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# Scratchers probability puzzlePrev TopicNext Topic

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• United States
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There's a broken lottery vending machine that sells only \$1 scratchers. If you put in \$1, it will randomly give you either one, three, or five scratchers with equal probability.  You can't put in more than \$1 at a time.

Someone plays at the machine for a while and stops when they have exactly \$10 worth of scratchers. What is the probability they put in only \$4?

• United States
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Here's a related problem with the same broken machine. Suppose you send your spouse to get some scratchers from the machine. You tell them to stop putting dollar bills in the machine once they have accumulated at least \$10 worth of scratchers. Assuming your spouse can follow instructions, what is the probability they come back with \$11 worth of scratchers?

Not sure if this is harder or easier than the first problem.

• Sugar Land
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There's a possibility that you end up with more than \$10 by chance. Example: first scratcher is \$5, next is \$3 (so you keep playing since you are under \$10 total), and the next one is \$5. Now you have \$13.

Assuming all possible outcomes of exactly \$10, and \$1/\$3/\$5 are equally probable, then the conditional probability of having spent \$4 to get exactly \$10 is 0.467 by my calculations. This is a result of \$5-\$3-\$1-\$1 or \$3-\$3-\$3-\$1 with any ordered partition of each.

• United States
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Quote: Originally posted by Orange71 on Aug 3, 2022

There's a possibility that you end up with more than \$10 by chance. Example: first scratcher is \$5, next is \$3 (so you keep playing since you are under \$10 total), and the next one is \$5. Now you have \$13.

Assuming all possible outcomes of exactly \$10, and \$1/\$3/\$5 are equally probable, then the conditional probability of having spent \$4 to get exactly \$10 is 0.467 by my calculations. This is a result of \$5-\$3-\$1-\$1 or \$3-\$3-\$3-\$1 with any ordered partition of each.

There's a possibility that you end up with more than \$10 by chance. Example: first scratcher is \$5, next is \$3 (so you keep playing since you are under \$10 total), and the next one is \$5. Now you have \$13.

This is a good hint for tackling the second question. For the first question, take it as a given the player got ten buck's worth and stopped there.

Given that the player received exactly \$10 worth of scratchers, the likelihood of the player having spent exactly \$4 works out to more than half. I can give a hint if you want. I also have a Python script to run simulations to estimate the probabilities, and it gives a consistent answer. It simulates a person putting \$1 in a machine until they reach a scratchers total of at least \$10. Then it discards all the instances where the total is more than \$10 so it can correctly estimate the conditional probability.

These problems are adapted from a set of problems about a child putting coins in a gumball machine that dispenses one, three, or five gumballs per coin.

• Sugar Land
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Indeed, I made an arithmetic mistake in the number of partitions (arrangements of order) for \$5-\$3-\$1(x2). It's 12 and not 6 as I originally calculated. I think the answer to your first question is p(\$10 return | \$4 spent) = 0.5832. I haven't tackled the second one yet.

There are the following possibilities to end up with exactly \$10:

\$5(x2), p=0.1111 (i.e. 1/9), with 1 partition

\$5-\$3-\$1(x2), p =0.1481, with 12 partitions (arrangements of order of picks)

\$5-\$1(x5), p=0.0082, with 6 partitions

\$3(x3)-\$1, p=0.0494, with 4 partitions

\$3(x2)-\$1(x4), p=0.0206, with 15 partitions

\$3-\$1(x7), p=0.0012, with 8 partitions

\$1(x10), p = 0.000017, with 1 partition

Out of these \$5-\$3-\$1(x2) and \$3(x3)-\$1 correspond to \$4 spent. So we take the sum of these two probabilities and divide by the total of the probabilities above (to get a conditional probability). That result is 0.5832.

• Sugar Land
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Technically I should have said p(\$4 spent | \$10 return) with the same value, not p(\$10 return | \$4 spent). This may seem subtle, but it is very important.

• Sugar Land
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Quote: Originally posted by cottoneyedjoe on Aug 3, 2022

Here's a related problem with the same broken machine. Suppose you send your spouse to get some scratchers from the machine. You tell them to stop putting dollar bills in the machine once they have accumulated at least \$10 worth of scratchers. Assuming your spouse can follow instructions, what is the probability they come back with \$11 worth of scratchers?

Not sure if this is harder or easier than the first problem.

The answer is p=0.246 or 24.6%. I worked this out by counting probabilities for all event combinations and partitions, not by random simulation. It is considerable harder than the first problem because there are many more possible events.

Furthermore, the average (Expected Value) of \$ won with these rules is \$11.41 and the average (EV) of \$ spent is \$3.80.

• United States
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Quote: Originally posted by Orange71 on Aug 6, 2022

The answer is p=0.246 or 24.6%. I worked this out by counting probabilities for all event combinations and partitions, not by random simulation. It is considerable harder than the first problem because there are many more possible events.

Furthermore, the average (Expected Value) of \$ won with these rules is \$11.41 and the average (EV) of \$ spent is \$3.80.

Nice solutions Orange. I agree the second was harder than the first. It's interesting that when you change the parameters a little bit (instead of {1, 3, 5} use a different set whose average is 3, such as {2, 3, 4} or {1, 2, 6}) the probabilities change a lot.

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