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Topic: One more spin? (Read 1882 times) |
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JocK
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One more spin?
« on: Dec 11th, 2005, 2:25am » |
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Three players compete for a price of $ 10,000. Each player takes turn spinning a wheel as often as (s)he wants. The wheel has an equal probability of stopping on every angle between 0 and 360 degrees. The score of an individual player is the sum of the angles of the wheel (s)he achieved from subsequent spins as long as this total is less than 360 degrees. If the sum for a player reaches or exceeds 360 degrees, that player is immediately disqualified. The player with the highest score wins the $ 10,000.
You are player one. What strategy do you follow?
Can you generalise to an arbitrary number N > 1 of players?
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| « Last Edit: Dec 11th, 2005, 2:27am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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mattian
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Re: One more spin?
« Reply #1 on: Dec 11th, 2005, 7:59am » |
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Blackjack - only round.
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towr
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Re: One more spin?
« Reply #2 on: Dec 11th, 2005, 8:33am » |
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I'm not seeing the similarity with black jack. There's no dealer, and only one person wins, to name but two things.
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towr
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Re: One more spin?
« Reply #3 on: Dec 11th, 2005, 8:51am » |
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Anyway, I'll keep spinning until the probably my next spin makes mego over 360 is greater than or equal to the probability of any of the others getting a better score.
Which is rather obvious.
I'm sure you'd like to know where the cutt-off point is, depending on N. And so would I 
Do only integer angles count? Or could you get 359.132435 degrees?
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JocK
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Re: One more spin?
« Reply #4 on: Dec 11th, 2005, 9:17am » |
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on Dec 11th, 2005, 8:51am, towr wrote:
Do only integer angles count? Or could you get 359.132435 degrees? |
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The angles are real numbers with 0 =< angle < 360.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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SWF
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Re: One more spin?
« Reply #5 on: Dec 12th, 2005, 5:18pm » |
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This game is very similar to the "showcase showdown" on the game show "The Price is Right". That game involves 3 players and spinning of a wheel, except the wheel is numbered discretely: 5, 10, 15, ... 100, with the goal being a total of 100.
For simplifying notation, I tried a slighly modified problem: instead of a number from 0 to 360, a number from 0 to 1 is spun. Player number 0 spins last, player number 1 2nd to last, ...
If player number 1 holds with a value x, the probability of Player 1 beating Player 0 turned out to be: g(x) = 1 - (1-x)*exp(x)
Player number N should obviously keep spinning until exceeding the best score previously obtained, and stop spinning if some number yN is exceeded. It turns out that yN is the solution of
g(yN)N = [integral] g(x+yN)N dx where limits of integration are from 0 to 1-yN.
The value of y1 is the solution to y+e = (3-2y1)*exp(y1), or approximately 0.570557. The value of y2 is 0.6879156.
For solution to the orginal problem multiply y by 360. This gives 247.65 as the value first player to spin should exceed in a 3 player game.
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JocK
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Re: One more spin?
« Reply #6 on: Dec 13th, 2005, 2:40pm » |
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Nice. Well done.
Clearly, for small number of players, the first player has a disadvantage compared to later players.
In the limit that the number of players N grows to infinity, what is the probability that the first player will win?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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rmsgrey
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Re: One more spin?
« Reply #7 on: Dec 14th, 2005, 4:41am » |
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on Dec 13th, 2005, 2:40pm, JocK wrote:Nice. Well done.
Clearly, for small number of players, the first player has a disadvantage compared to later players.
In the limit that the number of players N grows to infinity, what is the probability that the first player will win?
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I'd guess at 1/N but that's just a wild stab in the dark...
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SWF
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Re: One more spin?
« Reply #8 on: Dec 14th, 2005, 5:27pm » |
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The limit as N goes to infinity of the probability of the first player winning is clearly zero, but Jock must have wanted an asymtotic value.
The last player to spin should always have an advantage, and the first player's chances should be less than 1/N. Using equation given before, it is found that the first player's chances approach 1/(N*e). He should keep spinning until he gets a value of at least
1 - (1-(N*e)-1/N)/e
(That is for a wheel with 1 as maximum value. Multiply by 360 for the original problem.
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srn437
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Re: One more spin?
« Reply #9 on: Sep 1st, 2007, 3:09pm » |
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Although player n out of n players has the easiest strategy get more than anyone else has had. As player 1, I just need to get a high number they would have a hard time beating. They always have a way to beat it though.
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| « Last Edit: Sep 8th, 2007, 11:57am by srn437 » |
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Hippo
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Re: One more spin?
« Reply #10 on: Sep 8th, 2007, 9:48am » |
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A have thought a little bit now, so here I my results (exactly matching the old ones from SWF):
(with the angle measured in fullcircle unit).
Current strategy of a player can be characterized by his treshold ti.
When his score is less than ti he continues rolling.
Then after k rolls probability his score is less than x is for x<ti equal xk/k!.
(integrating, using induction).
For 1>=x>ti the probability score is less than x after at least k rolls is
tik/k!+(x-ti)\sum_j=0k-1ti j/j!.
Letting k go to infinity we get the probability (x-ti)eti.
Now the strategy for choosing ti. Number players in decreasing order n,n-1,...1,0.
Let si denotes best score obtained so far after i-th player ends.
Clearly ti>=si+1 otherwise i-th player cannot win.
t0=s1 and probability player 0 wins is (1-s1)es1.
Now think about choosing t1.
Probaility player 1 wins is et1intt11(1-(1-x)ex)dx that equals (1-t1-e)et1+(2-t1)e2t1, now let
ex1=(x1+e)/(3-2x1), than probability maximizes for t1=min(s2,x1)
(x1approx 0,570556528 and probability approx 0,424985743).
Now choosing t2.
Suppose t2>x1:
Probability palyer 2 wins is et2intt21(1-(1-x)ex)2dx
After integration (and derivation) we find numericaly the maximum at x2 approx 0,687915542 the probability is approx 0,285915671.
t2=min(s3,x2).
I don't thing there is better probability for t2<x1.
Choosing ti ... suppose ti>xi-1:
Probability player i wins is
etiintti1(1-(1-x)ex)idx
Maximizing the result we get xi and set ti=min(si+1,xi).
But the computation becames more and more complicated ...
BTW: I have made some more computation ... it is interesting how the integrals are related to errors of approximation of ex by Taylor polynom, but the effort to describe the result in a compact form failed.
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| « Last Edit: Sep 14th, 2007, 1:37pm by Hippo » |
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