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Problem 121

Posted: Sat Jul 05, 2008 11:46 am
by LarryC
Hi

Problem 121 looks easy but I am unable to verify that the probability of winning is 11/120 as stated in the question. I don't make it that at all and am pretty sure of my answer so it would be useful to see if my earlier answers agree with what is expected.

Can anyone verify:
Rounds | Probability Of A Win
1 | [frac]1,2[/frac]
2 | [frac]1,6[/frac]
3 | [frac]5,18[/frac]
4 | [frac]1,9[/frac]

That would be much appreciated...
Thanks,
Lster

Re: Problem 121

Posted: Sat Jul 05, 2008 12:02 pm
by Tommy137
5/18 for 3 rounds is not correct.

Re: Problem 121

Posted: Sat Jul 05, 2008 12:11 pm
by LarryC
Thanks! Are the first two rounds correct?

Re: Problem 121

Posted: Sat Jul 05, 2008 12:14 pm
by stijn263
Yes they are.

Remember that a player can win a game of three rounds if he takes:
- 3 blue disks (bbb)
- 2 blue disks and 1 red disk (bbr, brb, rbb)

Re: Problem 121

Posted: Sat Jul 05, 2008 12:22 pm
by LarryC
After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
I've just realized I may well be interpreting this in a way that is not intended.

Does the above statement imply that the second disk taking is also "noted" or not? It could be interpreted either way and my results assume that the disk is only removed.

Re: Problem 121

Posted: Sat Jul 05, 2008 12:29 pm
by Tommy137
Example:

bag contains: B , R
player takes B

now another red disc is added

bag contains: B , R, R
player takes: R

another red disc is added

bag contains: B, R, R, R
player takes: B

=> The player took 2 blue discs and 1 red, so he has won. (probability 1/2*2/3*1/4 = 1/12)

Re: Problem 121

Posted: Sat Jul 05, 2008 12:39 pm
by LarryC
Got it! Thanks for your help guys.
After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
This sentence is very ambiguous. It's primary meaning seems to imply that a disc is randomly removed after the turn and before the next. An example, similar to Tommy137's one, would go a long way! Or perhaps it should be rephrased.

Re: Problem 121

Posted: Sat Jul 05, 2008 1:05 pm
by Tommy137
lster wrote:Got it! Thanks for your help guys.
Gratz :D

lster wrote:
After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
This sentence is very ambiguous. It's primary meaning seems to imply that a disc is randomly removed after the turn and before the next. An example, similar to Tommy137's one, would go a long way! Or perhaps it should be rephrased.

This interpretation did never occur to me, but people may understand it that way.

Re: Problem 121

Posted: Sat Nov 22, 2008 11:37 am
by ImRe
After reading this topic, I still don't fully understand this problem.
How are the payouts done here? In the four turn game, is the player getting back £1 for every round (s)he won, or for every four turn (s)he won, or is it something completely different?

Thx

Re: Problem 121

Posted: Sat Nov 22, 2008 11:46 am
by Tommy137
The payout's done after the complete game is played. If the player wins the four turn game, he gets £10 including his stake of £1.

Problem 121

Posted: Fri Mar 13, 2009 5:58 pm
by Knut.Angstrom
How much is the player paid if he wins after n turns? After 4 turns he is paid £10 so much I understand :?

Re: 121

Posted: Fri Mar 13, 2009 6:15 pm
by Tommy137
Please specify your question... and you could read the already existing topic (viewtopic.php?f=50&t=997) for Problem 121 and see if it's helping you.

Re: Problem 121

Posted: Sat Mar 14, 2009 2:11 am
by rayfil
How much is the player paid if he wins after n turns?
If you read the problem description very carefully, THAT is the answer you have to find.

Statisticly, it would be the largest payout by the sponsor of the game such that no overall loss (and minimal gain) would be expected in the long run. (If the payout would be £11 when the probability of winning is 11/120, £121 would be given for every £120 of revenue.)

