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Topic: Game Of 100 Cups (Read 840 times) |
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Padfoot
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Game Of 100 Cups
« on: Mar 14th, 2007, 6:30pm » |
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Your friend has chalanged you to the folowing game:
He has layed out 100 colored cups onto a table and has hid a very rare diamond under one of them. For every cup you take off the table, you must pay him an amout of money progresivly going up by $100 (The first move costs $100, second costs $200, third move costs $300 etc).
When the diamond is found, the amount of money that the moves cost is totaled up, and you pay him that amount. The diamond is now yours to keep.
How much must this diamond be worth in order for you to be willing to play?
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| « Last Edit: Mar 16th, 2007, 5:53am by Padfoot » |
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"Don't blame me. I'm just an interpreter. I'm not supposed to know a power socket from a computer terminal." - C-3PO, Star WarsV
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #2 on: Mar 15th, 2007, 9:26am » |
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My guess: > 127500
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Barukh
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Re: Game Of 100 Cups
« Reply #3 on: Mar 15th, 2007, 10:46am » |
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I agree with towr .
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #4 on: Mar 15th, 2007, 10:47am » |
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on Mar 15th, 2007, 10:46am, Barukh wrote:
lol!
You don't need to hide the name... his answer is already hidden. 
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Random Lack of Squiggily Lines
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Re: Game Of 100 Cups
« Reply #5 on: Mar 15th, 2007, 11:21am » |
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it would have to be $100! + $0.01 at least so i could have ANY profit. ^_^
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You can only believe i what you can prove, and since you have nothing proven to cmpare to, you can believe in nothing.
I have ~50 posts to hack a "R" into a "D". Which one?
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Padfoot
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Re: Game Of 100 Cups
« Reply #6 on: Mar 15th, 2007, 2:33pm » |
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Isn't it strange that there is an equal chance of finding the diamond on any of your guesses. I know of many people that would say it is very unlikely to get it on the first guess, when it is more likely to find it on, say the 70th guess.
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"Don't blame me. I'm just an interpreter. I'm not supposed to know a power socket from a computer terminal." - C-3PO, Star WarsV
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towr
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Re: Game Of 100 Cups
« Reply #7 on: Mar 15th, 2007, 2:50pm » |
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on Mar 15th, 2007, 2:33pm, Padfoot wrote:| Isn't it strange that there is an equal chance of finding the diamond on any of your guesses. I know of many people that would say it is very unlikely to get it on the first guess, when it is more likely to find it on, say the 70th guess. |
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Yes, some people are strange. It is more likely to find it on a later guess, but not on a specific later guess (which is equally likely to the first).
It the same way people think HTTHTHHT is mor elikely than HHHHHHHH when you flip a coin 8 times. But what they're actually thinking is "a sequence like the first is more likely than a sequence like the second". The specific sequences in question are of course equally likely (for a fair coin).
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Grimbal
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Re: Game Of 100 Cups
« Reply #8 on: Mar 16th, 2007, 2:41am » |
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I also agree with towr.
Anyway you can improve the odds by realizing that there are considerable more chances to find it among the first 10 cups than to find it in any other single cup.
So, by being less greedy and concentrating your hope not on each cup as you lift it but on the group of the first 10 you can improve your chances of finding it there, and have a good chance of avoid the 90 more expensive cups.
But then, you still can be unlucky...
PS: 
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #9 on: Mar 16th, 2007, 9:24am » |
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on Mar 16th, 2007, 2:41am, Grimbal wrote:I also agree with towr.
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Lol!
You don't need to hide it. We know you always agree with towr  
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Barukh
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Re: Game Of 100 Cups
« Reply #10 on: Mar 16th, 2007, 10:18am » |
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on Mar 16th, 2007, 9:24am, Aryabhatta wrote:
Lol!
You don't need to hide it. We know you always agree with towr |
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Do you agree with towr, then?
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Grimbal
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Re: Game Of 100 Cups
« Reply #11 on: Mar 16th, 2007, 1:45pm » |
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It is towr who would agree with me if I were faster to reply.
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #12 on: Mar 16th, 2007, 1:57pm » |
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on Mar 16th, 2007, 1:45pm, Grimbal wrote:| It is towr who would agree with me if I were faster to reply. |
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Hard to do that isn't it? I don't think towr is human.
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irrational
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Re: Game Of 100 Cups
« Reply #13 on: Mar 16th, 2007, 3:06pm » |
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Can anyone please explain the math behind this.
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towr
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Re: Game Of 100 Cups
« Reply #14 on: Mar 17th, 2007, 3:12am » |
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on Mar 16th, 2007, 3:06pm, irrational wrote:| Can anyone please explain the math behind this. |
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Each cup is equally likely to have the diamond under it. But to get to the Kth cup, you have to open the previous K-1 cups as well, so you have to add all those costs.
