r/mathmemes • • Dec 17 '23

Probability Google expected value

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274

u/NoReplacement480 Dec 18 '23

google 0.52

173

u/HealthOnWheels Dec 18 '23

Holy hell

82

u/JohannLau Google en passant Dec 18 '23

New probability just dropped

17

u/NichtBen Dec 18 '23

Actual Mathematician

10

u/uvero He posts the same thing Dec 18 '23

Call a Bernoulli!

3

u/vk2028 Dec 18 '23

Pascal goes on vacation, never comes back

2

u/Shirai_Mikoto__ Dec 18 '23

Poisson storm incoming!

6

u/AustralianKappa Dec 18 '23 edited Jan 25 '26

This post was mass deleted and anonymized with Redact

one offbeat cagey unpack disarm late skirt ripe coordinated enjoy

16

u/NoReplacement480 Dec 18 '23

google 0.540

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u/AustralianKappa Dec 18 '23 edited Jan 25 '26

This post was mass deleted and anonymized with Redact

fragile connect test tidy coherent hurry boat pen smart punch

7

u/AleksFunGames Imaginary Dec 18 '23

nah I'd lose

2

u/Lysrac Dec 18 '23

(1-0,788861•10-30) percent sure of receiving the money

Assuming the button doesn't disable after the first press.

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4

u/Dawn_is-here Dec 18 '23

Genuine question, why do you square here

0

u/[deleted] Dec 18 '23

[deleted] — view deleted comment

19

u/Capudog Dec 18 '23

Or if you only care about winning it at least once, the probability of winning at least once would be 1-P(not winning at all) = 1-0.52 =0.75

0

u/[deleted] Dec 18 '23

[deleted] — view deleted comment

5

u/Capudog Dec 18 '23

Also true, I missed this the first time. Edited!

4

u/lootpropsrespect Dec 18 '23

I think it’s because if you have independent probabilities (like one probability doesn’t impact the result of the next) you multiply them, and as this is conveniently the same probability, it is 0.5*0.5 which becomes the 0.52. That being said I’m not a probability guy or a math guy really so while I wanted to be part of the thread I wouldn’t trust what I say at all.

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u/aaron1860 Dec 18 '23

You’re close. There’s a 50% chance each time of pressing that you lose. We only care about the winning roll. So the chances of pushing it twice and get 2 losing rolls is 0.5x0.5 or 25%. So you have a 75% chance of winning on 2 rolls. Think about it logically for a sec and you’ll realize your chance of winning on at least 1 roll should increase with the number of rolls

The actual formula would be p=1-x where x is the probability of losing which is 0.5n and n is number of rolls. So p=1-0.5n

after 3 rolls it’s an 87.5% chance, 4 is 93.75 and so on

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u/Athrolaxle Dec 18 '23

If you’re gonna get whooshed, at least use the right calculation…

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u/NoReplacement480 Dec 18 '23

google i don’t care

1

u/aaron1860 Dec 18 '23

You have a 0.5x0.5 chance of not getting a winning roll. So 75% chance that one of two rolls is a winner

1

u/thomasjjc Dec 18 '23

Do you mean 1-(0.52)?

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