Re: Problem 121

Posted: Thu Dec 31, 2009 2:50 pm
by estanford
This problem confuses me. How much of the prize fund does the player win as a consequence of a single win event? Does it vary according to some function or is it a constant?

Re: Problem 121

Posted: Thu Dec 31, 2009 8:56 pm
by TripleM
I'm not quite sure what you mean by 'single win event'; there is only one outcome. If the player has won after 15 rounds, they receive the whole prize.

Re: Problem 121

Posted: Fri Jan 01, 2010 10:51 am
by estanford
So am I understanding the question correctly if I read it like this?

~~~~~~~

Let there exist three piles of money:
1) The money the player has (assume infinite),
2) The money the player has given as entry fees,
3) The prize money.

Each time the player plays the game, they pay a fee of 1 pound. This money is moved from pile (1) to pile (2). Money is never transferred from pile (1) or (2) to pile (3). The size of pile (3) is an integer constant, set by the banker at the outset of the game. The player keeps playing until s/he wins. Given that the game is played for n turns, what is the smallest possible size of pile (3) such that the banker does not expect to lose money by the time the player wins?

~~~~~~

EDIT: After solving the problem, I see that the answer was 'yes'. Good times.

Re: Problem 121

Posted: Thu Apr 21, 2011 9:03 pm
by chiefsci
One quick question. I thought that I had calculated the answer with a pen and paper, but it is not being accepted. Am I correct in using combinatorics (like those in Problem 53) to calculate the numerator of the fractional form of the probability of success after 15 rounds? It works perfectly for the 4 round example, producing 11 for the 11/120, but my current answer isn't being accepted.

Re: Problem 121

Posted: Thu Apr 21, 2011 9:54 pm
by GenePeer
I did it by Paper/Pencil too but the calculations involved never required to find any binomial coefficient! The four turns example isn't enough, work on 5-7 turns to properly understand the question, for 7 turns the payout is 17. Enjoy.

Re: Problem 121

Posted: Thu Jun 09, 2011 7:32 am
by Francky
I can find the value for n=2000 in 1.3s
Expand
18947601191262360253803057378580697801329301076723684533165395191878298252411143714310186797902167711556489721419949786937810806579217329529972194678631638533442654482854398378334853149447673949069531831098625333410171883090632250214722128265332631618578193343037722320580778108792970960943917194802493735595124957823033085786518767053117759781727787481925319439991722568517552388959994651555002449463440443849286174361989270493279066133903871788110138556190411249273065724388009677095642044125626221408800076541520852501388731276343674563430133716553873119066915295087714681969427128364097287556503076753749261033901529628795705162638141601799421290765837486027446290819453816331147793260368555879505535195928639203774274367187886853060539294593371989299700558369155953188708103477719780340576440324888510899463266418391743864472803590157789734983924237329600328247299276816113120245038625988099165278269156944792572137261859304691144638028815873856145830794633779394516906153367302125054909056148357485705004747035359744295368111536404875360668494502460286401735834572359541235454426929525784334794585952373255214626158686354662359249543908790869827909963491808170242869435288820241447167109796298219300843432388827065285977845851960894674569017814246656143578649561309245724873016945463392800801043682030061270942226421597493557710046561744965436304327629486660152323148086882898259067769381711237127615966416208577517311635360353531519822934864594563448078129247043411597309885399882650589989621047431828838940981143748700014849004925829217807136929996400621310757067922608947488625826982185462782132866461977174839646598298568965269893792345617021149309841651847351814710878729499075558897197555924354883494223882772441154788098012216803163365150711443910760231605799867454150981804484389757065312639915716784035731430394938272616879342639822774294705177133888934991617227067614794193397828177325929851196033761663190136615668419549687325222773880256735019091199453667300030952332642141694819098158312082369911324910571469548115779839286716597787115318903991906781273606634785228815460532394616323919252904564713722838285424589389533414693842447046538894276472402672550529405533875

Re: Problem 121

Posted: Sat Apr 14, 2012 4:22 am
by enderw88
Seems like the 4 turn example considers drawing 2 Blues and 2 Reds as a win. Or am I reading it incorrectly?