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Two players play a simple game called

Odd Todd. They start with a pile of

an odd number of stones. Players take

turns alternately. On his turn, a

player may remove one or two stones

from the pile. The player who ends up

having taken an odd number of stones

wins. Under what conditions can the

first player force a win? When can the

second player force a win? What are

the winning strategies in both cases?

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Posted · Report post

Number of stones is n, if n+1 is divisible by 4 the second player wins, if not the first player wins.

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Posted (edited) · Report post

I had an idea for player 1 then after a few sample games i realised that its very difficult indeed for player 1 to win

strategy for player 2:

Whatever player one takes, player two need to keep the total number of stones removed odd, that way there is always an even number of stones left in the pile... and if i reckon right, that will mean the odd pebble will go to player two.

In fact if the pile has 7 stones in it i'm not certain player 1 can win if player 2 is trying to win (i.e. not just playing randomly) - if player two always does as I stated above he will always win.

Not had chance to think about what happens with very large piles of stones - limited my guestimation to 7,9 and 11 stones total.

going for coffee number 2 now...

Edited by RESHAW
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Posted · Report post

The total number of stones should be 19.

The play gives each player a free choice to take ...1,2,or 3 stones away.

the second player will win the game if he compliments the amount taken by the first player to 4

e.g. if the first player removes 2 stones,the second player should remove(4-2= 2 stones).

if 1...so the second will take 3 stones away..and so on.

but if the first player want to force the win,he should shuold raise the amount of stones to be taken to 4...i.e....1,2,3,or4

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If 1 coin: first person to play wins:

With three Tones, First player loses! If he takes 1, then second player takes 1, forcing first player to take last stone.Building from here, with 5 stones, first player wins by taking two, leaving three and second player in losing position. Continuing to 7 tones, player 1 would lose if he leaves opponent in winning position of 5 so must take only 1. Player 2 must take 2 or leave winning position for one. With 4 stones, player 1 takes 1 coin leaving player 2 in losing position. Now 9 stones, player 1 can only take one or present two with winning position. With 8, player 2 must take 2 or leave winning position of 7 coins for player 1. 6 stones remain and player 1 must take 2 to avoid giving the player 2 winning position of 5. Unfortunately this leave player 2 with the ability to leave player 1 with the losing position of 3. Following this same pattern, we see that player 2 wins when number of stones = 6n-3 where n is an integer greater than 0. For random number of stones greater than 3, player 1 should win 2/3rds if the time

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Posted · Report post

"With three sTones, First player loses! If he takes 1, then second player takes 1, forcing first player to take last stone" and then the first player wins because he "ends up having taken an odd number of stones". If the second player takes 2, then it is a draw.

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Posted · Report post

The total number of stones should be 19.

The play gives each player a free choice to take ...1,2,or 3 stones away.

the second player will win the game if he compliments the amount taken by the first player to 4

e.g. if the first player removes 2 stones,the second player should remove(4-2= 2 stones).

if 1...so the second will take 3 stones away..and so on.

but if the first player want to force the win,he should shuold raise the amount of stones to be taken to 4...i.e....1,2,3,or4

It is stated that "On his turn, a player may remove one or two stones from the pile."

I think there is no winning strategy.

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Posted · Report post

"With three sTones, First player loses! If he takes 1, then second player takes 1, forcing first player to take last stone" and then the first player wins because he "ends up having taken an odd number of stones". If the second player takes 2, then it is a draw.

it cannot be a draw because initially there is an odd number of stones

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Posted · Report post

that the very 1st post by James22 is correct. 1,3 & 5 have been explained. With 7 stones, player 2 is guaranteed a win by taking the same no. of stones that player 1 does. each time. So...sticking my neck out....

When (n-1) mod 4 = 0, P1 is guaranteed a win by taking 2 with his 1st go, and then taking same no. as P2 in subsequent goes.

When (n-1) mod 4 = 2, P2 is guaranteed a win by taking same no. of stones as P1 each time.

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Posted · Report post

the goal is not to take the last stone but simply to have an even number of stones at the end.

for 1 player 1 loses.

for 3 player 1 wins (takes 2.)

for 5 if player 1 takes 2, player 2 takes 2, player 1 loses.

if player 1 takes 1, player 2 takes 1, whatever player 1 takes next, player 2 takes 1 and forces player 1 to lose.

for 7, player 1 takes 2, and now we're in 5 stone state with it being player 2's turn, so same reasoning implies.

so i think fab pig his the right idea, just has the mods back words.

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Posted · Report post

bah! even i make mistakes apparently (goal to have an odd number of stones!)

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Posted · Report post

Player I can force a win if he takes 1 in the first round and then in the next turns take the same number of pebbles as Player 2 has taken.

Player 2 can win if Player 1 gets 2 first, and then Player 2 gets 1 in the first round and in the next rounds takes the same number of pebbles player 1 took.

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