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Alun science forum beginner
Joined: 19 Feb 2006
Posts: 11

Posted: Wed Jul 19, 2006 12:19 am Post subject:
Re: A mathematical game  probability questions



One further question on this.
Lets tweak the rules somewhat.We're now not interested in working out
the %age chance of each player winning.
What we want to know now is "what is the %age chance of the game ending
on a given number".
But its a slightly different game in this scenario. The rules are
tweaked like this....
The new board is
3 2 1 0 1 2 3 4 5 6 7 8 9 etc
We want to determine which number the game will end on.
In this new game, we keep a record of the highest point the counter has
been on. The end game figure is 3 below that highest point..
So when we start (at 0), the end game figure is 3. When the counter
lands on 3 then game over.
Yet should the counter move up to 2, this is the new highest point, and
so the end game figure becomes 1.
If the count drops from its high point of 2 to 1, the end game point
remains at 1 (still 3 below the highest point of 2).
And so on.
How do I work out the %age chance of the game ending on each number?
Thanks
Al
Alun wrote:
Quote:  Excellent. Merci beaucoup!
A bientot
Alain ;)
Patrick Coilland wrote:
Alun nous a récemment amicalement signifié :
Patrick
Forgive me, I'm not a mathematician. I'm almost grasping the equation
but in the formula
p(a) = 1  a/d
what does a represent. Is a the target for player a?
Yes :
In my explanation, I wrote :
Let d be the distance between the 2 targets.
Let p(n) be the probability that A wins when the coin is exactly at
the distance n of A.
So, when I write "p(a) = 1  a/d" (maybe p(n) = 1  n/d would have been
clearer), "a" is the "distance" between A and the coin.
So if player A's target was 6 and B's was 3 then the probability of a
winning is
1(6/9) = 0.33
is that right?
Yes
And so if Player A's target was 2 and Player B's was 8 then As chance
of winning is 80% because
1(2/10)=0.8
Yes
Many thanks for your assistance
"De rien" ;)

Patrick 


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Alun science forum beginner
Joined: 19 Feb 2006
Posts: 11

Posted: Mon Jul 17, 2006 3:27 pm Post subject:
Re: A mathematical game  probability questions



Excellent. Merci beaucoup!
A bientot
Alain ;)
Patrick Coilland wrote:
Quote:  Alun nous a récemment amicalement signifié :
Patrick
Forgive me, I'm not a mathematician. I'm almost grasping the equation
but in the formula
p(a) = 1  a/d
what does a represent. Is a the target for player a?
Yes :
In my explanation, I wrote :
Let d be the distance between the 2 targets.
Let p(n) be the probability that A wins when the coin is exactly at
the distance n of A.
So, when I write "p(a) = 1  a/d" (maybe p(n) = 1  n/d would have been
clearer), "a" is the "distance" between A and the coin.
So if player A's target was 6 and B's was 3 then the probability of a
winning is
1(6/9) = 0.33
is that right?
Yes
And so if Player A's target was 2 and Player B's was 8 then As chance
of winning is 80% because
1(2/10)=0.8
Yes
Many thanks for your assistance
"De rien" ;)

Patrick 


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Patrick Coilland science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197

Posted: Mon Jul 17, 2006 7:08 am Post subject:
Re: A mathematical game  probability questions



Alun nous a récemment amicalement signifié :
Quote:  Patrick
Forgive me, I'm not a mathematician. I'm almost grasping the equation
but in the formula
p(a) = 1  a/d
what does a represent. Is a the target for player a?

Yes :
In my explanation, I wrote :
Let d be the distance between the 2 targets.
Let p(n) be the probability that A wins when the coin is exactly at
the distance n of A.
So, when I write "p(a) = 1  a/d" (maybe p(n) = 1  n/d would have been
clearer), "a" is the "distance" between A and the coin.
Quote: 
So if player A's target was 6 and B's was 3 then the probability of a
winning is
1(6/9) = 0.33
is that right?

Yes
Quote: 
And so if Player A's target was 2 and Player B's was 8 then As chance
of winning is 80% because
1(2/10)=0.8

Yes
Quote: 
Many thanks for your assistance

"De rien" ;)

Patrick 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Mon Jul 17, 2006 5:49 am Post subject:
Re: A mathematical game  probability questions



David C. Ullrich wrote:
Quote:  On 16 Jul 2006 10:17:47 0700, "Proginoskes" <CCHeckman@gmail.com
wrote:
Alun wrote:
Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If the
counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?
Yes.
And what if we change the targets.
Google for "gambler's ruin" (with the quote marks).
I don't see what this has to do with that.

