Scratchers probability puzzle

Reply #7

cottoneyedjoe — Sat, 06 Aug 2022 18:35:05 GMT

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.

Reply #6

Orange71 — Sat, 06 Aug 2022 17:10:30 GMT

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.

Reply #5

Orange71 — Fri, 05 Aug 2022 14:19:58 GMT

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.

Reply #4

Orange71 — Fri, 05 Aug 2022 01:40:23 GMT

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... [ More ]

Reply #3

cottoneyedjoe — Thu, 04 Aug 2022 23:12:38 GMT

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... [ More ]

Reply #2

Orange71 — Thu, 04 Aug 2022 01:21:59 GMT

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.

Reply #1

cottoneyedjoe — Wed, 03 Aug 2022 20:18:48 GMT

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.

Scratchers probability puzzle

cottoneyedjoe — Wed, 03 Aug 2022 20:07:21 GMT

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