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At my local casino there is a one card poker game with the rules as follows:

The game is essentially WAR you get a card and the dealer gets a card. To beat the dealer you need a higher value card (in this case Ace is higher than King). There is a side bet which I believe gives the player an advantage.

Primary Bet:

Lose: -100% bet Tie: -50% Bet Win: +100% Bet

Tie Side Bet: 10x side bet

Now I ran a simulation with 5,000,000 hands played and learned this:

Player WINS: 42.75% Player TIES: 14.50% Player LOSES: 42.75%

Now for the betting:

PRIMARY BET: \$5 SIDE BET: \$5

If I calculate the expected value this is what I get:

(The win portion is 0 because you double your primary but lose your side bet breaking even)

E[V] = (\$0 * 0.4275)+(\$47.5 * 0.145)+(-\$10 * 0.4275) = $3.0875

Based on the calculation, this is a winning game? Right?

If not what calculations would prove otherwise?

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    $\begingroup$ I don't understand your simulation because the results look wrong. In a standard deck of 52 cards, with four of each rank, the chance of a tie when dealing two cards from a full randomized deck is only $3/51=5.88\%.$ Have you perhaps not given the full details of the game? $\endgroup$
    – whuber
    Commented Jul 7, 2023 at 19:44
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    $\begingroup$ The chance of a tie is 3/51 without replacement, or 4/52 with replacement. How do you find a roughly 1 in 7 chance of matching the dealer's card when there are 13 different card values? $\endgroup$ Commented Jul 7, 2023 at 19:44
  • $\begingroup$ My simulation is wrong because I was using blackjack deck values so all the face cards were the same value. Thank you for catching that! $\endgroup$
    – altheconda
    Commented Jul 7, 2023 at 19:51
  • $\begingroup$ The 47.5 profit should be a 40 profit. $\endgroup$ Commented Jul 7, 2023 at 20:00
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    $\begingroup$ Is the deck being reshuffled after a certain moves, or is there some way possible to count the cards? Can the side bet be any value or does it have to match your main bet? Are you allowed to place the side bet without playing the other hands? How many decks are being used? $\endgroup$ Commented Jul 7, 2023 at 20:02

2 Answers 2

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To accompany the probabilities described by Sextus, suppose the probability of a tie is $p$. Outside of ties, both players have the same probability $(1-p)/2$ of winning a hand.

For the main bet, winning and losing have equivalent loss/gain amounts and have equal probability, so they cancel out, leaving the expected value as $-p$ times the bet.

For the side bet, we have expected value $10p - (1-p) = 11p - 1$ times the side bet. The expected gain from this side bet alone is positive for $p > 1/11$.

Put together, if our side bet is $s$ times the main bet, we have expected gain $(11s-1)p - s = (11p-1)s - p$. As with the side bet, this can only be positive if $p > \frac{1}{11}$, in which case we require $s > \frac{p}{11p-1} = \frac{1}{11} \frac{p}{p-\frac{1}{11}}$.

In the given game, $p = \frac{3}{51} < \frac{1}{11}$, so making this a winning game is impossible. As expected, taking the side bet would only increase the house edge to be larger than $p$.

It is possible that card-counting could make the expected gain positive for certain rounds: changes in card composition only affect the outcome via the value of $p$, so once card-counting indicated that $p > \frac{1}{11}$, a large enough side bet could make the hand worth playing. If the side bet has to be the same size as the main bet, i.e. $s = 1$, then the expected gain is $10p - 1$, and we'd require $p > \frac{1}{10}$.

