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Subject: Minigame probabilities help rss

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Allison Macrae
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I'm working on a simple minigame to be a small part of a larger game that I'm designing, and I want to be able to tweak aspects of it to modify the probabilities of winning, but not being terribly good with mathematics I've been brute forcing the probabilities, which has been time consuming and possibly error filled. I would be grateful if anyone with a better head for these things could help me.

This is the game:

One player is designated the attacker, and another the defender. Each player is given two 2-value cards and two 3-value cards, and secretly chooses two of their cards to play face down. They are then flipped face up and each card of the attacker's that matches one of the defender's is discarded, and the attacker then scores the remainder in points.
So for example if the attacker places 3/3, he will score 0, 3 or 6 points, depending on if the defender played 3/3, 3/2 or 2/2.
This is repeated twice more, and by the end of the third round, if the attacker has 8 or more points he wins, otherwise the defender wins.

My current estimation of the probabilites, accounting for optimal play (if the defender plays 2/3, he can guarantee the attacker scores no more than 3 points each round, which for example makes winning impossible if the attacker scored 0 in the first round) is that the attacker has a 44% chance to win if they start with 3/3, a 27% chance if they start with 2/2 and a 24% chance if they start with 3/2.

I'm curious how this game changes with modifications such as: reducing the number of points needed to win to 7, or even 6, or guaranteeing the attacker gets at least 1 point each round.

If anyone has any suggestions or wants to offer some help with the numbers, it would be greatly appreciated. I am aware that this is not going to be a terribly interesting game in a vaccuum, but it serves its function as a simple sub-system. It doesn't need to become a perfectly fair game, I just want to increase the attacker's chances, and maybe have some bonuses that could be obtained in the rest of the game to make it easier.

Thanks for the help.
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Josh Taylor
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There's two ways to do this. You can either assume one of the players plays randomly and then work out the expected outcome of various plays, or you can use game theory and work out what the optimal strategy would be. (Since there's nothing random here, looking for a general probability of winning doesn't make sense.)

Using game theory, on round 1 the payout matrix would look like this. (Note that the defender here is trying to minimize the attacker's score and ignores his own score.)


Atk 2/2 2/3 3/3
Def

2/2 0,0 -3,3 -6,6

2/3 -2,2 0,0 -3,3

3/3 -4,4 -2,2 0,0


Given this, 2/3 would be the clear winner for defense, but this isn't a pure strategy equilibrium because if defense always played 2/3, attack could always play 3/3 and get 9 points in the end.

If I've done my math right, the best strategy (for a certain definition of best) for both players is to play 2/2 40% of the time and 3/3 60% of the time. If they do, the attacker will on average gain 2.4 points (so on average, a total of around 7 after 3 rounds) and the attacker will get 0 about half the time (52%). If either player deviates the other player would be able to take advantage of that to better their outcome.

IIRC, playing multiple rounds makes it more likely human players will converge on such an equilibrium. (Of course, my memory isn't that great.)

Edit: With a 24% chance of getting 6 points and a 24% chance of getting 4 points, the attacker has a 47% chance of getting 8 or more points after 3 rounds. (Here's a list of all the outcomes: http://anydice.com/program/1c69.)
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Andrew I
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I believe the correct way to determine optimal play is for each player to randomly use a combination of the available options to maximize their chance of winning, assuming the opponent is also playing optimally. However, we're playing to win, not to maximize/minimize points (unless there is some effect to scoring other than increasing your chance of winning the contest), so the payout matrix is dependent on the round we're in and how many points have been scored so far. In round 1, the defender will mostly play 3/3, but in round 3, they'll play 2/3 for sure if 0-4 points have been scored, play 2/2 and 3/3 with equal probability if 5 points have been scored, and play all 3 options with equal probability if 6 or 7 points have been scored.

In round 3, the payout is 1 for the player who wins, and 0 for the player who loses. In previous rounds, the payout for each player is the probability of that player winning, so you have to work backwards from round 3 to determine rounds 2 and 1.

FWIW, I'm getting a probability of the attacker winning at around 33.4%.
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Josh Taylor
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That makes sense, and I get the same strategies for round 3, but I'm getting a probability of about 40% when I do the rest of the calculations.
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Nate Wasser
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With optimal play on both sides I get the following chances for the attacker to win:

Three rounds and ...
... 8 points to win: 33.36%
... 7 points to win: 41.96%
... 6 points to win: 61.43%

Without changing the players options or results, the closest to 50% are:

One round and 3 points to win: 50% (also boring )
Two rounds and 4 points to win: 53.33%
Four rounds and 9 points to win: 48.54%

Of course all of that is based on (A) knowing the optimal strategy at any given moment and (B) being able to play that optimal strategy. (A) is going to be tough for certain situations and (B) is impossible for a human to do (provided the optimal strategy is a mixed strategy and not "always play 2-3"). So I wouldn't give it all that much weight.

BTW, guaranteeing the attacker one point gives the following close-to-50% games:

Two rounds and 5 points to win: 51.02%
Four rounds and 10 points to win: 49.60%

And the original three rounds and 8 points now gives the attacker a 39.42% chance to win.
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Allison Macrae
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Thank you all for your assistance. I could never come up with these numbers with anywhere near that speed, or any degree of certainty.

There are some good suggestions here that I might try to tweak the chances of success. Someone else suggested that another variant might be to make the positioning of the numbers matter (ie. card number 1 is a 2 and card number 2 is a 3), which would have the effect of making 3/2 a stronger attack strategy since it would only be blocked 50% of the time. I wonder if anyone would be able to help me discern the resultant probabilities from that? I think at the moment I might try 3 rounds with a target of 7, or 4 rounds with a target of 9.
I'm also okay with one or two variants that raise the attacker's chances well over 50%, as depending on testing that may be desirable.

Thank you again for your contributions.
 
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Nate Wasser
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If 23 and 32 are different plays, I get the following:

3 rounds & 7: 57%
3 rounds & 8: 44%

4 rounds & 10: 49%
 
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