# Problem with card game called “El Monte”

I'am analyzing this problem.

From an standard deck of 52 cards four cards are dealed:

-One card is for the house
-Three cards are for the players


The player wins if his card is greater than the house card.

So for example if we have:

House - 5 of clubs
Player 1 - 7 of spades
Player 2 - 3 of diamonds
player 3 - 5 of hearts


Player 1 wins the house, the house wins player 2 and player 3.

The problem is to calculate the probability to win, being a player or the house.

I made a brute force aproach with a Perl script. What I have made is to generate all posible permutations 52 nPr 4, and all the favorable permutations for the players (permutations are needed for this aproach, so it can be simulated a card dealed for the house and each player)

Analyzing the problem I realized that a good way to visualize the odds are with a matrix.

Please, look at the first column of the table below as a matrix. The numbers in this matrix symbolize the players cards compared to the house card. If the house card is greater than a player card, I put a Zero. If the house card is equal in number to the player card, I put a "2", and if the house card is lower than the player card, a "1" is put.

"Be the house card x,

if x < player 1 card, x < player 2 card, x > player 3 card", then we have this:

110


"Be the house card x,

if x > player 1 card, x equals player 2 card, x > player 3 card", then we have this:

020


Now, with this approach I have got all the combinations for each case:

000     1435200
001     420992
002     121056
010     420992
011     420992
012     54912
020     121056
021     54912
022     7488
100     420992
101     420992
102     54912
110     420992
111     1435200
112     121056
120     54912
121     121056
122     7488
200     121056
201     54912
202     7488
210     54912
211     121056
212     7488
220     7488
221     7488
222     312


At this point is possible to calculate the probability of a player to win. For player 1 it is:

Since the first row simbolyze player 1, we pick the rows with first column equal to "1" and we sum their ocurrences:

   Updated: 3057600


Next we divide this for the total of ocurrences:

3057600 / 6497400 = 0.47059.


This is the same probability to win for each player (see the matrix, the permutations are repeated).

Im not 100% sure about the reasoning made in this problem, so comments are welcome.

The next problem is that when I want to analyze the problem dealing five cards, one for the house and four for four players, my computer freeze because there are too much calculations involved. I would like to compute the odds for more players.

I have tried to develop a method of more intelligent counting with no sucess. I suspect, that the numbers of permutations above can be generate with a permutation formula that can be generalized.

Thanks for your help.

• Not sure I follow. Are you just asking for the probability that a given player's card has greater face value than the house card? – lulu Nov 10 '15 at 21:22
• Yes. Just like that. – Carlitos_30 Nov 10 '15 at 21:25
• Ok...but by symmetry that is just $\frac 12(1-p_{tie})$ where $p_{tie}$ is the probability of a tie. – lulu Nov 10 '15 at 21:27
• Thank you. Very simple indeed. – Carlitos_30 Nov 10 '15 at 23:17

## 1 Answer

You're missing one of the $54912$ terms. It should be $3057600/6497400 = 0.47059$ (approximately). That is equal to $8/17$.

That can be obtained by ignoring the other two players, and just focusing on the house and one player. There are only three possibilities: the player wins, the house wins, and they tie. The probability that they tie is the probability that they end up with the same rank. Once the house draws a card, the probability that they end up with the same rank is $3/51 = 1/17$.

Once that possibility is eliminated, the probability that they don't tie is $16/17$. By symmetry, the player and house both win with the same probability, which is $(1/2)(16/17) = 8/17$.

• Thank you very much. Very clear explanation. – Carlitos_30 Nov 10 '15 at 21:34