Infographic: in which Giant Bomb take 37 tries to guess a number between 1 and 25

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Clint

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No Caption Provided

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htr10

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This is the best. I love this amazing moment in the history of Giant Bomb lore.

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deactivated-5cdbdf5c5a6e1

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Lol didn’t know they still neither picked 14.

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kaungo

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Any source to this shenanigan?

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elmorales94

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#5  Edited By elmorales94
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transwave

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I believe the expected number of guesses to get the correct number (on average) by guessing purely randomly and not taking into account previous guesses is equal to the number of possible choices, so in this case 25.

In other words, the giant bomb team working together performed about 50% worse than a 25-sided die would be expected to. Hmm....

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hawkinson76

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GOTY

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briguyd

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@transwave: No, I think the average number of guesses is 25/2 (12.5).

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BeachThunder

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This is amazing.

I'm surprised Dan didn't just use his magic.

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Monkeyman04

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I was so sure that Jason would have gotten it when he said 5 and then 3. But I love that it took so long.

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Onemanarmyy

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Jason with the 2x 11 back to back :D

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ripelivejam

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ajamafalous

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@briguyd said:

@transwave: No, I think the average number of guesses is 25/2 (12.5).

It's been a while since I've taken probability/statistics, but:

It depends on whether they were guessing with memory of the previous guesses (which they clearly weren't).

Assuming no memory (i.e. completely random, can repeat guesses; think rolling a 25-sided die until it lands on 4):

There is a uniform probability of any given roll landing on 4: p = 1/n = 1/25 = 0.04

X~Geo(0.04) =>
E(X) = 1/p
E(X) = 1/0.04 = 25 guesses

Assuming memory (i.e. will not repeat guesses; think blindly picking from a bucket containing 25 balls numbered 1 through 25 and picking until you get 4 without replacing the ball you've picked out after each guess):

The probability of guessing correctly on the first attempt is 1/25. Given failure on the first attempt (probability 24/25), the probability of success on the second attempt is 1/24. Given failure on the first and second attempts (24/25 times 23/24), the probability of success on the third attempt is 1/23. This continues until the twenty-fifth attempt, where the probability of success is 1.

The expected value for X where X is the number of guesses before the correct number:

E(X) = (1/n) ((n(n+1))/2) = (1/25) ((25x26)/2) = 13 guesses

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Kidavenger

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reasonablesteve

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What a beautiful, dumbass thing. This video is how I'll have to introduce people to Giant Bomb from now on. It's a website about video games.

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GundamGuru

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#16  Edited By GundamGuru
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Redhotchilimist

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Oh lord.

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MundaneSoul

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This is just some quality, feelgood shit right here.

2018's coming up roses.

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WetRacoon

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Knifey_Spoony

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When did this happen? I'm a bit behind.

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htr10

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Knifey_Spoony

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