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You have a wallet with four kinds of coins with the value of \$1, \$5, \$10, and \$25. Every kind of coin is unlimited. Every time you take money from the wallet, you have an equal chance of getting one of these coins, i.e., \$1, \$5, \$10, or \$25.

Now you need to buy something that costs $x$ dollar, and you keep taking coins out of your wallet until you have enough money to buy. When $x$=30, what is the expected value of money left in your hand after your purchase?

i.e., if you take out a \$25 the first time, you have to take more coins since the price is \$30. And suppose you get a \$10 in the second time, you can stop. You pay \$30 and you have \$5 left in your hand.

How about when $x$=300, $x$=3000000?

kjetil b halvorsen
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LIU ZHIWEN
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  • I try to get the answer using simulation, I get \$7,75 for $x$=30, \$8.67 for $x$=300, and \$7.5 for $x$=3000000. While the simulation results may not be very precise. – LIU ZHIWEN Feb 12 '22 at 08:45
  • Cross-posted: https://math.stackexchange.com/questions/4380108/expect-coin-value-probelm-a-variant-of-coin-change-problem – Peter O. Feb 12 '22 at 12:19
  • @PeterO. Yes, I also put the same problem on [Math] Section. – LIU ZHIWEN Feb 12 '22 at 14:30
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    Closely related threads include https://stats.stackexchange.com/questions/191193 (this computes the expected number of coins) and https://stats.stackexchange.com/questions/145621 (which solves the present problem for a different set of coin denominations and suggests an asymptotic argument to obtain the answer to the last question). – whuber Feb 12 '22 at 16:09
  • @whuber Thanks a lot. These two questions and their answers are very helpful, especially the second one. But one thing is that in my problem the distribution of coin values is very asymmetry, compared with 6-sided dice. Maybe for a very large $x$ (i.e. 3000000) it's similar to the second Reference, but for $x$=30 and $x$=300 it may need a different method. – LIU ZHIWEN Feb 13 '22 at 06:07
  • @whuber Using the method mentioned in [questions/145621], I get ${355 \over 41}$, which is about 8.65. Comparing this result with simulation, I think $x$=300 is large enough to assume that $x−24$ to $x−1$ are nearly equally probable at time $\tau−1$. ($\tau$ is the smallest number of coins that total value is larger than $x$) – LIU ZHIWEN Feb 13 '22 at 07:19
  • They are not equally probable, though. You might need to run a longer simulation to see that. – whuber Feb 13 '22 at 15:20

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