[Guest]

100 people ( Player 1, Player 2, ... Player 100) are playing a game.

In the first round Player 1 wins.

In the second round Player 2 wins

In the third round Player 3 wins and so on till 100

so in total 100 rounds of the game are played.

If a person wins a round he has to give some money to the other 99 players. The winner must double the money of each of the other players (that they had before the start of the round).

Example; if when Player 5 won the 5th round, he doubled the money all other players at the start of round 5. In the 6th round, player 6 doubled the money other players had at the start of round 6 (same as end of round 5 after player 5 had distributed the money to other players).

At last after the 100 th player wins the amount of money which everone has is same and is "X"

Questions:

1) Which player(s) ended up richest? (i.e. richer meaning that the player had more money at the end of the game than he had at the beginning of the game)

2) Which player(s) lost the most money?

For math lovers:

3) How much money did the 63rd player had in beginning of the game in terms of "X"?

Edited July 2, 2009 by DeeGee

[Guest]

What do we know about the start capital of each player?

If they all have the same start capital, player one won't be able to pay the other players after the first round.

[Guest]

What do we know about the start capital of each player?

If they all have the same start capital, player one won't be able to pay the other players after the first round.

Well then, they all do not have the same amount of money at the beginning! The information in the question is enough...

the start capital of each player can be made a function of "x"

[Guest]

100 people ( Player 1, Player 2, ... Player 100) are playing a game.

In the first round Player 1 wins.

In the second round Player 2 wins

In the third round Player 3 wins and so on till 100

so in total 100 rounds of the game are played.

If a person wins a round he has to give some money to the other 99 players. The winner must double the money of each of the other players (that they had before the start of the round).

Example; if when Player 5 won the 5th round, he doubled the money all other players at the start of round 5. In the 6th round, player 6 doubled the money other players had at the start of round 6 (same as end of round 5 after player 5 had distributed the money to other players).

At last after the 100 th player wins the amount of money which everone has is same and is "X"

Questions:

1) Which player(s) ended up richest? (i.e. richer meaning that the player had more money at the end of the game than he had at the beginning of the game)

2) Which player(s) lost the most money?

For math lovers:

3) How much money did the 63rd player had in beginning of the game in terms of "X"?

just logically, that if it proceeds sequentially to each of the 100 players in turn, and all players start out with the same capital... well, none could manage to pay out anything.

Player 1 would have X, the same as everyone else. And he would have to increase the other 99 players' stashes to 2x... so he'd have to pay out 99 times his own starting capital.

SOOO, I guess they're paying in IOUs. In that case, I'm guessing that player 1 will have to pay out the least, because everyone else will be paying out 2^n*X to player 100 alone, where n is the player's number, and X is the starting capital. The question then, is whether what player 100 has to pay out will effectively negate all he's earned over the 99 previous 2^n*x iterations.

I'm not mathy enough to get it in my head (probably has something to do with those vaguely recalled integrals), though I have a sneaking suspicion that player 1 and 100 end up about the same, and player 50 or so ends up closest to the top.

[Guest]

Since player one has to pay the first time he has to have the largest start capital, and since player 100 only have to pay after he has been payed 99 times he has the smallest start capital.

This means that player 100 got the richest, and player 1 lost the most.

I tried to write a little program for this. It turned out to be that player one must have a startcapital of 5.9E24*X, and player 100 a start capital of 7.81E-29*X. Player 63 must have a start capital of 5.09E-9*X.

[Guest]

just logically, that if it proceeds sequentially to each of the 100 players in turn, and all players start out with the same capital... well, none could manage to pay out anything.

Player 1 would have X, the same as everyone else. And he would have to increase the other 99 players' stashes to 2x... so he'd have to pay out 99 times his own starting capital.

SOOO, I guess they're paying in IOUs. In that case, I'm guessing that player 1 will have to pay out the least, because everyone else will be paying out 2^n*X to player 100 alone, where n is the player's number, and X is the starting capital. The question then, is whether what player 100 has to pay out will effectively negate all he's earned over the 99 previous 2^n*x iterations.

I'm not mathy enough to get it in my head (probably has something to do with those vaguely recalled integrals), though I have a sneaking suspicion that player 1 and 100 end up about the same, and player 50 or so ends up closest to the top.

No IOUs were given and they didnt start out with the same amount of money... It is a logic question not a tricky worded puzzle

[Guest]

If the 100th player started with 1*X and players 99 through 1 would need 2^(99-n)*X where n is their number. With this set up, each time a player won, they would spend ALL their money doubling the rest, so at the end of the round they would have nothing. This means that the next round, the winner there would not have to pay anything to the people before him because they have $0. Following this pattern, the 100th player would end up will all the money.

