Marco can have 1 dollar less than he had before the 3 tosses only in one case, i.e., when there is "H" twice and "T" once.

THH = (1/2)^3
HTH = (1/2)^3
HHT = (1/2)^3

THH + HTH + HHT = 3/8

We need a WLL scencario. it will be 1/2*1/2*1/2 --> 1/8

Now WLL --> 3!/2!1! --> 3 - the number of possible arrangements of WLL. So 3*1/8 -> 3/8

It should be two tails and 1 head.

3 WAYS possible = HTT+TTH+THT

PROBABILITY = 3/ 2^3=3/8

Can only happen.. if we have two Heads and one Tails....

HHT = (1/2)^3 x 3!/2! (Arrange HTT) = 3/8

A combinatoric approach:

There are \(2^3\) total permutations of possible results.

The ones where he ends up with -1$ obviously consist of 2x head, 1x tails. We can calculate the number of ways this can happen (head, head, tails or head, tail, head etc.) with the formula
\(\frac{P^3_3}{2!} = \frac{3!}{2!} = 3\)
(\(P^3_3\) is the number of permutation of 3 different items. However, since 2 items are the same (head & head), we need to divide by the factorial of the number of equal items, so \(2!\))

This gives us a probabilitiy of
\(\frac{\frac{P^3_3}{2!}}{2^3}=\frac{3}{8}\)

Its 3/8

Probability of individual event : 1/2
No. of individual events : 3 therefore 1/2 X 1/2 X 1/2 = 1/8
No. of ways in which we can contain favorable result : 3 they are HTT, THT, TTH
thus 3X1/8 = 3/8
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Bunuel

Should not be Total # of outcomes = 2^3 instead of 2^8 ??
Tks

Would appreciate help with this!

Does number of ways matter here? We are just concerned with the probability of Marco being -1$ and that's 1/8 right?
I'm not clear about why we need to consider the other variations of HHT when we are dealing with a money situation.

Thanks!

I'd say around 600.

Cheers,
Brent
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