Comments on: “Higher or Lower” – Part II

By: Micky

Micky — Sun, 25 Oct 2009 11:47:21 +0000

hi, sorry for the late reply! of course you can, I’d be honoured

By: Polprav

Polprav — Thu, 22 Oct 2009 20:36:25 +0000

Hello from Russia!
Can I quote a post in your blog with the link to you?

By: Scott cleburne

Scott cleburne — Wed, 14 Oct 2009 21:18:01 +0000

Yes the card could be thought of as being put back into the deck and reshuffled. ‘replacement’ I think it’s referred to as?
I said infinite deck as the game I played was electronic. There was no mention of number of decks etc so I have just assumed the program picks any random card independently of the proceeding card. I have not played it long enough to notice a repeated card. I hope you have fun with it!

By: Micky

Micky — Wed, 14 Oct 2009 10:47:45 +0000

Very interesting problem that Scott, thanks. By an “infinite deck” I presume you mean a normal deck but, effectively, that between each revealing of a card the previous card is returned to the deck and the deck shuffled? I’ll have a go at this at some point! Mick

By: Scott cleburne

Scott cleburne — Wed, 14 Oct 2009 00:27:59 +0000

Hi. This article made a good read.
  I came across it when looking for a similar game.
  Are you interested in another challenge?
  In this game you pay 8 tokens to play.
  You are presented with a card from an infinite deck.
  Higher or lower?
  If you get it right you get +6 tokens, a second card presented to you
  and the chance to “let it ride, higher or lower?”
  It then repeats like this until you either lose or take your winnings.
  you then must pay 8 tokens to play again.
  The probability of winning each round is easily calculated.
  Whether or not it’s the best decision to take your winnings under each
  circumstance is what I have been trying to calculate.
  My instinct has told me to take the winnings whenever I am risking 18
  tokens and receive something close to 7 and to only play on when risking
  24 tokens if a 2 or queen is returned. It has been my undoing to play on
  risking 30 tokens for a queen before.
  In this game if the same card number is drawn in succession you do not
  lose, you are asked to pick again. (effectively you can assume that the
  same card never appears twice in succession in the maths).
  If you would like to take on this challenge, I would be very interested
  in your findings.
  The game just doesn’t feel like the house has the edge…
  -Scott

By: Malenfant

Malenfant — Thu, 01 Oct 2009 17:52:17 +0000

Hi Micky,
Well, i am interested in Maths and Physics but i am not a mathematician, so i recommend you better refer to the Wiki Page for elaboration. I became kinda obsessed with the DA since reading about it in Stephen Baxter’s Book ‘Manifold : Time’.
Maybe you post your take on it on your intereting site when you have studied it properly, as it seems the opinions about it’s validity differ greatly.

By: Micky

Micky — Thu, 01 Oct 2009 16:19:10 +0000

Hi Malenfant,
I’ll have to study the Doomsday Argument properly as it’s not something I’ve been exposed to before. On first consideration though, the only similarity I can identify between it and this post is that they are both calculated predictions of probabilities. I’ll study further when I have a moment, but in the meantime perhaps you’d elaborate on your own thoughts! Thanks for the heads up. Micky

By: Malenfant

Malenfant — Wed, 30 Sep 2009 19:34:36 +0000

Hmm, is this a variation of the Doomsday Argument ?
“magine two urns, one containing 10 balls numbered 1 thru 10; and a second urn containing a million balls, numbered 1 thru 1,000,000. If you pull a ball at random from one of the urns and you obtain a “3″, the likelihood, according to Bayesian prior/posterior probabilities, is much higher that you have pulled from the 10-ball urn than the 1,000,000-ball urn.
http://en.wikipedia.org/wiki/Doomsday_argument

By: John1559

John1559 — Mon, 14 Sep 2009 08:32:43 +0000

Very nice site!