Solved – Casino texas poker game: playing against dealer. Strategy

probability

I have some questions regarding this casino Texas Poker game. You can wikipedia the rules but basically the game goes as following:

You make a blind bet first .
The dealer deals three cards, and deals two cards to you and the dealer himself. Dealer does not look at his hand, but you can look at your hand.
If you decide to play, the you must make another bet worth twice your blind bet, oterwise you lose your blind bet and game ends.
If you decide to play, then the dealer deals anoter two cards before opening your hand.
If dealer beats you then you lose all your bet, if you beat dealer then you get paid the sum of your blind and subsequent bet.

However, there are two exceptions regarding the payout for (5):

If dealer has a weaker hand than a pair of 4, then this is a "no gae" and he pays you only the blind bet regardless of if you win or not.
If you have straight, full house or better then you are usually paid more than 1:1 of your bets.

My question is, call me naive, but I feel this game seriously favours the player? Thinking in a simple way — lets say the blind bet is 5 usd. As if the player plays every hand, then he and dealer has equal chance of winning 15 usd. Now the player even has an option to pay only a third of the total bet(i.e. pass at (3)) to stop the game — and I think it is reasonable to assume the player makes correct decision 50% of the time at least. (i.e. namely that if the player decides to pass 10 games then he would lose more than 5 if he did not stop) it just makes me think I would much rather be the player than the dealer. The only thing I am not sure about this analysis is the how the 'no game" rule affects the probability.Would be grateful if someone could point out

Best Answer

The Wikipedia entry for the game you are referring to has the answer to your question: the house edge (i.e. the average win percentage for the casino assuming perfect play from the player) is between 2.0 and 2.5%.

Note that perfect play does not have the exact same meaning you provide. Perfect play means calling (paying the extra \$10 in your example) whenever he has more than 50% chance of winning given the information available.

The casino's advantage comes from the risking \$15 only after he has seen all the cards, unlike the player that must decide whether to risk \$15 knowing only 5 cards.

Related Solutions

Solved – Count Outcomes in Three Card Poker

Here's a Python program that I believe is correct. It takes a little under 7 minutes on my machine to run through every pair of hands.

from itertools import combinations
from collections import Counter

JACK, QUEEN, KING, ACE = 11, 12, 13, 14

N_HANDS = 22100       # 52 choose 3

deck = frozenset((rank, suit)
  for rank in range(2, ACE + 1)
  for suit in range(4))
assert len(deck) == 52

evaluate_hand_cache = {}
def evaluate_hand(hand):
    if hand not in evaluate_hand_cache:
        evaluate_hand_cache[hand] = evaluate_hand_(hand)
    return evaluate_hand_cache[hand]

def evaluate_hand_(hand):
    ranks = tuple(sorted((rank for (rank, _) in hand), reverse = True))
    suits = [suit for (_, suit) in hand]
    if ranks[0] == ranks[1] == ranks[2]:
      # Three of a kind
        return (5, ranks[0])
    straight = (ranks[0] == ranks[1] + 1 == ranks[2] + 2
        or ranks == (ACE, 3, 2))
    flush = suits[0] == suits[1] == suits[2]
    if straight and flush:
        return (6, ranks[0])
    if straight:
        return (4, ranks[0])
    if flush:
        return (3,) + ranks
    if ranks[0] == ranks[1]:
      # Pair
        return (2, ranks[0], ranks[2])
    if ranks[1] == ranks[2]:
      # Also a pair
        return (2, ranks[2], ranks[0])
    return (1,) + ranks

results = Counter()
divisor = N_HANDS // 100

for i, dealer_hand in enumerate(combinations(deck, 3)):
    if i % divisor == 0:
        print "{}% complete".format(i // divisor)
    dealer_hand = frozenset(dealer_hand)
    dealer_e = evaluate_hand(dealer_hand)
    for player_hand in combinations(deck - dealer_hand, 3):
        player_e = evaluate_hand(player_hand)
        outcome = (
          'Dealer does not qualify'
              if dealer_e < (2, QUEEN) else
          'Player loses'
              if player_e < dealer_e else
          'Player wins'
              if dealer_e < player_e else
          'Tie')
        results[(player_e[0], outcome)] += 1

print "Player hand class,Outcome,Count"
for (hand, outcome), n in sorted(results.items()):
    print ','.join([str(hand), outcome, str(n)])

The output is:

Player hand class,Outcome,Count
1,Dealer does not qualify,264851952
1,Player loses,38038608
2,Dealer does not qualify,60280560
2,Player loses,8453520
2,Player wins,242784
2,Tie,2592
3,Dealer does not qualify,17630040
3,Player loses,1260596
3,Player wins,1298780
3,Tie,3288
4,Dealer does not qualify,11581584
4,Player loses,267804
4,Player wins,1392204
4,Tie,23688
5,Dealer does not qualify,836520
5,Player loses,3312
5,Player wins,118216
6,Dealer does not qualify,771024
6,Player loses,956
6,Player wins,112204
6,Tie,168

Solved – In simple Russian Roulette how many times should you shoot before handing it off to the other person

What are the optimal strategies for player 1 and player 2? How many times should Player 1 shoot before passing it for an optimal chance at winning?

Once the barrel is spun, the position of the bullet is fixed, which means that there is a $\frac{1}{6}$ probability to be shot at turn number $i$, $i \in [|1,6|]$. As a result, the optimal strategy for player 1 and player 2 is to shoot only once on their turn. Following this strategy, the probability that player 1 wins is the same as the probability that player 2 wins, i.e. $0.5$, regardless of who starts.

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