Question

You are given a box of 100 silver dollars, all facing heads up. You are instructed to shake the box 6 times; after each shake, you will remove all the dollars that are heads up before shaking again. You may keep all the dollars that are still tails up following the second shake. How many dollars will you most likely get to keep?

Answer 1

50+25+12.5+6.25+ and on to 6 terms...

Answer 2

Answer:

2 dollars ( approx )

Explanation:

Given,

Initial numbers of dollars = 100,

In each shake the number of removed dollars ( in which head appeared ) = 1,

We have to find the number of dollars after 6 shakes,

Also, the coins in which tail appear will be remained.

∵ Probability of getting tail in each trial = $\frac{1}{2}$

Thus, the total probability of getting tail in 6 trials = $(\frac{1}{2})^6$

Hence, the number of dollars remained = initial number × probability of tail in 6 trials

$=100\times (\frac{1}{2})^6$

$=\frac{100}{64}$

$=1.5625$

$\approx 2$