Split Prize Money for Head & Tails Game: 6 Wins vs 4 Wins Ratio

[h0llow]

Homework Statement

Game of Head & Tails between 2 people. The first one to 8 is the winner. If the game is interrupted: when one person has 6 wins and the other has 4 wins, how should the prize money be split?


2. The attempt at a solution

The obvious answer (at first) i got is 3:2
But it just seems too easy.. i think the part 8 wins should fit in somewhere but i have no idea how =/.

 

[tiny-tim]

welcome to pf!

hi h0llow! welcome to pf!

Game of Head & Tails between 2 people. The first one to 8 is the winner. If the game is interrupted: when one person has 6 wins and the other has 4 wins, how should the prize money be split?

if the game was interrupted just long enough to place bets on it,

what should the odds be? ​

 

[h0llow]

assuming it was 100$... person with 6 wins would have 75% chance of winning? and the other would have 50% chance..but if add 75+50 it adds up more than 100 =/

 

[tiny-tim]

... person with 6 wins would have 75% chance of winning? and the other would have 50% chance..

you mean, because 6 is 75% of 8, and 4 is 50% of 8 ?

no!

work out the probability that (starting from 0-0) there are 2 heads before there are 4 tails

 

[h0llow]

huh 0_o? can u rephrase that please =/?

 

[tiny-tim]

if you keep tossing a coin,

work out the probability that (starting from 0-0)

the 2nd head comes before the 4th tail​

(or, if you prefer the opposite, that the 4th tail comes when there are only 0 heads or 1 head)

 

[h0llow]

considering there's a 50% chance..during each toss.. 2nd head has 25% chance? and 4th tail has 6.25% chance?

 

[Ray Vickson]

considering there's a 50% chance..during each toss.. 2nd head has 25% chance? and 4th tail has 6.25% chance?

You need to account for the fact that there might not be a second head at all. Just "guessing" does not work in problems of this type; you need to write things down in detail.

Also: the answer will depend quite a bit on whether the two-heads player tosses first or second.

RGV

 

[tiny-tim]

considering there's a 50% chance..during each toss.. 2nd head has 25% chance? and 4th tail has 6.25% chance?

sorry, but that's no way to solve this

try these …

what is the probability that a coin tossed 4 times will come out all tails?

what is the probability that a coin tossed 5 times will come out 1 head and 4 tails, and that the head isn't the last one?​

 

[h0llow]

(I) 0.5 X 0.5 X 0.5 X 0.5 = 6.25% (1/2 chance it is tail)
(ii) 3.125%?? (im not sure how exactly to do it)

 

[tiny-tim]

what is the probability that a coin tossed 4 times will come out all tails?​

(I) 0.5 X 0.5 X 0.5 X 0.5 = 6.25% (1/2 chance it is tail)

yes

what is the probability that a coin tossed 5 times will come out 1 head and 4 tails, and that the head isn't the last one?​

(ii) 3.125%?? (im not sure how exactly to do it)

write out all the possible combinations

 

[h0llow]

0.5 x 0.5 x 0.5 x 0.5 x 0.5??

 

[Ray Vickson]

0.5 x 0.5 x 0.5 x 0.5 x 0.5??

Don't keep guessing; that is no way to learn. As I said before: write things out in detail. By this, I mean: start enumerating some of the possibilities, until you see a pattern starting to emerge.

The outcome strings can be lists of H and T. If the two-head player (A) starts, then passes the coin to player B, etc., a string such as HHTHT means A gets H on the first toss, then B gets H on the second toss (that is, on B's first toss), the A gets T on the third toss (that is, on A's second toss), etc. Player A wins on his second toss if the outcomes are HHH or HTH. Player A wins on his third toss if the outcomes are HHTHH, HHTTH, HTTHH, HTTTH, THHHH, THHTH, TTHHH, TTHTH, etc. You need to figure out all the probabilities and add them up.

Then, you need to repeat this type of analysis if A goes second.

RGV

 

[h0llow]

alright thank you

 

[h0llow]

so i should write out all the 120 possibilities? (5!)

 

[h0llow]

nvm.. i got it...can u tell me if i am right though?

TT THHT HHHH
HTT HTHT HTHHH
HHTT THHHT THHHH
HHHTT THTH HHTHH
THT HHTHT HHHTH

= 15 possibilities

chance player A (T) wins = 10/15 = 66.67%
chance player B (H) wins = 5/15 = 33.33%

am i right?

 

[h0llow]

nvm.. i got it...can u tell me if i am right though?

TT THHT HHHH
HTT HTHT HTHHH
HHTT THHHT THHHH
HHHTT THTH HHTHH
THT HHTHT HHHTH

= 15 possibilities

chance player A (T) wins = 10/15 = 66.67%
chance player B (H) wins = 5/15 = 33.33%

am i right?

 

[tiny-tim]

hi h0llow!

nvm.. i got it...can u tell me if i am right though?