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To do this without simulations use the following:

Let there be $n$ different types of cards and the number of cards of each type be $k_1, k_2, \dots, k_n$, then the probability that a draw of two cards results in two of the same type is

$$P(draw) = \sum_{i=1}^n \frac{k_i}{\sum_{i=1}^n k_i} \cdot \frac{k_i-1}{\left(\sum_{i=1}^n k_i\right)-1}$$

In the case of 4 two's, 4 three's, 4 fours, 4 fives, 4 sixes, 4 sevens, 4 eights, 4 nines, 4 tens, 4 knaves, 4 queens, 4 kings, 4 aces, you get

$$P(draw) = \frac{ \sum_{i=1}^{13} 4\cdot 3}{ 52\cdot 51} = \frac{3}{51}$$

In the case of 4 two's, 4 three's, 4 fours, 4 fives, 4 sixes, 4 sevens, 4 eights, 4 nines, 16 tens plus face cards, 4 aces

$$P(draw) = \frac{ 16\cdot 15 + \sum_{i=1}^{9} 4\cdot 3}{ 52 \cdot 51} = \frac{29}{221} \approx 0.1312 $$

And when you play with 6 times as many cards

$$P(draw) = \frac{ 96\cdot 95 + \sum_{i=1}^{9} 24\cdot 23}{ 312 \cdot 311} = \frac{587}{4043} \approx 0.1452 $$

Card counting strategy

In the comments you mentioned that the game works with 6 decks that are reshuffled after approximately 5 decks are played. In that case you can beat this game when you are allowed to increase the bet.

But, you have to wait for a long time in order to get a situation where the probability of a draw is above 10%, the level that is neccesary for the game to be in favour for the player, and you would have to wage an incredibly large amount of money beyond practical limits.

Here are the result from 1000 simulated games, in none of the cases did the draw probability go beyond 10%

example of draw probability during 1000 simulated games

### simulating 6 decks with 13 type of cards each 4 repetitions
### out of these 312 cards 
### we make 130 draws of pairs (= 5 decks) each time computing the probability of a draw
simulate = function() {
   cards = rep(4*6,13) 
   pdraw = rep(NA,130) ### variable to store the evolution of draw probabilities 
   for (i in 1:130) {
      sel = which(cards >0)
      sel2 = which(cards >1)
      pdraw[i] = sum(cards[sel2]*(cards[sel2]-1))/sum(cards)/(sum(cards)-1)
      draw = sample(1:13,1,prob = cards)
      cards[draw] = cards[draw] -1 
      draw = sample(1:13,1,prob = cards)
      cards[draw] = cards[draw] -1 
   }
   return(pdraw)
}

set.seed(1)

plot(simulate(), type = "l", ylim = c(0.055,0.10), 
     xlab = "turn", 
     ylab = "probability for draw", 
     col = rgb(0,0,0,0.1), 
     main = "simulation of evolution of draw probabilities in 1000 games")

for (i in 1:999) {
   lines(1:130,simulate(), col = rgb(0,0,0,0.1))
}

lines(c(-10,150),c(0.1,0.1), lty=2, col = rgb(0,0,0,0.5))
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  • $\begingroup$ This is correct. I was getting 14.5% because all the face cards were the same value. The game however is played where each consecutive face card is valued higher than the previous, so my simulation was completely wrong. The EV wouldn't be positive and in fact would be about -$1. $\endgroup$
    – altheconda
    Commented Jul 7, 2023 at 20:31
  • $\begingroup$ I have heard of a betting strategy that in theory could work, but I'd need to take into account the "Risk of Ruin" or Ruin Theory, since the betting strategy increases every time you tie or lose and in theory could increase until I have no money left. I'm trying to find a way to beat this game, but It may not be possible long term. $\endgroup$
    – altheconda
    Commented Jul 7, 2023 at 20:31
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    $\begingroup$ Are you allowed to change the betting value during the game? In that case you can play with pennies and bet a thousand dollars at the moment that the stakes are better. Although you will have to wait for a long time to get the possibility that the advantage is high. $\endgroup$ Commented Jul 7, 2023 at 20:34
  • $\begingroup$ So the minimum bet is $5 and I've thought about counting the cards, but it's pointless since the dealer has an equal chance of getting the high cards as you do, so there is no statistical advantage to counting the cards and betting higher when the count is high. $\endgroup$
    – altheconda
    Commented Jul 7, 2023 at 20:36
  • $\begingroup$ I believe the only strategy is increasing your bets when you lose then reducing it down to the minimum when you win, but the Ruin Theory eventually takes hold and bankrupts you long term. In the short term it's probably a viable strategy. $\endgroup$
    – altheconda
    Commented Jul 7, 2023 at 20:37

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