# 7 coins - probability over a dollar

1. Dec 27, 2012

### bloynoys

Alright my brother posed this question to me tonight. You have 7 coins (for this sake we will go with the "normal" coins, penny, nickel, dime and quarter) and we are trying to find the probability that a person has over a dollar in coins.

So basically I did it and just wanted to confirm my answer before telling him tomorrow morning.

You just do:

E(x)=((1*(1/4))+(5*(1/4))+(10*(1/4))+(25*(1/4)))
E(x)= 41/5 = 10.25
7*E(x)=71.75

Var(x)=E(x^2)-E(x)^2
Var(x)=187.75-105.0625
Var(x)=82.6875
7*Var(x)=578.8125

Now I believe that I can do normal Z score stuff to find probability with the continuity correction. So we are looking for P(Z≥100) so we'll flip it into 1-P(Z≤100). Then we will add the continuity correction to make it 1-P(Z>100.5) and calculate it from there. Am I right so far?

So I go:

(71.75-100.5)/(24.058)

Z≈-1.195

And thus, using R normal probability calculator we are looking at a final answer of .1160379. Did I do that all right? Thanks!

2. Dec 27, 2012

### chiro

Hey bloynoys.

I used R on the assumption that E[X] and Var[X} where calculated correctly (the method you used is good) and assuming the normality condition holds (which it hopefully should for 7 observations) then R gave me:

> 1 - pnorm(100,71.75,24.058)
[1] 0.1201483

However I would suggest you use continuity correction since you have discrete units instead of continuous ones (in terms of 1, 5, 25, etc units).

Are you familiar with continuity correction?

3. Dec 27, 2012

### bloynoys

Yeah I tried to apply the continuity correction to the z calculation, so it would basically move the mean to 100.5 which is what gave me the answer that I had in the first post. Does that look right to you?

4. Dec 27, 2012

### chiro

It does, but the only thing I am thinking about is whether the distribution isn't skewed enough to make it normal.

One way you could check this is to simulate the distribution of the sum and see if its normal.

It might sound anal, but it's just a nice thing to do for peace of mind.

5. Dec 27, 2012

### bloynoys

Is there an easy way to simulate this in R or similar? It would be interesting to see what kind of distribution it comes out as. I would hope the pennies are able to "normal" it out but hard to know.

6. Dec 27, 2012

### chiro

Are you aware of sampling techniques like EM or Metropolis Hastings?

7. Dec 27, 2012

### bloynoys

I am not. I am about to start my final semester of undergraduate stats education (have stats 2, stochastic processes and multivariate analysis this semester) but so far we haven't gotten to things like that. Any reading or hints you would suggest as I try to learn about these things? Not a big deal in terms of this question but seems like a good skill to understand going forward. They seem very cool from a quick perusing of the wikis.

8. Dec 27, 2012

### chiro

9. Dec 27, 2012

### bloynoys

Alright, I wanted to check this fast so what I did is just write a simple program in Java to generate random numbers between 1 and 4 and then if statements to create the cent amount after 7 coins. I ran that 50000 times and got this histogram. Do you think this is approximately normal? It doesn't look entirely great and maybe a slight skew right, what do you think?

10. Dec 27, 2012

### bloynoys

Yeah pretty clear it isn't close enough to normal. From just checking how many coin amounts are over 100 a few times it is always between 8 and 9 percent pretty far off from the 11-12 that it would be with the normal distribution.

11. Dec 27, 2012

### chiro

You might want to consider using a distribution with a heavier tail or simulate say 50,000 times and use the generated empirical distribution (i.e. the one you just simulated).

12. Dec 27, 2012

### D H

Staff Emeritus
In this case, a brute force approach works quite nicely. There are 47 or 16384 equiprobable outcomes, so just check each one. Here's a quick and dirty perl script:
Code (Text):
use strict;

my $N = 4**7; my$M = 0;
my @value = (1, 5, 10, 25);
for my $index (0 ..$N-1) {
my $sum = 0; for my$ii (0 .. 6) {
$sum +=$value[$index & 3];$index >>= 2;
}
$M++ if ($sum > 100);
}

printf "%d, %d, %g\n", $N,$M, $M/$N;
2101 of the 16384 outcomes have a total value of more than $1.00, so the probability is 0.128235. The approximate value from the original post (0.1160379) is good to one decimal place. 13. Dec 28, 2012 ### haruspex If it's just 7 coins you happen to have in your pocket, in the real world those are not equiprobable. People tend to give change as well as receive it, in the process avoiding a build up of of many coins of the same denomination. A more realistic model might be one where the number of pennies is only 0-4 (and would have been equally likely but for the constraint of 7 total) etc. 14. Dec 28, 2012 ### chiro That is another good reason to use simulation. 15. Dec 31, 2012 ### broccoli7 Someone (haruspex) on this thread mentioned the right thing. In the real world, those coins are not equally probable. Guess it all depends on whether the coins are selected at random or are they real "change" from transactions. 16. Jan 1, 2013 ### D H Staff Emeritus The OP assumed an equiprobable distribution. Some people tend to give change as well as receive it. If the question is about those who adroitly manage their, the answer is zero. There's no reason to ever have more than 99 cents in loose change on hand. Others pay with bills only, instead dumping all the change they receive into their pocket or handbag. Presumably it's this latter group that the question is about. To be realistic, one would have to model how merchants price their goods. For example, you are much more likely to find something priced at$9.99 than at $10.01. One would also have to factor in things like sales taxes (which in some states vary by locale; e.g., In store A I'll get hit with a sales tax of 8.25%, in store B a few hundred yards away I'll only pay 7.75%, and in store C a mile away I'll pay 8%). You are missing a key point by focusing on realism. That point is that a normal distribution is not a good model for this process. 17. Jan 2, 2013 ### bpet A good outcome considering that the Berry-Esseen error estimate here is around 0.25 18. Jan 4, 2013 ### haruspex In the attempt at solution maybe, but not in the statement of the problem. It is entirely unclear whether it is intended as equiprobable or 'suppose you have 7 coins in your pocket'. Hence my post. 19. Jan 4, 2013 ### 0xDEADBEEF In this type of questions priors must always be stated otherwise you produce something like the "two envelopes problem" as an extreme case. 20. Jan 6, 2013 ### Bacle2 Maybe you can simplify your calculations a bit by using the fact that you need at least two quarters to have a total to add up to$1.00 . That leaves one option : 5 dimes.
Then you can have 3 quarters , and 3^4=81 cases, or 4 quarters , and certainty.

So your winning cases of total larger than $1.00 are (using _exact_ number of quarters): 0 quarters: 0 ways , 3^7 non-ways 1 quarter : 0 ways 3^6 non-ways. 2 quarters: 1 way ( 5 10's) 3 quarters: only complicated cases 4 quarters: 3^3 ways 5 quarters: 3^2 ways, etc. So it comes down to having four non-quarter coins adding up to$0.25-or-more.