## Friday, July 1, 2022

Published July 01, 2022 by with 3 comments

# Mario Party's 'Hide and Sneak' is not Balanced

Are you consistently losing to a 6 year old and looking for an excuse. If so, you've come to the right place.
'Hide and Sneak' is one of the mini-games in Mario Party. The basic rules are that 3 people hide and 1 person finds them. In round 1, there are 4 hiding spots. In round 2, 3. In round 3, 2.

Each player on the hiding team picks a spot to hide and the seeking player picks one spot. If a player is hiding in the picked spot, he's out.

If a player is out he doesn't participate in later rounds.

If any players remain on the hiding team after 3 rounds, the hiding team wins.

Is this game 50/50? Working through the math, start with 1 hider:

• round 1, there's a 3/4 chance of not being found
• round 2, there's a 2/3 chance of not being found
• round 3, there's a 1/2 chance of not being found
Thus, for an individual player on the hiding team, there's a (3/4) * (2/3) * (1/2) chance of not being found. That's 1/4, or 25%.

Inverting that, the seeker has a 3/4, or 75% chance of finding a given player after 3 rounds.

Since there are 3 players and their hiding decisions are independent, the chance of the seeker finding all 3 players in 3 rounds is just (3/4)^3, or 27/64. That's only 42%. It isn't balanced at all.

Is there an obvious way to balance it? What if we did 4 rounds with 5 starting hiding spots. For one hider:

• round 1, there's a 4/5 chance of not being found
• round 2, there's a 3/4 chance of not being found
• round 3, there's a 2/3 chance of not being found
• round 4, there's a 1/2 chance of not being found
Multiplied out, that's a 1/5, or 20% chance of not being found so 4/5, or 80% success chance for the seeker. The chance that the seeker finds all 3 players in 4 rounds then is (4/5)^3, or 51%. 4 rounds is way more balanced than 3 rounds.

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2. Your math is correct, but only under certain critical parameters. It all comes down to whether or not one is playing against AI or against other humans capable of communicating.

Assuming independence of the hiding spot selection is fair, so long as the hiders are controlled by an AI. However, independence completely changes should three humans be controlling the hiders.

If at any point in the game there are the same number of hiding spots available as there are hiders remaining, the hiders can strategically force a win by communicating which hiding spot they are going to, and ensure no two characters hide in the same spot. Since the seeker can only select one spot at a time, there will always be one hider remaining at the end of three rounds. This concept will be referred to as the 'collusion strategy'. (Note: This strategy applies regardless of whether the seeker is controlled by a human or AI)

Therefore, the question stands: what are the odds of a round beginning with the same number of hiding spots as there are hiders? This can happen either in round two or round three.

During round one, hiding spot selection can be made independently. There are twenty different combinations of hiding spot selections with a 50% chance of no character being selected.
This means there is a 50% chance of being able to use the collusion strategy to win after the first round, ensuring victory for the hiders. However, there is still a 30% chance that the seeker finds one hider, a 15% chance to find two hiders, and a 5% chance of finding all three hiders, thus making the seeker victorious.

If one hider is found in the first round, we continue to the second round.

Now there are three hiding spots remaining and two hiders. There are six different combinations possible for the hiders. There is again, a 50% chance for no hider to be found, which would leave two hidings spots for two hiders in the last round ensuring a victory for the hiders through the collusion strategy. There is also a 33.3% chance of finding one hider, and a 17.7% chance of finding both of the remaining hiders, leading to a victory for the seeker.

However, if two seekers were found in the first round, there's a 66.6% chance of finding no one, and a 33.3% of finding the last hider, thus allowing the seeker to win.

If there is only one hider left in the final round with two hiding spots, both the hiders and the seeker each have a 50% chance of winning.

Here's the breakdown:

Round 1:
Odds of 0 hiders found = 50% (HIDERS WIN)
Odds of 1 hider found = 30%
Odds of 2 hiders found = 15%
Odds of 3 hiders found = 5% (SEEKER WIN)

Odds of 1 hider found in round 1, 0 hiders found in round 2 = 15% (HIDERS WIN)
Odds of 1 hider found in round 1, 1 hider found in round 2 = 10%
Odds of 1 hider found in round 1, 2 hiders found in round 2 = 5% (SEEKER WIN)

Odds of 2 hiders found in round 1, 0 hiders found in round 2 = 10%
Odds of 2 hiders found in round 1, 1 hider found in round 2 = 5% (SEEKER WIN)

Odds of 1 hider found in round 1, 1 hider found in round 2, 0 hiders found in round 3 = 5% (HIDERS WIN)
Odds of 1 hider found in round 1, 1 hider found in round 2, 1 hider found in round 3 = 5% (SEEKER WIN)
Odds of 2 hiders found in round 1, 0 hiders found in round 2, 0 hiders found in round 3 = 5% (HIDERS WIN)
Odds of 2 hiders found in round 1, 0 hiders found in round 2, 1 hider found in round 3 = 5% (SEEKER WIN)

Therefore, if hiders use the collusion strategy, the odds of the hiders winning the entire game is 75%. This is much higher than the 42% victory odds with no collusion strategy.

In conclusion, the odds of hiders winning in "Hide and Sneak" in Mario Party is contingent on whether the hiders are controlled by the AI (42%) or controlled by humans using the collusion strategy (75%).

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