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        riddles >> hard >> One more spin?
        
(Message started by: JocK on Dec 11^th, 2005, 2:25am)

Title: One more spin?
Post by JocK on Dec 11^th, 2005, 2:25am

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?

Title: Re: One more spin?
Post by mattian on Dec 11^th, 2005, 7:59am

Blackjack - only round.

Title: Re: One more spin?
Post by towr on Dec 11^th, 2005, 8:33am

I'm not seeing the similarity with black jack. There's no dealer, and only one person wins, to name but two things.

Title: Re: One more spin?
Post by towr on Dec 11^th, 2005, 8:51am

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 ;D

Do only integer angles count? Or could you get 359.132435 degrees?

Title: Re: One more spin?
Post by JocK on Dec 11^th, 2005, 9:17am

on 12/11/05 at 08:51:07, towr wrote:

Do only integer angles count? Or could you get 359.132435 degrees?

The angles are real numbers with 0 =< angle < 360.

Title: Re: One more spin?
Post by SWF on Dec 12^th, 2005, 5:18pm

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 y_N is exceeded. It turns out that y_N is the solution of
g(y_N)^N = [integral] g(x+y_N)^N dx where limits of integration are from 0 to 1-y_N.

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.

Title: Re: One more spin?
Post by JocK on Dec 13^th, 2005, 2:40pm

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?

Title: Re: One more spin?
Post by rmsgrey on Dec 14^th, 2005, 4:41am

on 12/13/05 at 14:40:31, JocK wrote:

I'd guess at [hide]1/N[/hide] but that's just a wild stab in the dark...

Title: Re: One more spin?
Post by SWF on Dec 14^th, 2005, 5:27pm

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.

Title: Re: One more spin?
Post by srn347 on Sep 1^st, 2007, 3:09pm

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.

Title: Re: One more spin?
Post by Hippo on Sep 8^th, 2007, 9:48am

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 t_i.
When his score is less than t_i he continues rolling.
Then after k rolls probability his score is less than x is for x<t_i equal x^k/k!.
(integrating, using induction).
For 1>=x>t_i the probability score is less than x after at least k rolls is
t_i^k/k!+(x-t_i)\sum__j=0^k-1t_i^j/j!.
Letting k go to infinity we get the probability (x-t_i)e^t_i.

Now the strategy for choosing t_i. Number players in decreasing order n,n-1,...1,0.
Let s_i denotes best score obtained so far after i-th player ends.
Clearly t_i>=s_i+1 otherwise i-th player cannot win.
t₀=s₁ and probability player 0 wins is (1-s₁)e^s₁.

Now think about choosing t₁.
Probaility player 1 wins is e^t1int_t1¹(1-(1-x)e^x)dx that equals (1-t₁-e)e^t1+(2-t₁)e^2t₁, now let
e^x1=(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 e^t2int_t2¹(1-(1-x)e^x)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 t_i ... suppose t_i>x_i-1:
Probability player i wins is
e^tiint_ti¹(1-(1-x)e^x)ⁱdx
Maximizing the result we get x_i and set t_i=min(s_i+1,x_i).
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 e^x by Taylor polynom, but the effort to describe the result in a compact form failed.