PDA

View Full Version : A probability puzzle

JTG51
10-23-2002, 10:30 PM
You have 50 coins that total \$1.00 in value. You drop one coin down the sewer. What are the chances you lost a quarter?

Bob T.
10-24-2002, 01:59 AM
1 out of 100.

Good luck,
Play well,

Bob T.

Jim Brier
10-24-2002, 03:00 PM
This is a very difficult problem to solve without the use of a computer. Here is how I might begin to set it up.

Let A = # of pennies.
Let B = # of nickels.
Let C = # of dimes.
Let D = # of quarters.
Let E = # of fifty-cent pieces

You have only two equations:

(1) A + B + C + D + E = 50
(2) A +5B +10C +25D + 50E = 100

You have some constraints like E = 0 or 1, D = 0, 1, or 2, and so forth.

That is about as far as my thinking takes me.

10-24-2002, 05:31 PM
Bob T's solution looks right... I ended up writing a short program to tell me the answer. There are two ways to satisfy the 50 coin, \$1 requirement.

One way involves a single quarter, the other way involves none. So, his 1 out of 100 answer is right. Not sure how you'd attack this without enumerating/brute-forcing the possible coin combinations w/ constraints.

lorinda
10-24-2002, 05:33 PM
I think the answer will lie in the number of ways you can make a quarter with the smaller sums, this should lead to a ratio of the possibilities.

Any way of making 25c with the smaller coins will naturally be a substitution for all the ways that involve quarters so we dont need to calculate everything, just the number of ways of making 25c.

You can probably take this further and work out how many ways there are of making 5c, 2c etc and get the solution that way.

The only case that wont be affected is the one with two 50c coins.

M.B.E.
10-24-2002, 05:41 PM
Not enough information is given to answer this puzzle without making a number of additional assumptions. First note that when I have fifty coins totalling \$1 in value, there are only two possible combinations for the coins:

1x25 + 2x10 + 2x5 + 45x1 = 100, or

2x10 + 8x5 + 40x1 = 100.

So first I will assume that each of these combinations is equally likely. In the second case there are no quarters, so the probability the lost coin is a quarter is zero. Now consider the first case. I will make the additional assumption that of the 50 coins, each one is equally likely to be dropped. This is not really a reasonable assumption because the coins are of different sizes, and I think the smaller a coin is the more likely it is to be dropped. Or maybe it's the other way around. But I don't think this was intended to be a question in the physics of carrying coins, so I'll make the assumption that each coin of the 50 is equally likely to be dropped. So for this case, the probability the lost coin is the quarter is 1/50 i.e. 0.02.

Since case one and case two are (by my assumption) equally likely, the overall probability that the lost coin is a quarter is

(1/2)*(1/50) + (1/2)*0 = 1/100.

10-24-2002, 05:49 PM
jim -

no need to be so baroque with the calculations. you can't have a 50-cent piece because then you need to have 49 coins totalling 50 cents, which you cannot do without a 2-cent piece or some pesos. similarly, you can have at most one quarter; otherwise you would need to get 48 coins totalling 50 cents.

now, can we use a quarter? using a quarter means that we need 49 coins totalling 75 cents. use a dime, we need 48 coins to total 65 cents. use another, we need 47 coins to total 55 cents. use a nickel, we need 46 coins to total 50 cents. use another, we need 45 coins to total 45 cents, and we know how to do that. so: we can use a quarter, meaning that we have at most a 2% chance of dropping a quarter.

now, can we do this without a quarter? sure. use a dime, we need 49 coins to total 90. use another, we need 48 to total 80. 8 nickels, and we need 40 coins to total 40 cents. we know how to do that.

so there's one way to do it with a quarter, and one way without, meaning a 1% chance that we've dropped a quarter.

btw, i am not a computer, but i am a computer scientist, so maybe that helps.

the club

M.B.E.
10-24-2002, 06:02 PM
You don't need a computer to figure out the answer. I did it in a few minutes with pencil and paper. Essentially all you have to do is solve for the two equations

50a + 25b + 10c + 5d + e = 100 and

a + b + c + d + e = 50,

where a, b, c, d, and e are nonnegative integers. It's easy to see that a=0, so then the two equations reduce to this one:

24b + 9c + 4d = 50

(recall that b is the number of quarters, c the number of dimes, and d the number of nickels).

From here there are a few different ways to proceed to arrive at the two solutions. Here's one. First observe that the terms 24b and 4d are both even, and since 50 is also even, the term 9c must be even. (Because two even numbers and one odd number can never sum to an even number.) Since 9c is even, c must be even.

But c cannot be divisible by 4. Because then each one of the terms 24b, 9c, and 4d would be divisible by 4, and the sum of the three terms would also be divisible by 4. But 50 isn't. Therefore c is an even number not divisible by 4. Possible values of c could be 2, 6, 10, 14 etc. But if c is 6 or greater then 9c would be 54 or greater, and we can't have that because then either b or d would be negative. So c=2.

Now that we know c=2, we have:

24b + 9*2 + 4d = 50, which reduces to

24b + 4d = 32, which further reduces to

6b + d = 8

Then since b and d are both nonnegative integers, we see that b cannot be greater than 1. So the only two solutions for (b,d) are (0,8) and (1,2).

Hence the only two solutions for (a,b,c,d,e) are (0,0,2,8,40) and (0,1,2,2,45).

Jim Brier
10-24-2002, 06:36 PM

I am going back to school to study math because I enjoy mathematics. Consider yourself officially inducted into the "Jim Brier Advisory Board On Mathematics" along with Uston, Bruce Z, and others who post on this forum.

Jim Brier
10-24-2002, 06:39 PM
(n/t)

lorinda
10-24-2002, 07:48 PM
Ooops, once again I made an error, didnt read the bit that says I have 50 coins....

Perhaps a holiday is in order /forums/images/icons/frown.gif

Bozeman
10-24-2002, 09:54 PM
Let me try.

Suppose you have x dimes, y nickels, z pennies.
You can't have 4 quarters, or you'd only have 4 coins.

If you have 3 quarters, x+y+z=49, 10x+5y+z=25. Since 25&lt;49, this has no all non-negative solutions.

If you have two quarters, x+y+z=48, 10x+5y+z=50. Thus 9x+4y=2, impossible for any integers x,y.

If you have one quarter, x+y+z=49, 10x+5y+z=75. 9x+4y=26. x=2, y=2 and then z=45 is the only integer solution.

0 quarters, x+y+z=50, 10x+5y+z=100. 9x+4y=50.
x=2, y=8, z=40 is the only one that works.

If the probability of dropping is independent of type of coin, and each combination is equally likely, then P=0.5*1/50=1%

For me, chance of dropping is inversely porportional to value, so P=0.5*c*1/25 where 1=c*sum(1/value), so c=1/45.64, and P~.044% /forums/images/icons/smile.gif

Craig

JTG51
10-25-2002, 12:45 AM
Of course, the answer many of you came up with is the correct one. There is a 1/100 chance of losing the quarter.

Ray Zee
10-26-2002, 12:17 AM
i assume too much. when i first did it i thought it natural that the question did assume you had a quarter in the 50 coins, so i had to look twice when i saw 1 in 100 for an answer. but then i realized that for some people the answer had to be zero. know why