TT THHT HHHH
HTT HTHT HTHHH
HHTT THHHT THHHH
HHHTT THTH HHTHH
THT HHTHT HHHTH

= 15 possibilities

that looks right

chance player A (T) wins = 10/15 = 66.67%
chance player B (H) wins = 5/15 = 33.33%

no, because not all of them have equal probability

for example, the first one, TT actually is 8 times as likely as THHHH,

because it includes TTHHH TTHHT TTHTH TTHTT THTHH THTHT THTTH and THTTT

(which are all 5 long, so are all equally likely)

ok now try to think how to work out the likelihood of all the others! ​

EDIT: do the Ts look bolder than the Hs to you, or is it just my weird eyesight?

 

[h0llow]

TT = 25% likely
TTHHH,etc.. = 3.125% likely (since there are 5(0.5^5))...X 8
HTT = 12.5%(0.5^3) X 2
HHTT = 6.25% X 4

everything adds up to 100% 0_o..what now..

the t's look fine

 

[h0llow]

is the answer by any chance 81.25% and 18.75% splits in prize?

 

[Ray Vickson]

nvm.. i got it...can u tell me if i am right though?

TT THHT HHHH
HTT HTHT HTHHH
HHTT THHHT THHHH
HHHTT THTH HHTHH
THT HHTHT HHHTH

= 15 possibilities

chance player A (T) wins = 10/15 = 66.67%
chance player B (H) wins = 5/15 = 33.33%

am i right?

There are lots of other possibilities---infinitely many, in fact. So, we need a systematic way to approach the problem. Let us define two functions:
A(a,b) = probability that player A wins the game, given that A goes first and A must win a rounds before B wins b rounds,
and
B(a,b) = probability that A wins the game, given that B goes first and A must win a rounds before B wins b rounds.
Note that both refer to Mr. A winning; the different letters correspond to who starts.

If A starts and wins the round, the coin passes to B; at that point, B will be starting and A needs to win a-1 rounds before B wins b rounds. If A starts and loses the round, A passes the coin to B, at which point B starts and A must win a rounds before B wins b rounds. Thus
A(a,b) = (1/2)*B(a-1,b) + (1/2)*B(a,b).

Next, suppose B starts and A must win a rounds before B wins b rounds. If B wins the starting round, A starts next and must win a rounds before B wins b-1 rounds. If B loses, then A must win a before B wins b. Thus
B(a,b) = (1/2)*A(a,b-1) + (1/2)*A(a,b).

We can start with low values of a and b, and use these to get higher values (illustrated below). You want to compute A(6,4) and B(6,4), which are the probabilities that A will win, depending on who starts

What is A(1,1)? If A wins the first round he wins the game and B does not even get to play a round. This has prob = 1 /2. If A loses the first round, B starts next and must lose his first round, thus passing the coin back to A, and we are back to A(1,1) again. The probability of this is (1 /2)(1 /2) = 1/ 4. Therefore, A(1,1) = 1 /2 + (1/ 4)A(1,1), so A(1,1) = 2/3.

Similarly, B(1,1) = 1/3.

Where does the value A(1,1) = 2/3 come from? Well, when A starts and wins one round before B does, the possible outcomes are W (stop), LLW (stop) LLLLW (stop), LLLLLLW (stop) ,..., which has total probability
$$A(1,1) = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{4}\right) + \frac{1}{2} \left(\frac{1}{4}\right)^2 + \cdots\\ = \frac{1}{2} \left[ 1 + \left(\frac{1}{4}\right) +\left(\frac{1}{4}\right)^2 + \cdots \right] = \frac{1}{2} \frac{1}{1 – (1/4)} = \frac{2}{3}.$$
This accounts for infinitely many possible outcomes, but is still pretty simple.

Note that the above calculation is the same as what we would get if we used A(0,b) = B(0,b) = 1 for b > 0, and A(a,0) = B(a,0) = 0 for a > 0.

Anyway, you can now get A(2,1), B(2,1), A(1,2), B(1,2) and then A(2,2) and B(2,2). Then you can get even more values in a similar way. Just keep going until you arrive at A(6,4) and B(6,4). This will involve a lot of simple calculations; you could, for example, do it in a spreadsheet.

RGV

 

[tiny-tim]

hi h0llow!

(just got up :zzz:)

TT = 25% likely
TTHHH,etc.. = 3.125% likely (since there are 5(0.5^5))...X 8
HTT = 12.5%(0.5^3) X 2
HHTT = 6.25% X 4

everything adds up to 100% 0_o..what now..

the t's look fine

is the answer by any chance 81.25% and 18.75% splits in prize?

(that's 13/16 and 3/16) yes

let's go over that again, to see how you got there, and so that you can do it more quickly next time

(but first, please don't use % or decimals, it's much easier to use fractions such as 3/32 … and it's easier to check it afterwards, for mistakes! )

ok, the game can't last more than 5 tosses,

so there are only 2⁵ = 32 possible equally-likely outcomes

so in this case you can just list all the winning outcomes, add them up and divide by 32

(or, if that's quicker, list all the losing outcomes, add them up and divide by 32, and then subtract from 1 …

in this case, 4 "fives" = 4*1/32, 1 "four" = 1*1/16, total 6/32 = 3/16)

of course, if the question was more complicated, listing and counting would take too long :yuck:, and so you'd need to use the systematic methods illustrated by Ray!​