Analysis of a simple Hi-Lo game
August 16th, 2011 at 2:18:12 AM
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I've been trying to analyze a very simple Hi-Lo game (that I couldn't find anywhere on the Wizard of Odds site) and it's turned out to be harder than I thought.
The game is played with a standard 52 card deck. One card is dealt, and the player has to guess whether the next card will be higher or lower. If the player guesses correctly, they get to guess again for the next card. The player loses if they guess wrong. The player also loses if the next card to come out is the same value (there is no option for ties, just high or low).
Aces are high.
After 2 correct guesses, the player has the option to "cash out" and accept a payout instead of continuing to guess.
The payout table:
2 correct guesses: push
3 correct guesses: win 1 unit
4 correct guesses: win 2 units
5 correct guesses: win 3 units
6 correct guesses: win 5 units
After 6 correct guesses, the player is automatically paid 5 units and the round is over.
So for example, I bet 1 unit and I'm dealt 2c. I guess high and get dealt 7d. I guess high again and get dealt Js. Now I have the option of "cashing out" (and receiving my original bet back) or guessing high or low. If I guess correctly, I can then cash out and win 1 unit. If I guess wrong, I lose.
Questions I have about this game and its analysis:
What is the house edge?
What is the likelihood of losing?
What is the optimal strategy (in terms of choosing when to cash out or when to continue guessing)? I imagine you would only want to continue if you had a very good chance of winning. Intuitively, I would think a good cash out strategy would be to choose to cash out rather than continue (if I have the option) on a 7, 8 or 9.
How would these change if the game were played with infinite decks instead of 1?
When trying to build the sample space, do you exclude events that are impossible given that you're using a certain strategy? For example, the event (J, K, 2) would only happen if you get dealt a Jack and then guess high, which is a losing strategy.
The game is played with a standard 52 card deck. One card is dealt, and the player has to guess whether the next card will be higher or lower. If the player guesses correctly, they get to guess again for the next card. The player loses if they guess wrong. The player also loses if the next card to come out is the same value (there is no option for ties, just high or low).
Aces are high.
After 2 correct guesses, the player has the option to "cash out" and accept a payout instead of continuing to guess.
The payout table:
2 correct guesses: push
3 correct guesses: win 1 unit
4 correct guesses: win 2 units
5 correct guesses: win 3 units
6 correct guesses: win 5 units
After 6 correct guesses, the player is automatically paid 5 units and the round is over.
So for example, I bet 1 unit and I'm dealt 2c. I guess high and get dealt 7d. I guess high again and get dealt Js. Now I have the option of "cashing out" (and receiving my original bet back) or guessing high or low. If I guess correctly, I can then cash out and win 1 unit. If I guess wrong, I lose.
Questions I have about this game and its analysis:
What is the house edge?
What is the likelihood of losing?
What is the optimal strategy (in terms of choosing when to cash out or when to continue guessing)? I imagine you would only want to continue if you had a very good chance of winning. Intuitively, I would think a good cash out strategy would be to choose to cash out rather than continue (if I have the option) on a 7, 8 or 9.
How would these change if the game were played with infinite decks instead of 1?
When trying to build the sample space, do you exclude events that are impossible given that you're using a certain strategy? For example, the event (J, K, 2) would only happen if you get dealt a Jack and then guess high, which is a losing strategy.
August 23rd, 2011 at 1:53:19 PM
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This could very easily be figured out with a brute force computer program analyzing every possibility, but I have no programming skills. I figured it out as if with every guess there were 51 cards left in the deck, 4 of every value except the one showing, of which there are only 3 (that is, any card no longer in play is shuffled back in the deck). I understand this is not how the game is played, but it made the math much easier. Here is the optimal strategy under these conditions, starting with the 3rd guess (that is, you've already guessed 2 correctly):
3rd guess -- guess on any card
4th guess - stay on 7-9
5th guess - stay on 6-10
6th guess - stay on 7-9
If you should guess and the card is 9-A, guess lower; if it's 2-7, guess higher. With an 8 you could guess either way with this set-up, although in the scenario you present, keeping track of previous cards would make one choice better than the other. It is possible that the strategy could change if 3 or 4 of one kind had already been played.
The average result will be a loss of 0.158368382584655 units, or a house edge of ~15.8%. Infinite decks would raise the house edge, as that would increase the chance of ties, which are player losses. My guess is the true house edge of the game you present is a bit lower, as you could make slightly higher percentage decisions based on the cards already played.
3rd guess -- guess on any card
4th guess - stay on 7-9
5th guess - stay on 6-10
6th guess - stay on 7-9
If you should guess and the card is 9-A, guess lower; if it's 2-7, guess higher. With an 8 you could guess either way with this set-up, although in the scenario you present, keeping track of previous cards would make one choice better than the other. It is possible that the strategy could change if 3 or 4 of one kind had already been played.
The average result will be a loss of 0.158368382584655 units, or a house edge of ~15.8%. Infinite decks would raise the house edge, as that would increase the chance of ties, which are player losses. My guess is the true house edge of the game you present is a bit lower, as you could make slightly higher percentage decisions based on the cards already played.
