By poker_boy | 3 Comments
Last updated: Monday, June 21, 2010 | 79 Views Tags: bets, poker dice, stake
A poker question that should definitely interest you all this week
Question: 5 dice displaying A,K,Q,J,10,9
Punter bets on 2 of the 6 selections
all 5 dice thrown
if at least 1 of each of the punter’s selections are showing they are paid 2/1 and receive 3 times their stake .
Any other outcome and stakes are forfeited.
What is the house percentage?
The book answer is 12.3% to the house!
I think it is about 1.5% to the house.
Can someone clarify please?
Thank you.
Punter bets on 2 of the 6 selections
all 5 dice thrown
if at least 1 of each of the punter’s selections are showing they are paid 2/1 and receive 3 times their stake .
Any other outcome and stakes are forfeited.
What is the house percentage?
The book answer is 12.3% to the house!
I think it is about 1.5% to the house.
Can someone clarify please?
Thank you.
Answer: So if the punter bets on 2 selections, he’ll loose if non of the 5 dice shows them.
For each dice the probability of not showing the punters predictions is 4/6 = 2/3.
So the probability that none is shown is (2/3)^5 = 32/243
Let’s call A and B Punter’s bets.
The probability that only A shows up and not B is
P( B never shows up) - P(A and B never show up) = (5/6)^5 - (4/6)^5 =
(3125 - 1024)/7776 = 2101/7776
So the probability of losing is 2(2101/7776) + 32/243 = (2101 + 512)/ 3888 = 2613 / 3888.
The probability of winning is 1275/3888 and the expected return is 1275/1296 = 425/432.
So the house takes 7 / 432, which proves the book messed up and you are right.
Clearly the author never gambles. No player would risk a bet against a house taking 12.3%
Comments
3 comments
Mike Smith
June 21, 2010
No
June 21, 2010
Sorry, Ken, j’ai toujours été nulle en maths mais j’étoile ta question.
Cheers !