By poker_boy | 1 Comment
Last updated: Thursday, July 14, 2011 | 84 Views Tags: ace king queen jack, bet, game, images, king queen, poker dice, probability question, three times, wager
We’ve had a reader question come in on the subject of poker. Let’s take a look:
Question: A particular game is played with 5 poker dice.
Each die displays an ace,king,queen,jack ,ten or nine.
Players may bet on two of the six images displayed.
When the dice are thrown and the bet-on images turn up,the player receives three times the amount wagered.
In all other cases,the amount of the wager is forfeited.
Is this game advantageous for the player?
……………………………………………………
The book claims that there is a house percentage of 12.3%!
Can anyone show me how the author arrived at that figure,please?
Any help appreciated,thank you.
Hello Tom,it was the house percentage wanted.
The author means,I think,that the player will be paid out if there are 1 of each selected and then some more of them.
I get E=3×0.5391 -1×0.46089 which I think gives the player 15.6%.
Hello M3,the question is just as typed but I agree it is most odd.
I think the second line is wrong,
1/6 *(4/6)^4 *2*5
Each die displays an ace,king,queen,jack ,ten or nine.
Players may bet on two of the six images displayed.
When the dice are thrown and the bet-on images turn up,the player receives three times the amount wagered.
In all other cases,the amount of the wager is forfeited.
Is this game advantageous for the player?
……………………………………………………
The book claims that there is a house percentage of 12.3%!
Can anyone show me how the author arrived at that figure,please?
Any help appreciated,thank you.
Hello Tom,it was the house percentage wanted.
The author means,I think,that the player will be paid out if there are 1 of each selected and then some more of them.
I get E=3×0.5391 -1×0.46089 which I think gives the player 15.6%.
Hello M3,the question is just as typed but I agree it is most odd.
I think the second line is wrong,
1/6 *(4/6)^4 *2*5
Answer: revised ans
—————-
P[win] = 1 - P[lose]
= 1 - P[neither of the chosen 2 turns up] - P[ only one of the chosen 2 turns up]
= 1 - (4/6)^5 - 2( sum C(5,k)*(1/6)^k *(4/6)^(5-k) for k = 1 to 5)
= 425/1296
http://www.wolframalpha.com/input/?i=+1+-+%284%2F6%29%5E5+-+2%28+sum+C%285%2Ck%29*%281%2F6%29%5Ek+*%284%2F6%29%5E%285-k%29+for+k+%3D+1+to+5%29
—————-
P[win] = 1 - P[lose]
= 1 - P[neither of the chosen 2 turns up] - P[ only one of the chosen 2 turns up]
= 1 - (4/6)^5 - 2( sum C(5,k)*(1/6)^k *(4/6)^(5-k) for k = 1 to 5)
= 425/1296
http://www.wolframalpha.com/input/?i=+1+-+%284%2F6%29%5E5+-+2%28+sum+C%285%2Ck%29*%281%2F6%29%5Ek+*%284%2F6%29%5E%285-k%29+for+k+%3D+1+to+5%29
taking that "three times the amount wagered" is inclusive of the wager,
house edge % = 100(1296 - 3*425)/1296 = 1.62% <———–
Comments
1 comment
Mike Smith
July 14, 2011
This one was difficult to work with so I’m not sure if it’s the right answer. But I’ll try.
A die has a 1/6 probability of rolling on any particular face. So for argument’s sake, let’s see what the probability is of landing TWO dice on the bet that we want (say a king and a queen). The order in which the king and queen appear is irrelevant and the probability of landing on a king or a queen is the same for each die.
So to get a king and a queen it is a 1/36 probability (because 1/6 probability for one die and 1/6 probability for the other; probabilities of sequential events are determined by multiplicity).
Now that we know the probability of landing two dice on our bet is 1/36, we have to contend with multiple dice (namely, 5).
Our chance of landing any two-event sequence with 5 dice is much higher than working with just 2 dice because this time probability is additive. The more dice we use to get any two-event combination we so desire, the higher our probability of winning the bet.
There are 5 dice, so the chance that any two dice appear as we expect them to appear is proportional to the number of possible pairs. In otherwords, how many different pairs are there in a set of 5 dice that can produce the effect we want? Let’s count them.
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ]
The brackets above represent the dice. I numbered them for convenience. The following is a list of all possible combinations in which at least two dice land on the bet we want.
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
There are a total of 10 possible combinations. So that is 1/36 (the probability that a pair combo lands on the bet) multiplied by 10 (the number of possible combinations).
So our answer is 10/36 = 27.78 %
This is markedly different from the house percentage you gave, so I don’t know if I left something out or if the book is wrong (or perhaps I’m wrong). You should seek out a different answer and get a second opinion before settling on just one answer.