# coin tossing

• Jan 30th 2010, 09:39 AM
alexandrabel90
coin tossing
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?

my working:

a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6

b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.

so, (9 choose 7) x ( 0.5)^10

but i didnt get the answer :(
• Jan 30th 2010, 10:02 AM
Quote:

Originally Posted by alexandrabel90
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?

my working:

a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6

b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.

so, (9 choose 7) x ( 0.5)^10

but i didnt get the answer :(

hi alexandrabel90,

for the first one, they both must win 3 times each,
so that's the probability of either winning exactly 3 times and losing exactly 3 times.

for the second one, you must add the probability for the 2 players.
this is different from the first one because one or other could win the game on the 10th go, having won 7 of the preceding 9.
• Jan 30th 2010, 10:19 AM
Plato
As I read the question we have two players each having $6 to play with. Each player either wins$1 or looses $1 on each toss of the coin. So it is possible for one player of loose all in six tosses- loses six straight. So for both to breakeven in six tosses each would wins three and lose three. Is that a misreading of the question? If not, the answer to part a) is $\frac{\binom{6}{3}}{2^6}$. If it is a misreading, please clarify the question. • Jan 30th 2010, 11:24 AM alexandrabel90 it is not a misreading of the question. your answer is correct for part (a)..i never think that we would need to count then chances of wining since there is only a 0.5 chance of a win and lose. for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? • Jan 30th 2010, 12:31 PM Soroban Hello, alexandrabel90! Quote: Two gamblers bet$1 each on the successive tosses of a coin.
Each has $6. .What is the probability that: a. they break even after 6 tosses? They each win 3 games and lose 3 games. There are $_6C_3 \:=\:\frac{6!}{3!\,3!} \:=\:20$ ways this can happen. Answer: . $(20)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\ri ght)^3 \:=\:(20)\,\frac{1}{64} \:=\:\frac{5}{16}$ Quote: b. one player wins all the money on the tenth toss? my working: In order to win all the money of the other player, he needs to win 8 games and loss 2 games, where the 10th toss has to be a win. So: . $_9C_7\left(\tfrac{1}{2}\right)^{10}$ . . . . correct, so far $_9C_7\left(\tfrac{1}{2}\right)^{10} \:=\:\tfrac{36}{1024} \:=\:\frac{9}{256}$ This is the probability that $A$ wins all of $B$'s money on the 10th game. It's also the probability that $B$ wins all of $A$'s money on the 10th game. Therefore: . $P(one\text{ player wins all the money}) \:=\:\frac{9}{256} + \frac{9}{256} \;=\;\frac{9}{128}$ • Jan 30th 2010, 01:28 PM Archie Meade Quote: Originally Posted by alexandrabel90 for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? Yes, that was it, because you calculated the probability of only one out of the two winning on the tenth throw. Soroban has shown that really nicely. Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. In this situation, one could break even on one's own throws but make a profit if the other player does badly, or make a loss if the other player does well, "if" losing a toss loses a dollar and winning a toss wins a dollar. Or, if the two players lost every time, "player A" wins "player B"'s 6 dollars and "player B" wins "player A"'s 6 dollars, so they break even. They could also both win 1 and lose 5 each, win 2 and lose 4 each, win 4 and lose 2 each, win 5 and lose 1 each, win 6 and lose 0 each. That's the thing with probability questions, quite often open to alternative interpretations. The answer book will distinguish given it's answers. We've been interpreting this as..... If one player loses on a throw, then the second player wins the first player's dollar. In this case then, both would have to win 3 times and lose 3 times to end up with the 6 dollars they began with. It is probably safest to assume there is just 6 tosses rather than 6 each. • Jan 30th 2010, 02:27 PM Plato Quote: Originally Posted by alexandrabel90 2 gamblers bet$1 each on the successive tosses of a coin. each as $6. what is the probability that b. one player wins all the money on the tenth toss? Quote: Originally Posted by Archie Meade Yes, that was it, Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. Look at the way part b) is put. It seems to me that in this game on any toss player A gets a dollar from player B or gives a dollar to player B. That is, A wins or looses on any toss. Moreover, the game is over when one player is out of money. If player A wins in each of the first six tosses she has all of the money in six tosses. If B wins five of the first six tosses he has 11 dollars and she has 1. So the question becomes “how can the game be over in exactly ten tosses?” • Jan 30th 2010, 02:35 PM Archie Meade Brilliant analysis, Plato! it's not over yet. • Jan 30th 2010, 03:08 PM Archie Meade I reckon that, in order for one player to win 8 to 2, the score had to stand at 5-1, at which point one player has$10 and the other has \$2 (win 5 and lose 1... 6+5-1=10, win 1 and lose 5.... 6+1-5=2).

Then the score can go to 6-1 or 5-2.

From 6-1 the score must go to 6-2, since 7-1 would end the game.
From there to 7-2 and then to 8-2.

From 5-2 the score goes to 6-2, to 7-2, to 8-2.

Beginning at 5-1, there are two ways for the leading player to win on the 10th throw.

So, we count the number of ways of getting to a 5-1 lead and double that.

$2\binom{6}{5}=2\binom{6}{1}$

$P=\frac{2(6)}{2^{10}}=\frac{3}{2^8}=\frac{3}{256}$

Then double this to get the result for either player winning by 8 to 2.