[Guest]

Since player one has to pay the first time he has to have the largest start capital, and since player 100 only have to pay after he has been payed 99 times he has the smallest start capital.

This means that player 100 got the richest, and player 1 lost the most.

I tried to write a little program for this. It turned out to be that player one must have a startcapital of 5.9E24*X, and player 100 a start capital of 7.81E-29*X. Player 63 must have a start capital of 5.09E-9*X.

if after 100 rounds everyone has X money, after 99 rounds everyone had 0.5*X money except for player 100 who had (X + 99*.5X) money. So after 98 rounds everyone would have half of what he has after 99 rounds, except for player 99. This is an iterative process for which i can't find a general solution, but it is programmable.

[Guest]

If the 100th player started with 1*X and players 99 through 1 would need 2^(99-n)*X where n is their number. With this set up, each time a player won, they would spend ALL their money doubling the rest, so at the end of the round they would have nothing. This means that the next round, the winner there would not have to pay anything to the people before him because they have $0. Following this pattern, the 100th player would end up will all the money.

Not quite... the question says that at the end of the game, they all had the same amount of money!

Secondly, if someone is left with zero, he will never get paid anything when someone else wins!

Edited July 2, 2009 by DeeGee

[Guest]

Not quite... the question says that at the end of the game, they all had the same amount of money!

Secondly, if someone is left with zero, he will never get paid anything when someone else wins!

Ok, I missed the the part about the end game setup, but on your second note, the whole point of someone having zero left at the end of their win is so the following people will not have to pay them, leaving all the money in the hands of the last player, who would not have to pay anything for winning.

[Guest]

Ok, I missed the the part about the end game setup, but on your second note, the whole point of someone having zero left at the end of their win is so the following people will not have to pay them, leaving all the money in the hands of the last player, who would not have to pay anything for winning.

In that case, the last person to win would have all the money and the others zero which is against the condition that they all end up with the same amount of money

[Guest]

In that case, the last person to win would have all the money and the others zero which is against the condition that they all end up with the same amount of money

then they all must have started with $0, so all of the paying and doubling means nothing.

[Guest]

Each player in thier round has to have more than all the other players combined. to be a little more exact in any round N, the current player must have > (100/2^n)x by that logic, player 1 -4 all lose but player 1 loses the most. The rest all gain but player 100 gains the most.

Player 63 has some amount y which is doubled 62 times in the first 62 rounds so he has (2^62)y before the 63rd round. during the 63rd round player 63 gives away the total minus what he has or mathmatically, 100x - (2^62)y so after the 63rd round, player 63 has (2^63)y-100x. this quantity is multiplied by 2 for each of the last 37 rounds or mathmatically, (2^100)y- 25(2^39)x = x this can be simplified to Y= ((25(2^39)+1)x)/2^100 or player 63 starts out with around (1.084x10^-17)x

knowing me though, most likely i messed up that algebra or something somewhere so maybe thats wrong. lol.

[bonanova]

Player 100 doubled the money that players 1-99 had.

The result was all players ended up with x.

Work backward to get the previous amounts.

Before game 100:

Players 1-99 had x/2.

Player 100 had 99x/2 [which he gave to the others] + x [which he kept].

Before game 99:

Players 1-98 had [x/2]/2, and Player 100 had [99x/2 + x]/2;

Player 99 had [98x/2 + (99x/4 + x/2)]/2 [which he gave to the others] + x/2 [which he kept].

etc.

Tedious.

But try doing the calculation for fewer players - say four.

[Guest]

Player 1 lost the most. He lost 50 X - X = 49 X.

Player 100 gained the most. He gained X - 7.9675E-29 X = X (approx.)

Player 63 started with 1.0842E-17 X

[Guest]

Player 100 doubled the money that players 1-99 had.

The result was all players ended up with x.

Work backward to get the previous amounts.

Before game 100:

Players 1-99 had x/2.

Player 100 had 99x/2 [which he gave to the others] + x [which he kept].

Before game 99:

Players 1-98 had [x/2]/2, and Player 100 had [99x/2 + x]/2;

Player 99 had [98x/2 + (99x/4 + x/2)]/2 [which he gave to the others] + x/2 [which he kept].

etc.

Tedious.

But try doing the calculation for fewer players - say four.

You logic is correct. Just some minor calculation error.

