Math Behind Crown and Anchor Game
(Mods, If this post is better suited for the commune please feel free to move, wasn't sure where to post this)
So I just came from my buddie's Jack and Jill (or Stag and Doe or whatever) and they had a pretty common game there called Crown and Anchor. For those unfamiliar: http://en.wikipedia.org/wiki/Crown_and_Anchor
We played with a spinning wheel though, not dice. The wheel looks like this:
http://mccullochscostume.com/blog/wp...4/IMG_0168.jpg
http://www.google.ca/url?sa=i&rct=j&...71451665530486
The game is simple, people can play a minimum of $1 to a maximum of $5 per turn (at least that's what we did) You then spin the wheel and payouts are determined by how many symbols are on the winning spin:
1 symbol pays 1:1
2 symbols pay 2:1
3 symbols pay 3:1
First, I was just trying to derive the house edge for this game, but since I was drinking a bit (yes I was trying to do this AT the party) I came up with this:
there are 12 winning spins with 3 identical symbols
there are 12 winning spins with 2 identical symbols and 1 differing symbol
there are 4 winning spins that have no identical symbols
there are 28 total options for a winning spin
Then I imagined to come up with the house edge for the game, I pretended what would happen if I were to bet $1 on every symbol (yes I realize I said the house max was 5$, but just go with it)
based on this, I came up with the expectation of that play to be:
(12/28)(4)+(12/28)(5)+(4/28)(6)= $4.70
so whenever we bet $1 on each symbol, our expectation is $4.70. but this is not the house edge that I found when I googled it (house edge = 7.5% apparently)
This leads me to believe that there is a way to bet this game optimally, but I do not know how to find that out.
Please help me figure this out, and let me know if any of my math/assumptions are wrong.
Thanks!
Caddie444
