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15 replies to this topic

#1 bonanova

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Posted 18 February 2008 - 09:42 AM

There's an old game children play that goes like this.

1. Two players take turns saying a number.
2. The first number called must be 10 or smaller.
3. Each number is greater than the previous number, but by no more than 10.
4. The player who first calls 50 wins.

There is a coin toss to see who goes first. You win the toss.
What's your strategy?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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#2 Aatif

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Posted 18 February 2008 - 10:29 AM

There's an old game children play that goes like this.

1. Two players take turns saying a number.
2. The first number called must be 10 or smaller.
3. Each number is greater than the previous number, but by no more than 10.
4. The player who first calls 50 wins.

There is a coin toss to see who goes first. You win the toss.
What's your strategy?


Ahh .. I would never lose :)

Spoiler for Strategy specialist

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#3 brhan

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Posted 18 February 2008 - 09:17 PM

Spoiler for my strategy

Edited by brhan, 18 February 2008 - 09:18 PM.

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#4 jremoletesilerio

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Posted 18 February 2008 - 09:39 PM

Spoiler for Solution

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#5 Aatif

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Posted 19 February 2008 - 03:40 AM

Spoiler for my strategy


Brhan, the only even no. that you can start with to make sure you win this game is 6. Else you WILL lose.
Spoiler for What's the mistake

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#6 brhan

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Posted 19 February 2008 - 12:41 PM

Say You and I are playing. And you start with an even no. NOT 6.
U - 4
I - 6
U - 10
I - 17
U - 19

I - 28
U - 30
I - 39
U - 40
I - 50!
I Win B)) !


Aatif, you are violating rule number 3. Besides, when you choose an odd number, my next number will be odd -- this gives me the advantage that the number is always even before you make your choice ... sort of cornering you. Just to look at my procedure based on the above example:
B : Brhan and A: Aatif
B-4
A-6
B-16 (here, it should be greater than four)
A-23 (you choose 7)
B-32 (my choice is 9)
A-40 (you choose 8)
B-50 (I win)

Actually 2 is the perfect number to start with.
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#7 Aatif

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Posted 19 February 2008 - 02:00 PM

Aatif, you are violating rule number 3. Besides, when you choose an odd number, my next number will be odd -- this gives me the advantage that the number is always even before you make your choice ... sort of cornering you. Just to look at my procedure based on the above example:
B : Brhan and A: Aatif
B-4
A-6
B-16 (here, it should be greater than four)
A-23 (you choose 7)
B-32 (my choice is 9)
A-40 (you choose 8)
B-50 (I win)

Actually 2 is the perfect number to start with.

Maybe I am missing something. If the puzzle says: to choose by no more than 10, then 10 is inclusive. Right?? If thats true, then my only aim is to choose the number 6,17,28,39 and 50 (independent of what you choose). So once I can choose any one of them I can never lose.
So in the above stated case, after you choose 32, I'll simply choose 39 and hence I WIN! Simple.
In the case that 10 is NOT inclusive, so you have to choose numbers between 1-9, then the player who plays first can NEVER win (for the second player will always choose 10,20,30,40 and 50)! I don't see any confusion in this.
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#8 Aatif

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Posted 19 February 2008 - 02:38 PM

Maybe I am missing something. If the puzzle says: to choose by no more than 10, then 10 is inclusive. Right?? If thats true, then my only aim is to choose the number 6,17,28,39 and 50 (independent of what you choose). So once I can choose any one of them I can never lose.
So in the above stated case, after you choose 32, I'll simply choose 39 and hence I WIN! Simple.
In the case that 10 is NOT inclusive, so you have to choose numbers between 1-9, then the player who plays first can NEVER win (for the second player will always choose 10,20,30,40 and 50)! I don't see any confusion in this.

Ahhhh .. I see why you are confused .. In fact, after seeing that I am a tad bit confused too. So you interpret rule no. 3 as choosing number between 1-10, but the way I am interpreting it is that it's just the difference between your choice and your competitors choice.
Actually if, you 6 on your first choice and I choose 10 and then you choose 17 and then I choose 18, then my second choice of number is still greater than yours (17) by no more than 10. Thus the choice of the "Number" is to be greater than the last said number by no more than 10 and not the difference between them.

But now that I see you interpretation, I understand that that puzzle is also quite interesting. Which is:
You choose a number between 1-10 (inclusive), bigger than your last choice. Add it to the Sum written in the center. Then your opponent does the same. One who reaches 50 first wins.
What should you strategy be?! (I think Brhan need not reply to this! :) )
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#9 roolstar

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Posted 19 February 2008 - 03:42 PM

since if I reach 50 I win
==> If I reach 39 I win ( no matter what he will say, he can't reach 50 and I win)
==> If I reach 28 I win ( no matter what he will say, he can't reach 39 and I win)
==> If I reach 17 I win ( no matter what he will say, he can't reach 28 and I win)
==> If I reach 6 I win ( no matter what he will say, he can't reach 17 and I win)

I'll ha ve to agree with Aatif on this one and start with 6
then I will go to 17, 28, 39 then 50 as he choses his numbers...

Basically, the one who starts can always win!

Not the case if the number to reach is 110!!

In this case the one who starts will always lose!!!!!

NOW ALL THIS ASSUMING YOU CANNOT CALL 0!!! < The game will be silly in this case as it can be stalled indefinitely))

Edited by roolstar, 19 February 2008 - 03:46 PM.

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#10 brhan

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Posted 19 February 2008 - 03:46 PM

Ahhhh .. I see why you are confused .. In fact, after seeing that I am a tad bit confused too. So you interpret rule no. 3 as choosing number between 1-10, but the way I am interpreting it is that it's just the difference between your choice and your competitors choice.
Actually if, you 6 on your first choice and I choose 10 and then you choose 17 and then I choose 18, then my second choice of number is still greater than yours (17) by no more than 10. Thus the choice of the "Number" is to be greater than the last said number by no more than 10 and not the difference between them.

But now that I see you interpretation, I understand that that puzzle is also quite interesting. Which is:
You choose a number between 1-10 (inclusive), bigger than your last choice. Add it to the Sum written in the center. Then your opponent does the same. One who reaches 50 first wins.
What should you strategy be?! (I think Brhan need not reply to this! :) )


Exactly ... But I can see your point now.
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