If you're not joking, you should check out
http://mathworld.wolfram.com/GamblersRuin.html (the first link that
Google finds, BTW).
 Christopher Heckman
Quote: 
What if the counter has to land on left 6 for A to win, but only on
right 3 for player B to win. Does that means player B has a 66.67%
chance of winning, or is that too simplistic a calculation?
The way I'm viewing it is this................
Player A's chance of Winning is (Player B target) / (Player A Target +
Player B Target)
Am I missing something? Or is this formula correct?
Many thanks for any help
Al
************************
David C. Ullrich 


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Alun science forum beginner
Joined: 19 Feb 2006
Posts: 11

Posted: Mon Jul 17, 2006 1:09 am Post subject:
Re: A mathematical game  probability questions



Patrick
Forgive me, I'm not a mathematician. I'm almost grasping the equation
but in the formula
p(a) = 1  a/d
what does a represent. Is a the target for player a?
So if player A's target was 6 and B's was 3 then the probability of a
winning is
1(6/9) = 0.33
is that right?
And so if Player A's target was 2 and Player B's was 8 then As chance
of winning is 80% because
1(2/10)=0.8
I think I'm understanding this correctly, but I've done these examples
so that if I'm not you can correct me.
Many thanks for your assistance
Al
Patrick Coilland wrote:
Quote:  Alun nous a récemment amicalement signifié :
Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If
the counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?
Yes
And what if we change the targets.
Let d be the distance between the 2 targets.
Let p(n) be the probability that A wins when the coin is exactly at the
distance n of A.
You have :
p(0) = 1 The coin is on A target !
p(d) = 0 The coin is on B target !
p(a) = (1/2)p(a1) + (1/2)p(a+1)
The last expression gives
p(a+1)  p(a) = p(a)  p(a1) = c
So p(a) = p(0) + a*c = 1 + ac
And p(d) = 1 + dc = 0 ==> c = 1/d
So p(a) = 1  a/d

Patrick 


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David C. Ullrich science forum Guru
Joined: 28 Apr 2005
Posts: 2250

Posted: Sun Jul 16, 2006 7:45 pm Post subject:
Re: A mathematical game  probability questions



On 16 Jul 2006 10:17:47 0700, "Proginoskes" <CCHeckman@gmail.com>
wrote:
Quote: 
Alun wrote:
Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If the
counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?
Yes.
And what if we change the targets.
Google for "gambler's ruin" (with the quote marks).

I don't see what this has to do with that.
Quote:   Christopher Heckman
What if the counter has to land on left 6 for A to win, but only on
right 3 for player B to win. Does that means player B has a 66.67%
chance of winning, or is that too simplistic a calculation?
The way I'm viewing it is this................
Player A's chance of Winning is (Player B target) / (Player A Target +
Player B Target)
Am I missing something? Or is this formula correct?
Many thanks for any help
Al

************************
David C. Ullrich 

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Patrick Coilland science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197

Posted: Sun Jul 16, 2006 5:33 pm Post subject:
Re: A mathematical game  probability questions



Alun nous a récemment amicalement signifié :
Quote:  Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If
the counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?

Yes
Quote:  And what if we change the targets.

Let d be the distance between the 2 targets.
Let p(n) be the probability that A wins when the coin is exactly at the
distance n of A.
You have :
p(0) = 1 The coin is on A target !
p(d) = 0 The coin is on B target !
p(a) = (1/2)p(a1) + (1/2)p(a+1)
The last expression gives
p(a+1)  p(a) = p(a)  p(a1) = c
So p(a) = p(0) + a*c = 1 + ac
And p(d) = 1 + dc = 0 ==> c = 1/d
So p(a) = 1  a/d

Patrick 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 5:17 pm Post subject:
Re: A mathematical game  probability questions



Alun wrote:
Quote:  Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If the
counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?

Yes.
Quote:  And what if we change the targets.

Google for "gambler's ruin" (with the quote marks).
 Christopher Heckman
Quote:  What if the counter has to land on left 6 for A to win, but only on
right 3 for player B to win. Does that means player B has a 66.67%
chance of winning, or is that too simplistic a calculation?
The way I'm viewing it is this................
Player A's chance of Winning is (Player B target) / (Player A Target +
Player B Target)
Am I missing something? Or is this formula correct?
Many thanks for any help
Al 


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Alun science forum beginner
Joined: 19 Feb 2006
Posts: 11

Posted: Sun Jul 16, 2006 12:25 pm Post subject:
A mathematical game  probability questions



Hi
Here is the game.......
There are two players.
The board looks like this
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
The counter starts each game at Zero
Player A is to the Left and Player B is to the right.
A coin is tossed. Every time the coin is Heads the counter moves one
space to the left. Every time the coin is Tails it moves one space to
the right.
If the counter lands on the left 5, then player A is the winner. If the
counter lands on the right 5, player B is the winner.
Am I right in assuming therefore that Player A has a 50% chance of
winning and Player B has a 50% chance of winning?
And what if we change the targets.
What if the counter has to land on left 6 for A to win, but only on
right 3 for player B to win. Does that means player B has a 66.67%
chance of winning, or is that too simplistic a calculation?
The way I'm viewing it is this................
Player A's chance of Winning is (Player B target) / (Player A Target +
Player B Target)
Am I missing something? Or is this formula correct?
Many thanks for any help
Al 

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