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# Simple money probability

## Question

I have 19 coins in my pocket which total \$1.00. You have 15 coins in yours that also total \$1.00. Who has a greater probability of randomly pulling a dime out of their pocket?

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Not only I forgot about half dollar coin in my previous post, I also made few arithmetic errors in calculating probabilities.

Although, the answer who has a better probability is the same.

The tables of all coin collections with their respective weights are as following.

15 COINS:
PENNY NICKEL DIME QUARTER HALF WEIGHT
10 3 0 1 1 60060
10 0 4 0 1 15015
5 9 0 0 1 30030
10 1 1 3 0 60060
5 7 1 2 0 1081080
5 4 5 1 0 3783780
0 13 1 1 0 210
5 1 9 0 0 30030
0 10 5 0 0 3003
DIME PROBABILITY 0.268674895

19 COINS
PENNY NICKEL DIME QUARTER HALF WEIGHT
15 2 0 1 1 46512
10 8 0 0 1 831402
15 0 1 3 0 15504
10 6 1 2 0 23279256
10 3 5 1 0 46558512
5 12 1 1 0 2116296
10 0 9 0 0 92378
5 9 5 0 0 23279256
0 18 1 0 0 19
DIME PROBABILITY 0.205359807

15 coins still looks better.
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You must have 9 dimes and 10 pennies, giving you a probability of pulling a dime as 9 / 19 = 47.36...%

I must have 5 dimes and 10 nickels, giving me a probability of pulling a dime as 5 / 15 = 33.33...%

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You must have 9 dimes and 10 pennies, giving you a probability of pulling a dime as 9 / 19 = 47.36...%

I must have 5 dimes and 10 nickels, giving me a probability of pulling a dime as 5 / 15 = 33.33...%

It's not a complete answer, as the 15 coins could also be 9 dimes, 1 nickel, and 5 pennies.

I think you're on the right track, but the final solution must factor in all the possible combinations of coins that meet the 19 and 15-coin criteria.

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“Simple” is a relative term. To me it would be a simple programming code, if it was feasible to traverse all variations. Alas, I am not going to live long enough to see my laptop running through 419 possible arrangements of coins. (My laptop is old and slow.)

Nonetheless, I’ll go beyond the OP question and give the exact probability.

The following are the tables of all variations to make a dollar with 15 and 19 coins. (Unless, I missed some.) The likelihood of each collection is different, hence the weight must be figured into the probability calculation.

For 15 coins:
PENNY NICKEL DIME QUARTER WEIGHT
10 1 1 3 6651216
5 7 1 2 279351072
5 4 5 1 2933186256
0 13 1 1 162792
5 1 9 0 116396280
0 10 5 0 11639628
DIME PROBABILITY 0.252480672

For 19 coins:
PENNY NICKEL DIME QUARTER WEIGHT
15 0 1 3 15504
10 6 1 2 23279256
10 3 5 1 46558512
5 12 1 1 2116296
10 0 9 0 92378
5 9 5 0 23279256
0 18 1 0 19
DIME PROBABILITY 0.207250786

So, 15-coin collection offers higher probability of randomly drawing a dime out of your pocket. Although, I must note, there is no random drawing of a dime, since coins are of different distinguishable size.

Oops, I forgot about 50-cent coin. Back to the drawing board.

Edited by Prime
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Since my last post has not been recognized as the answer, there must be something else to this problem. I suppose, it could be that different coins are encountered with different frequency. E.g., a penny is a lot more common than a half-dollar coin.

I don’t have any data on how many of each coin type there are in circulation. However, I doubt an adjustment for each individual coin probability would change the final answer to the OP. I believe a 15-coin collection would still have a higher probability of spotting a dime.

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My apologies

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