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Started By
Message
re: Solve this OT
Posted on 7/24/18 at 4:41 pm to ell_13
Posted on 7/24/18 at 4:41 pm to ell_13
Put it this way:
“Picking” two boxes at the beginning means absolutely nothing in a random draw to open cases. It’s no different than simply opening random cases from the start until you get down to 2.
“Picking” two boxes at the beginning means absolutely nothing in a random draw to open cases. It’s no different than simply opening random cases from the start until you get down to 2.
This post was edited on 7/24/18 at 4:42 pm
0
Posted on 7/24/18 at 4:43 pm to TigerstuckinMS
quote:
If your chances of having the million case is 1/26, what are your chances of having any other case less than a million?
If the host will NEVER throw away the $1MM case, you should switch. It's essentially getting to pick 25 cases (switching) or 1 case (not switching). Either you picked the $1MM (3.85% chance), or you didn't, and it's in those 25 remaining cases. Because the host won't throw it away, we know the odds of it being that last case left are the odds you didn't pick the winner (25/26) times 100%, since it's a guarantee it will remain if you didn't pick it.
If the host doesn't know, and you just so happen to end up with $1MM and some other number left in the two remaining cases, the probability the remaining case is the $1MM is the probability you didn't pick the winner (25/26) times the probability the host threw away 24 of 25 boxes and still had $1MM remaining, which is 1/25. (25/26) * (1/25) = 1/26 = 3.85%, the same as your chance of picking the $1MM off the bat.
This post was edited on 7/24/18 at 4:45 pm
Posted on 7/24/18 at 4:59 pm to Fe_Mike
quote:
I believe you are the idiot.
Yep. I've thought about it more and I'm definitely the idiot on this one. It's 50/50. My apologies to TheMailman. Now, quit fricking up my packages.
The OT is still fricking stupid, though, and I helped!
This post was edited on 7/24/18 at 5:02 pm
Posted on 7/24/18 at 5:04 pm to slackster
quote:
roughly .3%.
That can't be correct.
Posted on 7/24/18 at 5:18 pm to Scooba
quote:
That can't be correct.
Idk, I tried to lay out the rationale.
You're asking for two specific cases ($.01 and $1MM) out of the 26. You've got to pick one of the with your first choice, which is clearly a 2/26 chance of occurring. Then you have to pick 24 straight cases without picking the remaining case. Alternatively, you can say you have to pick the second case correctly, which has a 1/25 chance of happening.
The probability of doing both of those events is (1/13)*(1/25), which is roughly .3%
Posted on 7/24/18 at 5:20 pm to Sao
quote:
67% is now added to the lexicon joining 288 and 350
yep. absolutely insane that this thread is 40 pages
Posted on 7/24/18 at 6:57 pm to Carson123987
I thought 1/2 at first, but realized it was 2/3 by this. Two of the three boxes have the same colored balls. So if the 1st ball is gold, there is a 2/3 chance the 2nd ball is gold too. Just like if the 1st ball is silver, there is a 2/3 chance the second is silver.
Posted on 7/24/18 at 6:58 pm to Rwt41
quote:
I thought 1/2 at first, but realized it was 2/3 by this. Two of the three boxes have the same colored balls. So if the 1st ball is gold, there is a 2/3 chance the 2nd ball is gold too. Just like if the 1st ball is silver, there is a 2/3 chance the second is silver.
Glad you're all caught up.
Posted on 7/24/18 at 7:01 pm to ThatMakesSense
Thanks buddy.
Posted on 7/25/18 at 2:50 am to Rwt41
Ok, let’s say that one accepts that you have a 67% chance of pulling another gold ball out of the same box. That means you only have a 33% chance of pulling a gold ball out of the other box.
How do you rationalize this?
How do you rationalize this?
Posted on 7/25/18 at 6:46 am to white perch
No. You have a 33% of pulling a silver.
Posted on 7/25/18 at 7:46 am to ell_13
This is almost my best thread ever. 5 more pages to go 
Posted on 7/25/18 at 8:07 am to Fe_Mike
let's keep this party going
No.
You are confusing a wager placed on a set of coin flips, vs a single coin flip.
Your chances of flipping heads 10 times in a row are .5*.5*.5*.5*.5*.5*.5*.5*.5*.5
Your chances of heads/tails on a single coin flip is 1/2.
Each coin flip has no memory of the previous flips. If you have already flipped heads 9 times in a row your next flip is still 50/50 to come up heads. Due to changing probability, your chances of completing that set of 10 heads flips in a row have gone from ~1/500 to 1/2.
quote:
he odds of flipping a coin in getting heads 10 times in a row is very small. If I flip a coin nine times in a row and get heads every time are you telling me I should pick tails for the 10th flip because the odds for it are much greater since at the beginning of the 10 flips the odds for 10 heads in a row were very small?
No.
You are confusing a wager placed on a set of coin flips, vs a single coin flip.
Your chances of flipping heads 10 times in a row are .5*.5*.5*.5*.5*.5*.5*.5*.5*.5
Your chances of heads/tails on a single coin flip is 1/2.
Each coin flip has no memory of the previous flips. If you have already flipped heads 9 times in a row your next flip is still 50/50 to come up heads. Due to changing probability, your chances of completing that set of 10 heads flips in a row have gone from ~1/500 to 1/2.
Posted on 7/25/18 at 8:27 am to glorymanutdtiger
quote:
But the next ball also being gold also is 100% for the one with two gold balls and 0 for one with gold and silver since you already picked the only gold
You don't know which box you took the gold from. That is the flaw in your reasoning.
You are assuming the gold came from the mixed box, but the problem did not state this.
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