It should be -

Before game 99:

Players 1-98 had [x/2]/2, and Player 100 had [99x/2 + x]/2;

Player 99 had [98x/2 + (99x/2 + x)]/2 [which he gave to the others] + x/2 [which he kept].

My answer is -

Player 1 started with 50X.

Player 100 started with 7.9675E-29X.

Player 63 started with 1.0842E-17X.

Agree?

Edited July 4, 2009 by ZinY

[Guest]

then they all must have started with $0, so all of the paying and doubling means nothing.

Yes. For me, I can agree with your answer as well.

[Guest]

I learnt how to use a spoiler only just now.

I mistakenly thought that there might be an administrator to hide those things.

Please pardon me! I am sorry.

I won't do that again.

I suppose I could not edit my posts anymore.

[Guest]

Player 100 doubled the money that players 1-99 had.

The result was all players ended up with x.

Work backward to get the previous amounts.

Before game 100:

Players 1-99 had x/2.

Player 100 had 99x/2 [which he gave to the others] + x [which he kept].

Before game 99:

Players 1-98 had [x/2]/2, and Player 100 had [99x/2 + x]/2;

Player 99 had [98x/2 + (99x/4 + x/2)]/2 [which he gave to the others] + x/2 [which he kept].

etc.

Tedious.

But try doing the calculation for fewer players - say four.

Exactly... that is the way to do it... once you do it for 3 or 4 rounds starting from the end, you will notice a pattern.. that would be enough to answer all 3 questions

Edited July 4, 2009 by DeeGee

[Guest]

You logic is correct. Just some minor calculation error.

It should be -

Before game 99:

Players 1-98 had [x/2]/2, and Player 100 had [99x/2 + x]/2;

Player 99 had [98x/2 + (99x/2 + x)]/2 [which he gave to the others] + x/2 [which he kept].

My answer is -

Player 1 started with 50X.

Player 100 started with 7.9675E-29X.

Player 63 started with 1.0842E-17X.

Agree?

There seems to be some error in the calculation. Keep it simple. At each step try getting the amount in terms of x for each player. You will see the pattern soon enough.

Edited July 4, 2009 by DeeGee

[Guest]

1) Which player(s) ended up richest?

Player 100 ended up richest.

He started with 50/2⁹⁹ X.

He gained quite close to X.

2) Which player(s) lost the most money?

Player 1 lost the most money.

He started with 50X.

He lost 49X.

3) How much money did the 63rd player had in beginning of the game in terms of "X"?

Player 63 started with 50/2⁶² X.

P.S. In general, the i^th player started with 50/2^(i-1) X.

[Guest]

Well ZinY, its still not correct!

[Guest]

Well ZinY, its still not correct!

[Guest]

you said that player 100 started with 50/2⁹⁹x which would mean he'd have 50x before the very last round and since he's giving away 100- what he has, that would mean he's giving away 50x leaving him with 0 which isnt necisarilly x. he actually has 50.5x before round 100 and therefore starts with 50.5/2⁹⁹x. by similar reasoning you answer to number 3 is wrong. remember, they have to have just a little more than 50x at their round so that after giving the rest away they have just enough left to be doubled enoujgh times to eventually reach x. That should help you with the rest of your answer.

by the way, was my answer on page 2 right?

Edited July 4, 2009 by sparx1

[Guest]

Thanks, sparx1!

[Guest]

1=> (50+¹/₂)=> (50/2+¹/₄)=> (50/4+¹/₈)...

1=> (¹/₂) => (50+¹/₄)=> (50/2+¹/₈)...

1=> (¹/₂) => (¹/₄) => (50+¹/₈)...

1=> (¹/₂) => (¹/₄) => (¹/₈)...

1=> (¹/₂) => (¹/₄) => (¹/₈)...

1=> (¹/₂) => (¹/₄) => (¹/₈)...

(I gave up typing in the sequence due to HTML formatting problem.)

The general formula for the i^th player, after j^th calculation -

(50/2^(i+j-101) + 1/2^j) X

1) Which player(s) ended up richest?

Player 100 ended up richest.

He started with (50/2⁹⁹+1/2¹⁰⁰) X.

He gained quite close to X.

2) Which player(s) lost the most money?

Player 1 lost the most money.

He started with (50+1/2¹⁰⁰) X.

3) How much money did the 63rd player had in beginning of the game in terms of "X"?

Player 63 started with (50/2⁶²+1/2¹⁰⁰) X.

P.S. In general, the i^th player started with (50/2^i-1+1/2¹⁰⁰) X.

edited: adding some brackets.

Edited July 4, 2009 by ZinY