Is Extra Information Helpful? A Probability Puzzle
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- Опубліковано 24 лип 2021
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I have two envelopes, one with $100 and one with nothing. You can choose one. But before you do, I will offer you a little more information. I have a canister with five balls in it, three marked with the correct envelope and two marked with the wrong envelope.
How much would you be willing to pay to look at one of the balls? And how much more would you be willing to pay to look at a second?
The first question is pretty straightforward. The second one is deceptive. Be careful!
Paying $10 for the first ball gives you the exact same expected winnings as taking the blind 50-50 chance: (100 - 10) x 60% - 10 x 40% = 54 - 4 = 50
It would be nice to see a full break down of third ball.
Very neat! I had a cool moment at my whiteboard when I finished the cost-benefit for the second question.
Great video.
The answer to the first question is 10$ and the answer to the second $0.
However, I am not completely convinced by the "new information is not pivotal" argument.
For me, the key in this example is not in weather the new information makes you chance your decision but in the expected earnings in both scenarios.
After "buying" one ball, expected earnings = $60. After buying the second ball, expected earnings = $60. It wouldn't matter if the information provided by the second ball suggested a change in my decision.
Maybe we can prove that as long as the new information is not pivotal, there are no increases in expected payoff and viceversa?
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I’m loving these puzzles! I always have a…really good time working through them. Thanks again.
Awesome. Keep up the good work.
The second and third ball were interesting
It would be nice to see a single offer to draw two balls (for one extra payment).
Same, 4/10: 3:1 that they are directing to the money.
Different, 6/10: No benefit
Overall: 12/40.
Not worth it(?).
Ah! I really expected this to be different with replacement, my math says it wasn't, but now I'm pretty sure my math is bunk. Help?
To me intuitively it stands to reason that with replacement you will always pay *a* price for an additional ball, but one that will decay exponentially. Here's my contradictory math. I probably fucked up on the permutations if anything? I thought it would just be permutations.
probabilities for 2 draws
2-0: 36% (.6*.6)
1-1 * 2: 48% (.6*.4*2)
0-2 wrong: 16% (.4*.4)
adds up to 100 so seems like method is working.
It looks like the EV of two balls is only 36* 48/2 = 60% so still providing no new value?
3 balls:
1-2: .096 * 3 combos = .288
0-3: .064
35.2% chance of being misled by draws with no chance of randomly guessing.
Now the choice after draw with replacement is worth 64.8$ with that 4.8$ new to the third ball.
I wanted to see if this was just a general case of only odd numbered balls helping.
2-2: 12 combos .6912
1-3 4 combos: .1536
0-4: .0256
.6912/2 + .1536 + .0256
Now for some reason I have a 52% chance of being misled by the balls. This can't be right. Wtf did I do wrong?
It’s the same with or without replacement. I can’t comment on whether your justification was right, but your conclusion is!
Okay, I skimmed that real quick and there wasn’t anything obviously wrong with your math. And this is a general problem with the even piece of information as you were speculating.
Okay I think I finally fixed the math, even though I can't figure out how it's not just 4 choose 2 on permutations. It's *six* combinations of 2-2 ties not 12 which still gets us the .352 probability of guessing wrong case.
As you alluded to it's just the general case of the information not being actionable when it can lead to a tied result.
If the A case was already ahead, drawing another good ball will still put it ahead and is not actionable.
If the A case was behind, now they're maybe tied and we just guess randomly (semi actionable since it improves our EV)
but that EV improvement is exactly cancelled out by the case where we draw B and A is now tied leading us to guess randomly.
So it's a general case of even numbered balls never mattering, with or without replacement.
If I understood what you were trying to do correctly, probabilities for 2 draws are wrong. The chances of 2-0 aren't 0.6*0.6, they are 0.6*0.5 (3/5 * 2/4)
2-0 is 30%, 1-1 is 60%, 0-2 is 10%
3-0 is 10%, 2-1 is 60%, 1-2 is 30%
3-1 is 40%, 2-2 is 60%
3-2 is 100%
I'm surprised with or without replacement didn't matter for this
I win!! When should I expect my $100?
Let's play a game. We'll both roll a dice, and whoever gets the higher roll wins.
...
I rolled a six. I promise I didn't cheat. In fact, I can even take a picture of the dice and show it to you as proof.
If the option to buy all the balls was possible, you'd pay $50 for all of them. If the balls were replaced after each time you took one out, you'll spend $50 in the limit as the number of balls you take out goes to infinity.