So the cost of opening the Kth cup is 100 + 200 + 300 + ... + K*100, that is to say K*(K+1)/2 * 100 (100 times the Kth triangular number)
Now we know the cost of each cup, and we knew the probability (1/100), so we can get the expected cost:
sum of K=1 to 100 over (K*(K+1)/2 * 100) * (1/100)
=
sum of K=1 to 100 over K*(K+1)/2
=
171700
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Grimbal
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Re: Game Of 100 Cups
« Reply #15 on: Mar 17th, 2007, 4:24am » |
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I was about to say that.
Well, actually my idea was that you take
cup 1: 100/100 times for $100
cup 2: 99/100 times for $200
cup 3: 98/100 times for $300
This gives sumk=1..100 (101-k)·k = 171700.
But if I were to play the game, I would try the first cup and stop if it didn't show the diamond.
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| « Last Edit: Mar 17th, 2007, 4:27am by Grimbal » |
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towr
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Re: Game Of 100 Cups
« Reply #16 on: Mar 17th, 2007, 4:37am » |
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on Mar 17th, 2007, 4:24am, Grimbal wrote:| But if I were to play the game, I would try the first cup and stop if it didn't show the diamond. |
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It'd still have to be worth more than 10000 to give a positive expectation. Of course that's less than 1/17th it should be worth if you keep playing till you have it.
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| « Last Edit: Mar 17th, 2007, 4:38am by towr » |
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ThudnBlunder
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Re: Game Of 100 Cups
« Reply #17 on: Mar 17th, 2007, 5:42am » |
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kFor the sake of completeness  r(r + 1)/2 = k(k + 1)(k + 2)/6 r=1
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #18 on: Mar 17th, 2007, 11:24am » |
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The reasoning for my guess:
If you pick 50 random cups, the probability that the diamond is in one of them is 1/2.
Thus if the cost of the diamond is more than the cost of checking 50 cups, I will play.
Cost of 50 cups = 127500.
This is wrong reasoning though (assuming you cannot give up in the middle)...
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| « Last Edit: Mar 17th, 2007, 11:34am by Aryabhatta » |
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #19 on: Mar 17th, 2007, 11:33am » |
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on Mar 16th, 2007, 10:18am, Barukh wrote:
Do you agree with towr, then? |
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I think I do
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Aryabhatta
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Re: Game Of 100 Cups
« Reply #20 on: Mar 17th, 2007, 11:42am » |
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on Mar 17th, 2007, 11:24am, Aryabhatta wrote:The reasoning for my guess:
If you pick 50 random cups, the probability that the diamond is in one of them is 1/2.
Thus if the cost of the diamond is more than the cost of checking 50 cups, I will play.
Cost of 50 cups = 127500.
This is wrong reasoning though (assuming you cannot give up in the middle)...
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In fact this is wrong even if you are allowed to stop in the middle.
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| « Last Edit: Mar 17th, 2007, 11:42am by Aryabhatta » |
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irrational
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Re: Game Of 100 Cups
« Reply #21 on: Mar 19th, 2007, 11:47am » |
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Thanks..
My thinking was:
If you have paid $100 for cup 1 and that is removed from the table the probability for the second cup becomes 1/99 instead of 1/100.
So the summation for cost becomes.
(1/100)*100+(1/99)*200 ....
Any flaw with the thinking or am I thinking too much
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| « Last Edit: Mar 19th, 2007, 11:49am by irrational » |
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SMQ
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Re: Game Of 100 Cups
« Reply #22 on: Mar 19th, 2007, 12:31pm » |
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on Mar 19th, 2007, 11:47am, irrational wrote:| Any flaw with the thinking |
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You need to account for the probability (99/100) that you even get to the second cup. Thus:
(1/100)*100 + (99/100)(1/99)*200 + (99/100)(98/99)(1/98)*300 + ...
= (1/100)*100 + (1/100)*200 + (1/100)*300 + ...
--SMQ
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--SMQ
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irrational
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Re: Game Of 100 Cups
« Reply #23 on: Mar 19th, 2007, 8:40pm » |
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Ahaa!!..
Thanks .. Now I know why I never win anything in Las Vegas! 
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Grimbal
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Re: Game Of 100 Cups
« Reply #24 on: Mar 20th, 2007, 9:55am » |
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on Mar 19th, 2007, 11:47am, irrational wrote:Thanks..
My thinking was:
If you have paid $100 for cup 1 and that is removed from the table the probability for the second cup becomes 1/99 instead of 1/100.
So the summation for cost becomes.
(1/100)*100+(1/99)*200 ....
Any flaw with the thinking or am I thinking too much |
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Besides the flaw pointed out by SMQ, this assumes you only pay the last cup.
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