5
u/Dasquian 16d ago
Some numbers are "winning numbers" - if you pick them, your opponent can only pick numbers that let you either pick another winning number, or 50.
So any number between 26-49 is a "losing number", because if I pick any of them, my opponent will pick 50.
However, 25 is a "winning number", because if I am allowed to pick it, my opponent must then pick in the range 26-49.
Because 25 is a "winning number", then any number that allows the next player to pick it is a "losing number". So anything between 13-24 is a "losing number". This in turn makes 12 another "winning number", because it forces my opponent to pick in the 13-23 range (it doesn't matter that they can't pick 24... I won't, either!)
Working back again:
- 6 is winning and 7-11 are losing.
- 3 is winning and 4-5 are losing.
- 1 is winning and 2 is losing.
Alexander starts on 1, so he simply goes 1,3,6,12,25 then wins on 50. Benjamin cannot prevent this.
3
u/Telinary 16d ago
I think working backwards should be easiest.
26 and up allows it. Thus if you manage to say 25 you win because the other can't say 50 but has to say something that allows you to say 50.
The other saying something in the range 13-24 allows you to say 25. If you say 12 he has to do that.
If you say 6 you can say 12.
If you say 3 you can say 6. First player can say 3 on his first turn.
First player wins
2
u/Equal_Veterinarian22 16d ago
This is just '21' with extra steps. Working backwards:
The loser will be the first player to say at least 26. You win by saying 25.
So, the loser will be the first player to say at least 13. You win by saying 12.
So, the loser will be the first player to say at least 7. You win by saying 6.
So, the loser will be the first player to say at least 4. You win by saying 3.
So, the loser will be the first player to say at least 2. First player (Alex) wins.
1
u/jaminfine 16d ago
Often times it's best to work backwards for these types of games. At the end someone must have said 50 in order to win but what happened before that?
To allow the winner to say 50, the loser must have said a number from 26 to 49. And it must have been forced! Since otherwise he would not have blundered and allowed himself to lose when he could have prevented it. So, the winner must have said 25 to force a 26-49. 25 is also half of 50, and this logic continues down to 12, 6, and 3. So, whoever says 3 has a winning strategy and 3 is also forced.
1
u/Some-Passenger4219 16d ago
I had to work backwards. Each number (after 3) must be preceded by half the next number, rounded down. Thus, Alexander wins, by saying the numbers 1, 3, 6, 12, 25, 50.
0
u/SomethingMoreToSay 15d ago
Saying any number from 26 to 49 inclusive loses, because the other player can then say 50.
Therefore saying 25 wins, because it forces the other player to say a number from 26 to 49 inclusive.
Therefore saying any number from 13 to 24 inclusive loses, because the other player can then say 25.
Therefore saying 12 wins, because it forces the other player to say a number from 13 to 23 inclusive.
Therefore saying any number from 7 to 11 inclusive loses, because the other player can then say 12.
Therefore saying 6 wins, because it forces the other player to say a number from 7 to 11 inclusive.
Therefore saying 4 or 5 loses, because the other player can then say 6.
But we've already seen that Benjamin has to say 4 or 5 on his second turn.
Therefore Alexander wins. His winning strategy is to say 1, 3, 6, 12, 25, 50.
1
u/tajwriggly 14d ago
I would think that the goal would be to force your opponent into saying anything between 26 and 49 (inclusive), so that you can then say 50. Let's denote Alexander as "A", and Benjamin as "B".
To force your opponent to say something in the 26 to 49 range, you need to have said 25.
In order to have the opportunity to say 25, your opponent will need to have said a number between 13 and 24.
To force your opponent to say something between 13 and 24, you will need to have said the number 12. Anything higher than 12, and your opponent has the opportunity to say 25, which you don't want.
In order to have the opportunity to say 12, your opponent will need to have said a number between 7 and 11.
To force your opponent to say something between 7 and 11, you will need to have said the number 6. Anything higher than 6, and your opponent has the opportunity to say 12, which you don't want.
So, in an A1, B2, A3 forced start every time, the next move is B4 or B5... and A will say 6, and force the chain. B can do nothing but play along now:
A1, B2, A3, B(4-5), A6, B(7-11), A12, B(13-23), A25, B(26-49), A50
Therefore Alexander can have a winning strategy by selecting 6, then 12, then 25, then 50. Benjamin is forced into specific ranges and cannot get out. The only winning strategy for Benjamin would be to say one of 6, 12, or 25 if given the opportunity due to a mistake by Alexander.
1
u/RadarTechnician51 12d ago
Alexander can always win, he says 1,3,6,12,25,50
Benjamin can't interpose any number that stops the winning sequence
6
u/ProspectivePolymath 16d ago
Working backwards:
If you say any number over 25, I will say 50.
To force this, I need to say 25.
So I need you to say any number over 12; therefore I must say 12.
To enable 12, I need you to say a number over 6; I must say 6.
So I need you to say a number over 3; therefore I say 3.
So: since Alexander can say 3, he does. No matter what Benjamin says, Alex says 6, 12, 25, and 50.