Solve it!

>>8863211
>Do my homework hurrr durrr
What's in it for me?

>>8863211
2/40+1/39

>>8863215
No, it's not homework, I just want to know, if
66% is right or not.

Preparing for A-level btw

>>8863221
Explain, please.

>>8863229
the probability of the first person you deal to getting a 1 or the second given the first did not get a 1 is 1/20 (2 1s in the deck out of 40 cards)
^probability that someone gets one 1

the probability of the second person getting a 1 given the first person got a 1 is 1/39 (one 1 out of 39 cards left)
^probabilitynthey both get a 1

add them and that's the probability that at least one 1 shows up (assuming you only deal one card to each person, OP never specified)

>>8863252
thinking about it now I'm actually wrong
it's 1/39 + 2/40 + (1/39)*(1/20)

>>8863225
Alright

P(A or B) = P (A) + P (B) - P(AnB)
P(A or B) = 2/40 + 2/40 - P(A)*P(A/B)
P(A or B) = 4/40 - 2/40 *1/39
P(A or B) = 4/40 - 2/1560
P(A or B) = (156 - 2) / 1560
P(A or B) = 154 / 1560
P(A or B) = 0.0987179487
P(A or B) = ~0.10 or 10%

>>8863221
This is wrong, A and B could happen simultaneously thus P(AnB) =/= 0

>>8863285
P(only second 1 or only first 1 or both 1)
*all are mutually exclusive therefore
= P(only second 1) + P(only first 1) + P(first 1 and second 1)
= 1/39 (no replacement) + 2/40 + (1/39)*(2/40)

*P(first and second 1)= P(1|one 1)*P(first 1)
^ by rearrangement of the conditional probability formula

>>8863309
It's simpler of you think of
A = First person getting a 1
and
B = Second person getting a 1

While P(B/A) is the chance B gets a 1 *after* A has gotten one

B is dependent on A
Neither is mutually exclusive

Thus: >>8863289

>>8863315
>>8863289
Looking closer, I wrote it >>8863289 backwards, with A being dependent on B but it's the same result if you consider A the second person and B the first.

>>8863309
P(only first 1) = P(first 1 and no second 1) = (2/40)*(38/39)

fuck,
1/39 + (2/40)*(38/39) + (1/39)*(2/40)

>>8863315
but it's also the probability of them both getting a 1 (at least one 1)

>>8863327
>but it's also the probability of them both getting a 1 (at least one 1)
Yes, the question is what's the probability that *at least* one of them gets a 1. This includes the possibility that both do.

>>8863330
B can't be 2/40 if there's no replacement but OP didn't specify
I assumed no replacement and you give one card to each person

>>8863327
Fuck you are right I got what you mean now.

It should be P(A or B) + P(A and B)

Fuck I'm done with this shit

>>8863327
wait fuck me,
P(only first 1) + P(only second 1) + P(first and second 1) = (2/40)*(38/39) + (38/40)*(2/39) + (2/40)*(1/39)

final answer

>>8863341
>>8863289
Alright alright my final answer
Previously I did not take into account card replacement and the chance of both happening

A = First person getting a 1
B = Second person getting a 1
B depends on A
Both are not mutually exclusive

So...
P(A or B) + P(A and B) =
P(A) + P(B) - P(AnB) + P(A) * P(B/A)
P(A) + P(B) - P(A) * P(B/A) + P(A) * P(B/A) =
P(A) + P(B) = 2/40 + 2/39 = (78+80) / 1560 = 0.1012820513
Or
10% probability lel

I'm not sure this is correct.
This guy seems to know his shit >>8863353

So if his answer and mine are the same you know that's it.

And if they are different... well... don't trust mine... I haven't done this shit in a while

>>8863211
100%
since you've mixed all of them and dealt them

>>8863411
Justified by https://en.wikipedia.org/wiki/Pigeonhole_principle

So if you've dealt all of the cards and the set contains two ones, you can guarantee that one of these people has a one

The cards are dealt, does at least one person have one 1?
Yes, 100%

Now, did you mean to ask what the probability of a 1 is for each card dealt as they are dealt?
his appears to be the approach being taken by many posters.

https://pastebin.com/MZ04Ecmt

With N=100000 I get ~0.755.

Don't know, if I understood the question right.

>>8863211
depends who dealt the cards

>>8863408
>>8863430

What now?

>>8863453
There is no way the chance is 75%
Put aside the math and think logically
It's a deck with 40 cards of which only 2 are winners. For the first guy alone that's a 5% chance of winning. For the second guy the chance is higher or lower depending on whether the first guy won, but only marginally different.

Without using math you can tell 75% is just wrong.

>>8863511
So you think that >>8863430 is wrong?

>>8863527
Yes.

>>8863511
https://pastebin.com/raMyk4mW

Even if I'm paranoid and mixing the cards, I get the same result.

>>8863553
But why?