Annoying Math Problem

[Jam-Fly 8]

Ugh, just can't figure this out.Basically, lets say you are 50-1 to win a given tournament. You play these tournaments in groups of 4s. What are the odds you will win 3 of the tournaments, and not win the other?I'm sure it's really obvious but I just can't get an answer I feel is right

[CBass1724 1]

125,000 to 1

[Jam-Fly 8]

125,000 to 1

nah I don't think it is. That's winning 3 tournies in a row, but does not factor in the fourth in tournament in which I don't win.

[CBass1724 1]

I was just kidding anyway. I am really good at math but I don't have the slightest clue as how to begin answering this problem.122,500 to 1?

[Suited_Up 2]

0%trust me.

[mx957 0]

Doesn't the standard 50% rule apply here?You either succeed or fail. 50/50.

[Mercury69 3]

125000 to 1 to win 3

[dna4ever 2]

am i the only one that is sure this is an elaborately set up brag post waiting to be delivered?

[bbgun 0]

<<<<<<Insert obligatory brazillion to 1 post here>>>>>>

[Azwethinkweiz 0]

125,000 to 1 to win 3 of the 4 tournaments.You're still 50 to 1 against winning the 4 tournament no matter how well or poor you did in the other three.So the odds you won't win the 4 tournament are 50 to 1.

[copernicus 0]

33,825:1 using the stated 50:1 (which != 1/50)If OP meant 49:1 which does = 1/50 then the answer is 31887:1For the latter its (1/50)^3 * 49/50 *4 because the losing tourney could be any one of the 4

[zimmer4141 0]

125000 to 1 to win 3

No, it isn't. By this logic if he had 1 million tries he's still 125000 to 1.It's kind of a tree problem. First tourney he wins 1/50 and loses 49/50. I think it's 4 * 49/(50^4), or 196/6,250,000 (.00003136) where 1/125,000 = (.000008)

[bbgun 0]

Lots of confusing formulas and numbers...

FYP

[copernicus 0]

No, it isn't. By this logic if he had 1 million tries he's still 125000 to 1.It's kind of a tree problem. First tourney he wins 1/50 and loses 49/50. I think it's 4 * 49/(50^4), or 196/6,250,000 (.00003136) where 1/125,000 = (.000008)

GMTA

[zimmer4141 0]

FYP

Well, he has to win 3 and lose 1, so it's (1/50)(1/50)(1/50)(49/50) because those are the probabilities of winning 3 and losing 1. He can lose any of the tournaments, so you multiply this by 4 because there are 4 possible combinations.

[checkymcfold 0]

33,825:1 using the stated 50:1 (which != 1/50)If OP meant 49:1 which does = 1/50 then the answer is 31887:1For the latter its (1/50)^3 * 49/50 *4 because the losing tourney could be any one of the 4

this is correct.

[Mercury69 3]

No, it isn't. By this logic if he had 1 million tries he's still 125000 to 1.It's kind of a tree problem. First tourney he wins 1/50 and loses 49/50. I think it's 4 * 49/(50^4), or 196/6,250,000 (.00003136) where 1/125,000 = (.000008)

Einstein cat has left teh buildingz

[copernicus 0]

Einstein cat has left teh buildingz

Good thing its Einsteins and not Schroediners cat..the box is open so there are no quantum issues any longer....in this universe anyway.

[Suited_Up 2]

33,825:1 using the stated 50:1 (which != 1/50)If OP meant 49:1 which does = 1/50 then the answer is 31887:1For the latter its (1/50)^3 * 49/50 *4 because the losing tourney could be any one of the 4

[Mercury69 3]

$100?

[davezz5 0]

this is correct.

Standard

[Merby 3]

Ugh, just can't figure this out.Basically, lets say you are 50-1 to win a given tournament. You play these tournaments in groups of 4s. What are the odds you will win 3 of the tournaments, and not win the other?I'm sure it's really obvious but I just can't get an answer I feel is right

I haven't read the other replies yet. I noticed that this thread already has two pages, so no doubt the correct answer has already been supplied.Nevertheless, here it is:1) Your odds of winning any one tournament is 50:1 or 1/51 = 0.0196...2) Hence your odds of losing any one tournament is 50/51 = 0.980...3) The outcome in any one tournament is independent of the outcome in the other three tournamentsTherefore, the final equation is:(1/51)*(1/51)*(1/51)*(50/51)*4 = 200/6,765,201So you will win exactly 3 out of a block of 4 tournaments 200 out of every 6.77 million attempts (or approximately 1 in every 33,826 times).Good luck!

[Poker Addict 0]

this thread already has two pages

You need to increase the number of posts per page imo

[Merby 3]

You need to increase the number of posts per page imo

I'm well aware of that option, but ...<--- Too lazy to change that option.<--- Feels that the two best words of the english language are "DE" and "FAULT"^^^ Above are the two fundamental reasons why I have not changed my preferences from the default settings ^^^

[Jam-Fly 8]

Came across another problem similar to this recently. My friend was playing roulette and was betting on four numbers only on each spin. In 10 spins, one of his four numbers came up 6 of the ten times.Using what I think is the correct formula, it was(4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (33 / 37) * (33 / 37) * (33 / 37) * (33 / 37) * 10 = 1.0101793 × 10^-5 likely for this to happen? This is almost 100,000-1, yes?