Amazon uses this lottery puzzle as a brain-teaser for interviewees

Comment #16

KY Floyd — Thu, 05 Mar 2020 20:10:45 GMT

I'd say, Go look for another job? Amazon can't pay that much per hour any ways?

Of course I'm only guessing, but maybe they don't ask that question in interviews for every job position.

= 74.26%

There you go. That single blue ball barely reduces the probability of selecting a red ball, but makes it 3 times as likely that you'll get to pick from that urn.

The answer is the same, but I used the same method of averaging the probabilities as mentioned in the article. With 100 red... [ More ]

Comment #15

kao1632 — Thu, 05 Mar 2020 06:26:22 GMT

Thinking on this overnight.

If..

Tip all blue balls into urn1 and all red into urn2

Pred = (Pred1(0/100) + Pred2(100/100))/2

= ( 0% + 100%)/2 = 50%

but..If you take one blue ball from urn1 and place it in urn2..

Pred= Pred1(0/99) x .25 + Pred2(100/101) x .75

= 0% + 99.01% x .75

= 0% + 74.26%

= 74.26%

not as good as the previous conditions 74.87%... but close...

Comment #14

Think — Thu, 05 Mar 2020 03:19:43 GMT

I instantly thought of that solution as I read the problem but then I came up with another solution.

It depends on the urns and if you have to mix the balls up.

If the urns are shaped right and you don't have to mix up the balls you could put half the blue balls in the bottums of both urns and then put half the red balls on top in both the urns. If the balls cant move around then you have a 100% chance of picking a red ball off the top of each urn!

My answer would depend on the... [ More ]

Comment #13

Ranett — Thu, 05 Mar 2020 02:46:37 GMT

I'd give you a hundred up-votes if I could.

Comment #12

ohiopick3 — Thu, 05 Mar 2020 01:19:07 GMT

I'd say, Go look for another job? Amazon can't pay that much per hour any ways?

Just my thought.

Comment #11

KY Floyd — Wed, 04 Mar 2020 17:17:39 GMT

= 25% + 37.31% = 62.31%

You're absolutely right, if you use the same 1 and 199 split, but you can mix the balls any way you want. Do you suppose that for a non-random selection of urn there might be a different mix that give the best results? And do you suppose that even with that 3:1 bias towards one urn you might still be able to get about the same probability as with the Amazon method?

I'll reserve the answer, too, but see if you can find one that includes this string: 257425. -- That... [ More ]

Comment #10

kao1632 — Wed, 04 Mar 2020 07:48:20 GMT

if mixed 50 red + 50 blue per urn.. prop(red) = 50%

if 1 red in urn 1 and 99 red + 100 blue in urn 2

p(red) = p1(1/1 x .25) + p2(99/199 x .75)

= 25% + 37.31% = 62.31

Comment #9

oneaweek — Wed, 04 Mar 2020 02:29:29 GMT

No applause. Just throw.

Comment #8

oneaweek — Wed, 04 Mar 2020 02:26:49 GMT

Well. Todd...... Who? S got. The balls

Comment #7

SweetRed83$ — Wed, 04 Mar 2020 01:35:13 GMT

219

0219

1958

21958

Comment #6

cottoneyedjoe — Tue, 03 Mar 2020 23:36:58 GMT

Good find, Todd. I heard a very similar one for a tech or investment banking job. It's the same set up with 100 balls of each color and two urns, and the same goal of maximizing your probability of getting a red ball by coming up with an optimal distribution. But instead of letting you choose an urn at random, a computer program randomly selects the urn for you.

If the urns have the same number of balls, the program will choose either urn with equal probability (i.e., one half). But if one ur... [ More ]

Comment #5

cottoneyedjoe — Tue, 03 Mar 2020 23:30:37 GMT

The average of the two cases is

(1/2)*(1/1) + (1/2)*(99/199)

= 1/2 + 99/398

= 199/398 + 99/398

= 298/398

Reduce the fraction 298/398 by halving the numerator and denominator to get 149/199.

Comment #4

KY Floyd — Tue, 03 Mar 2020 23:08:34 GMT

I'm not seeing any logical reason to start with 149, so I'm pretty sure it comes from working backwards to get the effective ratio if you only had one urn to choose from.

The 199 makes sense, but only as applied to the urn that has 199 balls in it. With only one red ball in one of the urns you're left with 199 balls for the 2nd urn so the numerator in the equation for that urn is 199, and the denominator is the number of red balls : 99/199. That gives us a 99/199, or 49.7487437185929648241206... [ More ]

Comment #3

destinycreation — Tue, 03 Mar 2020 22:18:29 GMT

As you are choosing an urn at random, you are equally likely to get one of these probabilities, so your winning percentage is the average of these two cases, which is 149/199, or 74.87%

I'm not very good at statistics. Where does the 149 come from in this explanation

Comment #2

VOLVO2LUV — Tue, 03 Mar 2020 20:20:44 GMT

0888

Comment #1

Sharon1000 — Tue, 03 Mar 2020 19:57:29 GMT

I need to hit a number. I haven't hit since 2019. Do someone have any good no. They want to share

Amazon uses this lottery puzzle as a brain-teaser for interviewees

Lottery Post — Tue, 03 Mar 2020 16:38:35 GMT

Includes video report

Tech giant Amazon reportedly poses a series of logical and mathematical puzzles during the interview process in order to select the best candidates, including the following lottery scenario. Take your best shot, then scroll down for the solution.

You have 100 red balls, 100 blue balls, and 2 urns. You distribute all of the balls between both urns, placing at least 1 ball in each, you cannot place them all in the same urn. You then pick one of the urns at random, and s... [ More ]