25,401.17 grams and the weights of coins are:
Penny: 2.5g (Wp)
Nickle: 5g (Wn)
Dime: 2.268g (Wd)
Quarter: 5.6g (Wq)

Then the total number of coins should be (T):
P + N + D + Q = T
P = T(Rp)
N = T(Rn)
D = T(Rd)
Q = T(Rq)

T(Rp)(Wp) + T(Rn)(Wn) + T(Rd)(Wd) + T(Rq)(Wq) = 25,401.17
Solve for T = (about)6,935 coins

P = 2951 pennies
N = 590 nickles
D = 1180 dimes
Q = 2213 quarters

Total value: $730.41
Total after 8.9% rake (no maximum): $665.30

I didn't check to see if anyone else already guessed this amount.
--
Scott

[/ QUOTE ]

This is how I did it as well, but I assumed half dollars as well. Where did you get your random distribution? I used the number of coins in circulation and mine was much different.

[AndysDaddy]

I didn't count halves or dollar coins, as one almost never gets these back as change. Except, of course, when one spends a considerable portion of one's time in a casino. Hmmmm...

I based my proportions on my second assumption:
- The cents-part of total purchase prices are random

By which I mean that the pennies amount for the thing(s) you are buying are randomly distrubuted between 00 and 99, resulting in random change per transaction of 99 - 01 (I discard 00 as an uninteresting case). I also used an unspecified assumption that such change would be given in the most coin-efficient manner possible. So 99 cents in change results in 3 quarters, two dimes and four pennies. Adding up all the possible change distributions and dividing each coin amount by the total gives my amounts.

Your method of using number of coins in circulation is an interesting variance I hadn't thought of. I assume you got your figures from some tresuary web page? I can think of a few reasons why circulation numbers wouldn't match with what a person puts into a coins-to-be-exchanged-for-dollars jar, but then I can think of several reasons why my method would be inaccurate, too.

Good luck!
--
Scott

[jakethebake]

[ QUOTE ]
Your method of using number of coins in circulation is an interesting variance I hadn't thought of. I assume you got your figures from some tresuary web page? I can think of a few reasons why circulation numbers wouldn't match with what a person puts into a coins-to-be-exchanged-for-dollars jar, but then I can think of several reasons why my method would be inaccurate, too.

Good luck!
--
Scott

[/ QUOTE ]

Yea. That's what I did. Data from the Treasury page. I know what you mean. People probably keep out quarters or whatever for various reasons (parking meters, laundry, etc).

[felson]

Nice. I thought about that approach too, but was too lazy. Since we're talking about ways to estimate it, my method was to take the first 30 or so guesses (except for the $1 or $2 guesses) and average them, since there's research saying that this approach works well. It'd be better to wait until the end, of course, and get more samples, but again I was too lazy. My guess (posted earlier), not too far from yours, was $639.62.

[utmt40]

639.51

[slamdunkpro]

$471.77