Poll of the Day > Statistics question

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Lokarin
01/19/20 11:02:51 AM
#1:


Answers?



You require 12 special event items to complete an objective

You have two opportunities:

  1. You can pay $2(gold/gems/doubloons/whatever) for a 33% chance at one
  2. You can pay $3 for a 50% chance at one.


The goal is to get all 12 for the lowest cost. The items are interchangable so you don't need to get a specific item, just a total of 12.

The naive impression is that both situations are equal, it should take an average of $72 in either case, but what's the real answer?


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MrMelodramatic
01/19/20 11:07:32 AM
#2:


same
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wolfy42
01/19/20 11:15:08 AM
#3:


$2 is better because:

It is possible to get all 12, for $24 using the $2 method. That is not possible with the $3 method (lowest cost possible is $36.

Since you don't end up actually having a higher average number of tries, but you have a lower possible minimum, you are better off with the $2 method, even if your chance of actually completing it for only $24 (or for that matter less then $36 dollars) is very low.

If you need 12 wins, and it's only $2 a pop, you can get all of them for $24, but you can also fail 4 times (or 33% of the time) and still pay less than the $3 method.

While on average you will fail more than that, it's still a fairly decent chance of actually paying less with the $2 method.

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Sahuagin
01/19/20 11:16:06 AM
#4:


the $2 one is very slightly worse unless 33% means 1/3

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wolfy42
01/19/20 11:22:27 AM
#5:


Sahuagin posted...
the $2 one is very slightly worse unless 33% means 1/3


Good point, I did figure that 33% was 1/3 and 50% was 1/2. If you only have a 33% vs a 50%...that would lean towards the 50% version being better on average, if only by a tiny amount, but..that being said, you still have (a very small) chance of getting all 12 for less than is possible with the 50% method.


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Sahuagin
01/19/20 11:24:15 AM
#6:


wolfy42 posted...
$2 is better because: It is possible to get all 12, for $24 using the $2 method. That is not possible with the $3 method (lowest cost possible is $36.
that is interesting.

there's only a 1/4096 chance of being able to pay only $36 with the $3 option

there's only a ~1/600,000 chance of being able to pay only $24 with the $2 option
I don't know how to calculate $26-$36 with the $2 option

even so, all else being equal, it is an advantage

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OniRonin
01/19/20 11:28:50 AM
#7:


it depends on your risk tolerance. both have the same expected value bue the 2$ option has higher variance

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wolfy42
01/19/20 11:28:51 AM
#8:


Sahuagin posted...
that is interesting.

there's only a 1/4096 chance of being able to pay only $36 with the $3 option

there's only a ~1/600,000 chance of being able to pay only $24 with the $2 option
I don't know how to calculate $26-$36 with the $2 option

even so, all else being equal, it is an advantage


Yeah, as long as it's actualy 1/3 and not 33%...as the .3333333% would make a different that would offset the small chance to basically get all 12 with the $2 method in 18 or less tries (which is impossible to beat with the $3 method).

As long as it's 1/3 though, that actually comes out to be a somewhat possible chance considering you only end up needing 2/3rds of them to succeed. I am not doing the math in my head no way, but it's a good enough chance to make it better (but again only if it's actually 1/3).

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OniRonin
01/19/20 11:29:41 AM
#9:


wolfy42 posted...
$2 is better because:

It is possible to get all 12, for $24 using the $2 method. That is not possible with the $3 method (lowest cost possible is $36.

Since you don't end up actually having a higher average number of tries, but you have a lower possible minimum, you are better off with the $2 method, even if your chance of actually completing it for only $24 (or for that matter less then $36 dollars) is very low.

If you need 12 wins, and it's only $2 a pop, you can get all of them for $24, but you can also fail 4 times (or 33% of the time) and still pay less than the $3 method.

While on average you will fail more than that, it's still a fairly decent chance of actually paying less with the $2 method.

this post is making me angry. you also have a higher chance of losing more money with the 2$ option, for the very reasons you are pointing out (the 2$ option has a chance of saving more money but the expected values are equal)

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wolfy42
01/19/20 11:32:51 AM
#10:


OniRonin posted...
this post is making me angry. you also have a higher chance of losing more money with the 2$ option, for the very reasons you are pointing out (the 2$ option has a chance of saving more money but the expected values are equal)


That was not the question posited initially, he wanted to know which could get the 12 for the lowest cost, and the answer is the $2 method for the reasons I stated.

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OniRonin
01/19/20 11:36:28 AM
#11:


wolfy42 posted...
The lowest cost way to make a billion dollars is to just hope you find it on the street


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Judgmenl
01/19/20 11:40:55 AM
#12:


Is the random number generator, entropy based, real random, or does it have some kind of deterministic outcome?

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Shark8637
01/19/20 4:55:49 PM
#13:


OniRonin posted...
it depends on your risk tolerance. both have the same expected value bue the 2$ option has higher variance

This.
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zebatov
01/19/20 5:48:20 PM
#14:


If its anything like CoD WW2 supply drops then you could spend any amount of money since dupes exist.

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