some peeps at work arguing over this odds question, so im coming to the :CF brain trust.
There are 10 names in a hat. You pick out a name, read it, and put it back in the hat so there are still 10 names in the hat. What are the odds of picking the same name out 3 times in a row?
1000 to 1.
What color is the hat?
Sus, are you sure or just throwing a guess out there? That's what I came up with too.
Obviously, the odds on the first pull would be 10 to 1.
Multiply by 10 on the second pull and it's 100 to 1.
Multiply by 10 on the third pull and it's 1000 to 1.
What they said.
I'd still like to know what color the hat is though.
the hat color is a touch of Mauvelous splashed with Cerulean.
thanks Sus.
One important distinction: while the probability of pulling the same name out of the hat 3 times in a row is 1/1000, the probability of pulling a given name on the 3rd try (or nth try) is still 1/10. Past performance does not predict future results, and all that...
i got pocket aces 3 times in a row the other night playing poker. someone tell me those odds. i don't even know how to start.
Quote from: phattymatty on December 19, 2006, 12:18:06 PM
i got pocket aces 3 times in a row the other night playing poker. someone tell me those odds. i don't even know how to start.
1,300:1
Quote from: Sgt PSN on December 19, 2006, 10:54:47 AM
Sus, are you sure or just throwing a guess out there? That's what I came up with too.
Obviously, the odds on the first pull would be 10 to 1.
Multiply by 10 on the second pull and it's 100 to 1.
Multiply by 10 on the third pull and it's 1000 to 1.
That's exactly the way it's figured; I just didn't bother spelling it out. (BTW, it's the same odds as drawing the Daily Number in PA, which pays 500-1. Cheap bastiches.)
how did you get that?
if you go by the same logic from murp's question the answer would be 10,648,000 to 1 but i know that can't be right.
how many people at the table at the deal factors into it i believe
no that shouldn't matter at all. there's still 4 of each card in the deck regardless if they're being dealt or staying in the deck.
But you're talking about drawing from a pool of available cards. If it's just you pulling from the deck, the first ace is a 1-in-52 chance. The second is a 1-in-51, and so on. But with other players at the table, the calculation gets a little more complicated because you have to assume whether they have drawn aces or not, as well.
I have no idea. I suck at math and just did what Sassy did (13 X 10 X 10 = 1,300).
nah, the chances of getting pocket aces just once are 220 to 1. (52/4) x (51/3)
so i guess 3 times in a row would be 220 x 220 x 220?
Well you dont know anyone elses cards, so you can count that.
The first time around you have a 4/52 chance of getting and ace. The second time a 3/51 chance. which equates to a 0.452% chance or about 1/200, of getting two aces. Your chance of getting that hand three times in a row is like 1 in 9 million. I wonder if thats right.
Odds of getting hit by lightning 1/600,000
Odds of winning lottery 1/60,000,000
That's about right. I forgot that suit didn't matter, so you're right about 4-52 and 3-51.
If it's only you drawing from the deck (not even the dealer in this example), it's like this:
52/4 = 13
51/3 = 17
Odds of getting 2 aces from the deck with the first two cards drawn: 13 X 17 = 221 to 1.
Doing it 3 times in a row is 221 X 221 X 221, which is 10,793,861 to 1.
ok, thats the number i came up with first.
i should play the lottery.
Quote from: phattymatty on December 19, 2006, 01:06:52 PM
ok, thats the number i came up with first.
i should play the lottery.
You'd be a sucker not to.
this is actually one hell of a stumper
if there are 4 people at the table, and you are dealt first the odds on ace #1 are 4/52, the second ace would be very complex. you have to factor the 3/51 for player #2, the 3/50 for player #3, and the 3/49 for player #4. your next card is out of a pool of 48 cards.
Quote from: Wingspan on December 19, 2006, 01:08:04 PM
this is actually one hell of a stumper
if there are 4 people at the table, and you are dealt first the odds on ace #1 are 4/52, the second ace would be very complex. you have to factor the 3/51 for player #2, the 3/50 for player #3, and the 3/49 for player #4. your next card is out of a pool of 48 cards.
Right. There are multiple different scenarios, depending on how many people are at the table... but the 1/221^3 is the simplest way to come to a nearly-correct answer.
Quote from: Wingspan on December 19, 2006, 01:08:04 PM
this is actually one hell of a stumper
if there are 4 people at the table, and you are dealt first the odds on ace #1 are 4/52, the second ace would be very complex. you have to factor the 3/51 for player #2, the 3/50 for player #3, and the 3/49 for player #4. your next card is out of a pool of 48 cards.
i think you're thinking too much into it. you don't know whether other people get an ace or not, so you don't have to factor them in at all. whether there are 10 people at the table, or you're just pulling from a deck by yourself, the odds should be the same.
Quote from: phattymatty on December 19, 2006, 01:13:34 PM
i think you're thinking too much into it. you don't know whether other people get an ace or not, so you don't have to factor them in at all. whether there are 10 people at the table, or you're just pulling from a deck by yourself, the odds should be the same.
Agree - in a random shuffle the probability of the aces being the first 3 cards is the same as the probability of the ace being every nth card.
The odds of drawing pocket aces in hold 'em 3 hands in a row are better than the odds of the Flyers and/or Sixers making the playoffs this year.
I think the answer to the first one would only be 1/100, because you don't have to get a specific name 3 times in a row; you just have to get whichever one you pick the first time, 2 more times. So I don't think the odds factor in on the first try, but I could be wrong
Quote from: PhillyandBCEagles on December 19, 2006, 04:46:35 PM
I think the answer to the first one would only be 1/100, because you don't have to get a specific name 3 times in a row; you just have to get whichever one you pick the first time, 2 more times. So I don't think the odds factor in on the first try, but I could be wrong
Good point. Looking back at how the question is worded, it seems to suggest randomly picking a name from the hat and then trying to draw it 2 more times rather than picking a predetermined name on 3 consecutive pulls.
Kudos to you.
so what the hell is the real answer. ha.
spaghetti
Quote from: MURP on December 20, 2006, 09:45:43 AM
so what the hell is the real answer. ha.
I would say 100 to 1 since the first pick is random. You've just got to pull the same twice.
If you want to pull a specific name 3 times in a row, then it's 1000 to 1.
BTW, when you're done with this little experiment, can I have the hat? Mauve really compliments my skin tone.
i like the new theory 100-1. it makes sense.
you guys think too much
oh, and what ever MMH puts down is the answer i'm going with, it got me through high school math, why buck the trend now.
Ghey.
Quote from: phattymatty on December 20, 2006, 10:19:52 AM
i like the new theory 100-1. it makes sense.
Well, it's simple:
The odds of drawing a specific predetermined name 3 times in a row are 1 in 1000.
The odds of drawing any name 3 times in a row are 1 in 100.
It's kind of like how the odds of getting pocket Aces are 1 in 221, but the odds of getting any pocket pair are 1 in 17.