# Two wrongs can make a right

Can you construct two single-player games A and B, such that:

• Every time you play game A you are guaranteed to lose.
• Every time you play game B you are guaranteed to lose.
• Every time you alternate turns between playing game A and game B you are guaranteed to win. There is a shared state that is carried over between the two games, for example money in the bank. As you swap games, you also swap the rules you are playing with.
• if you've seen this before then please don't answer it and give others a chance. Mar 24, 2022 at 10:19
• It's very unclear what you're looking for here - you need some sort of shared state between the "games", right? What counts as a loss/win, in that case?
– Deusovi
Mar 24, 2022 at 16:22
• @Deusovi I understand it to mean "If you play A at least once, you'll lose some game at least once; same for B; but if you alternate them, you'll win some game at least once". Mar 24, 2022 at 18:15
• @Deusovi I tried to provide more information. Is it more clear now? Mar 25, 2022 at 4:30
• @msh210 not quite. Every time you play game A you lose. Every time you play game B you lose. But if you alternate them then you always win. Let me update the text. Mar 25, 2022 at 4:31

Game A being a rigged game (let's take a simple D20 die roll). Game B being a "money" game.

Both games are supposed to make you lose if you only play one of them.
For Game A, that means guessing a value on the loaded die (and losing because the game is rigged for you to lose). That game will slowly decrease your balance, and you will lose all your money.
For Game B, the game is simple, if the balance you have is odd, you win, if the amount is even, you lose.
The requirements for this to work would be:
- The price of Game A has to be smaller than the gain of Game B.
- The price of Game B has to be even, and its gain, odd.
- The price of Game A has to be odd.
- The gain of Game B has to be bigger than the price of both games added together

Example:

If I decide to play Game A, I'd pay 5 coins to play, with a potential payout of 20 coins (which I will never reach because the game is rigged). Playing this game over and over, if the dice is properly loaded, will make me lose all my money after 200 games. (If the dice is loaded but you still have a chance, let's assume the chance of winning is 1/50 rolls, which statistically still makes me lose all my money)
If I decide to play Game B, I'd pay 20 coins, and a payout of 27 coins. If I only play Game B, my balance will never be odd, and I will always lose.
However, if I decide to lose at Game A, then go to Game B, I will have won 2 coins. Netting me a profit of 2 coins.

• This answer is the closest to what I had in mind. Mar 25, 2022 at 23:14
• The shared state being money made think of this yea Mar 26, 2022 at 11:08
• the puzzle came from this fascinating paradox: en.m.wikipedia.org/wiki/Parrondo%27s_paradox Mar 29, 2022 at 11:28

Ignoring some obscure corner cases and attempts at sneaky interpretation:

Game A

The "Turn the light on" game, you need to do it twice to win.

Game B

The "Turn the light off" game, you need to do it twice to win.

• I don't understand what is the "turn the light on" game and why is it losing? Mar 24, 2022 at 12:22
• Flip the light switch in a room, the light is now on. You won't be able to turn it on a second time unless it gets turned off first. If you don't consider being deadlocked as losing, just impose a time limit. Mar 24, 2022 at 13:04
• I think you are on the right track, but you need to give more details. What is a losing state in each game? How does alternating games give you a win? Mar 24, 2022 at 13:19
• This feels pretty obvious. If you only play game A you'll never get to the point of turning the light on a second time. If you only play game B you'll never get to the point of turning the light off a second time. If you alternate between playing game A and B you not only win both games, you can even broadcast a morse code message to your neighbors. :-) Mar 24, 2022 at 13:35
• A variation would be the game to turn on two lights. In game A you can only act on light 1, in game B you can only act on light 2. The shared state is the lamps. As stated the question is too vague and allows too many solutions. Mar 25, 2022 at 7:16

This might seem like a childish answer, but I thought of

A seesaw, which has two positions that cannot be changed unless a weight is placed on the opposite end.

Therefore, in Game A

You must change the position of the seesaw such that end A is in the downwards position, by dropping a weight onto side A, propelling a weight placed on the opposite end up and off of the board.

And in Game B

You must do the same as in Game A, but with side B of the seesaw.

The failure states being

If the seesaw is unable to be moved, or if there is no weight on the opposite end of the seesaw to propel.

In this system, if either Game A or Game B are done repeatedly

The seesaw will remain in the position that it was last in, and there will not be a weight on the other end to propel off of the board.

But by alternating between Game A and Game B

The position of the seesaw will be constantly changing, and a weight will always be placed on the other end to be propelled.

Not too bad, though I wouldn't say these are fun games.

Starting conditions for Both games: Shared Variables [Money in the Bank] = 50 Dollars, [Money in hand] = 50 Dollars. Win conditions are checked at the beginning of the turn, but allow the turn to still be performed with the "Win" already achieved.

Game A: Robber Retirement. Rules: "On your turn, you take all dollars out of the bank and place into your hand" Win Condition: "Have 100 dollars in the bank. Otherwise, you lose."

Game B: Greedy Saver. Rules: "On your turn, you take all dollars out of your hand and place it into the bank" Win condition: "Have 100 Dollars in hand. Otherwise, you lose."

Both games are impossible to win on their own, but by alternating from A > B or B > A after losing the first game played, you always win the following games played immediately.

Example:

Game A Begins, 50 in bank, 50 in hand. You can't win, but you move all your money from the bank and into your hand, ending with 100 dollars in hand. You then move to Game B, winning immediately as you have 100 in hand, then on your turn, move it all the the bank. Then you play Game A again, starting with 100 in the bank, winning and moving it all to your hand once more etc. Similar interaction when starting with Game B

Conditions met:

Every time you play game A you are guaranteed to lose.

Yes

Every time you play game B you are guaranteed to lose.

Yes

Every time you alternate turns between playing game A and game B you are guaranteed to win. There is a shared state that is carried over between the two games, for example money in the bank. As you swap games, you also swap the rules you are playing with.

Yes, key phrase being emphasised, as we only care about the game state after alternating the games.

• Your setup would allow Game B to win every time in the alternating scenario, but you would always lose Game A in this setup, as you would never be able to achieve the win condition. Mar 25, 2022 at 15:25
• I think he is assuming that a "Win" in one game resets the conditions. Mar 25, 2022 at 15:43
• Question states that the shared state does not revert at the end of one game or the start of the next. So if you start with B, you can move all the money to the bank, lose, then play Game A, winning with 100 in the bank. Changed the turn rules to allow all money to be moved, but I didn't intend for this to be an endless cycle of winning per se, more that you must lose the first game to allow you to win the second. I don't believe we must start with game A. If the win state doesn't stop the player from then performing their turn after winning, this is an endless win cycle after loss 1 Mar 25, 2022 at 15:53
• Altered the rules to allow the endless win cycle post loss 1 Mar 25, 2022 at 16:00
• @Bewilderer What is preventing you from winning Game A immediately after losing or winning Game B? Both Games are designed not to be winnable only if it's the first game you play. Mar 25, 2022 at 21:42

Game $$A$$:

Choosing even numbers

Game $$B$$:

Choosing odd numbers

Winning:

The two chosen numbers add up to an odd number.

Guaranteed win:

Then, the sum of two even or two odd numbers is even, so one could never win this way. The only way to win would be to choose one even number and one odd number.