Tiger is overpowered, everyone knows this @Dice_Box. 50% of the time it wins 100% of the time and that's like a 75% win rate.
Manaless Dredge loves to lose the die roll.
If you aren't interested in that deck, how about trying some non-dice methods of randomization, like flipping a coin? It's hard to believe your stats claim, but if it's actually that unfavorable, then try something else. Any random method ought to be fine as long as your opponent agrees. I like Dice_Box's suggestion. I personally keep a deck of poker cards with me and like to put a card face down and have my opponent guess whether it's a red card or a black card. With more casual games or poker fans, it's fun to do a quick poker hand and have the winner of that choose whether to be on the play or on the draw.
Play miracles. Prob great on the draw.
-rob
The Monty Hall problem is framed as follows:
The game show presents you with three doors, and you can pick one. Behind one is a car, behind both of the others is a goat. You arbitrarily pick one door. The game show host will *always* then open a door with a goat. You then get the choice of staying with the door you picked (which had a 1/3 chance to be the car), or switch to the other closed door (which has a 1/2 chance to be the car, since you now have knowledge that the open door is not a car).
Monty Hall doesn't apply, because what makes it "work" is the change in knowledge from initial pick to the second pick. Assuming you *each* have a set of paper/scissors/rock - you have no knowledge of the other players choice and equal odds to win. If you try and make it work using only one set, and randomly pick onecard for each player, then each players still has equal odds of winning, but those odds are only 35% (the remaining 35% is a draw and forces the players to redraw).
But, there will *never* be a game where your odds are 35% and his are 50%, because it's a zero zum situation. Either he wins the roll or you win the roll.
You both have a 50% chance of winning.
He presents you the 3 cards face down. He picks one at random (e.g. Rock). He presents you with the remaining 2 choices. One of them loses to it (e.g. Scissors) and the other one beats it (e.g. Paper), so you have a 50% chance of winning and 50% chance that he beats you. It's perfectly fair.
The Monty Hall problem is different. It involves having the option to switch your choice after Monty eliminates one of the other options. There's no switching with the Rock Lobster game. Also the key difference is Monty doesn't remove one at random. As the show host he already knows the real answer, and he always removes a losing choice (for the drama of opening a door and revealing a dud). His choice adds information (he never eliminates the car) and that's why the odds change. If Monty removed one at random, it wouldn't matter whether you switch or not.
Monty Hall applies to any statistical problem where Choice 1 is invalidated by Choice 2. In the classic problem it's because Choice 1 was made at different odds, and in this problem it's because Choice 1 serves only to seed the definition of Choice 2.
If you're worried about it I recommend adding a Goat Token to get the true Zonk experience.
Ok sure the problem can be generalized to other things. Still the odds only change because both of these occur:
1) someone with perfect information removes one of the options nonrandomly, revealing information about the remaining choices
2) you're allowed to change your choice after the information is revealed
Even in the classic problem, if you couldn't change your choice after he opens the door, there's no trick. It's a fair game.
If Monty eliminated a door at random, again there's no trick. It's a fair game.
There's no Monty Hall issue playing with Rock Lobster and friends. You're both picking at random. Each player has a 50% chance to win.
Last edited by FTW; 03-06-2019 at 12:41 AM.
Probably easier to make games best-of-two
And will be a fairer metric..
If you cant “break their serve”, u dont win..
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