What is the probability that 10 coins selected randomly from 100 coins add up to $1?

May 29, 2010 - 1:21 am 11 Comments

Assume the 100 coins consist randomly of pennies, nickels, dimes, and quarters. Each coin has an equal probability of being selected.
Just as a side note, I got the idea for this question at toll booth. So I suppose this does have some type of real-world application.

There are several ways to make $1 out of 10 coins:
A
quaters: 3
dimes 2
penies 5
B
quateres 2
dimes 2
nickels 6
C
quaters 1
dimes 6
nickels 3
D
qaters 0
nickels 10

The number of ways to construct selections is
nA: 10!/(3!2!5!) = 2520
nB: 10!/2!2!6! = 1260
nC: 10!/1!6!3! = 840
nD: 10!/10! = 1

Total volume of desirable configuration space is
nA+nB+nC+nD = 4621 out of
volume of entire configuration space 4^10

Answer:
4621/4^10 = 0.440 692 901 611 328 125 %
(decimal fraction above is exact)

11 Responses to “What is the probability that 10 coins selected randomly from 100 coins add up to $1?”

  1. The Corinthian Says:

    You would need to know the number of each coin to accurately predict
    References :

  2. jamo796 Says:

    not enuf info i would say…it doesn’t say how many of each coin there is
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  3. Jon Says:

    If each coin has an equal probability of appearing, that means that, assuming size doesn’t matter, that getting a certain type of coin is 1/4.

    penny(1), nickel(2), dime(3), quarter(4). 25 of each, out of 100, 25/100 or 1/4.

    If you took 10 coins out, then theoretically, 2.5 of each type of coin will appear.

    10coins/4options=2.5per type of coin

    This is the hard part. I can’t help you there, but then you’ll have to find all the possible combinations in 10 coins.

    4 coins, each coin you pick will not affect the next pick. 4 more coins for each type of coin. 4 * 4=16 (two picks), 4 more after that, 16 * 4=64 (three). Or, 4^10, FOUR to the power of TEN. 40 000 000 000(forty billion). There are 4^10 possible combinations for ten picks from your bag, but I can’t say the exact probability of making the total >= $1. Here, you’re on your own. Sorry.
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  4. Cat M Says:

    There are several ways to make $1 out of 10 coins:
    A
    quaters: 3
    dimes 2
    penies 5
    B
    quateres 2
    dimes 2
    nickels 6
    C
    quaters 1
    dimes 6
    nickels 3
    D
    qaters 0
    nickels 10

    The number of ways to construct selections is
    nA: 10!/(3!2!5!) = 2520
    nB: 10!/2!2!6! = 1260
    nC: 10!/1!6!3! = 840
    nD: 10!/10! = 1

    Total volume of desirable configuration space is
    nA+nB+nC+nD = 4621 out of
    volume of entire configuration space 4^10

    Answer:
    4621/4^10 = 0.440 692 901 611 328 125 %
    (decimal fraction above is exact)
    References :

  5. me Says:

    1/100000000000000000000 Because you do 1/100 to the tenth power. Because you do 1/100 for one coin and do that ten times. Then the result(1/100000000000000000000) is the probability of all 10 coins adding to be $1.
    References :

  6. holdm Says:

    since the coins have equal probability of being selected (at the start) there are 25 of each.

    10D 25C10
    1Q6D3N 25*25C6*25C3
    2Q2D6N 25C2 ^2 * 25C6
    3Q2D5P 25C3 * 25C2 * 25C5

    sum these over the sample space which is 100C10
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  7. meme m Says:

    not enough info
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  8. Merlyn Says:

    I get about 0.44%.

    The following code is run in R. It creates 2000 collections of 100 coins using a random multinomial vector with equal probability of penny, nickels, dimes and quarters.

    each of the collections goes through 1000 simulations where a sample of 10 coins is selected without replacement and if the sum of the 10 coins is equal to 1.00 then the count variable is indexed.

    the probability is estimated as count / (sims * collections)

    ===

    collections <- 10000
    sims <- 1000
    count <- 0

    x <- rmultinom(collections, 100, c(.25, .25, .25, .25))

    for(i in 1:collections)
    {
    t <- x[,i]
    set <- c( rep(0.01, t[1]), rep(0.05, t[2]), rep(0.10, t[3]), rep(0.25, t[4]))
    for(j in 1:sims)
    {
    if(sum(sample(set, 10)) == 1.0) count <- count + 1;
    }
    }

    count / (sims * collections)
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  9. A J Says:

    You would need to know the number of each coin to accurately predict
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  10. ATG Says:

    10/100
    References :

  11. Myself Says:

    Am sure there will be accurate ways to calculate this if you ask math geniuses.
    but if you ask nob head like me (which you did :)

    25+10+5+1+25+10+5+1+10+5 will be average pick. which is equal 97cents
    I’ll give ya a discount… forget the 3 cents and lets move on :) )
    References :

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