Coin probability problem?

Question

Three coins are thrown. If there are either 1 or 2 heads, half of the time, two coins selected at random are thrown again. This process is repeated until either there are 0 or 3 heads, or when half of the time it's left alone. What is the probability of 0 heads (or 3 heads) being the final state?

Thanks ksoileau, for restating the problem, that's what I meant to say. Whenever there's 1 or 2 heads, there is a 50% chance it would be left alone, otherwise 2 out of the 3 coins are chosen at random and tossed again.

ksoileau, the could theoretically repeat endlessly. It doesn't necessarily halt every time after a certain number of throws.

Answer 1

Binomial factors for 3: 1,3,3,1
Binomial factors for 2: 1,2,2,1

Probability of all heads or all tails first roll (Use binomial factors for 3 and divide by 8)
(1 +1)/8= 1/4

Remaining probability
1- 1/4=3/4

Probability of stopping after first round
1/2 * 3/4 = 3/8

Of the remaining 3/8th, probability of matching the coin remaining (use binomial factors for 2 and divide by 4) : 1/4 of remainder

And probability of ending at that point 3/8th of remainder.

It is a simple repeating function. Therefore,

Let X be our normalizing factor

1/X= 1/4+3/8 = 5/8
X=8/5

Therefore probability of either 0 heads or 3 heads being final state

= 1/4 *8/5 = 2/5

Probability of 0 heads

2/5 * 1/2 = 1/5 = 20%

Answer 2

Is the following an equivalent statement of your problem?

Three coins are thrown. If all are heads, 3 heads is the final state. If none are heads, 0 heads is the final state.
If there is 1 head, then with 50% probability, two coins selected at random are thrown again, otherwise the end state is 1 head.
If there are 2 heads, then with 50% probability, two coins selected at random are thrown again, otherwise the end state is 2 heads.
What is the probability of 0 heads or 3 heads being the final state?

If so:
If the original throw produces 0 or 3 heads, we are done and that will happen with probability 1/4.

Now assume the original throw has produced 1 head. Then let's consider the probability of reaching the end state of 3 heads on the second throw. This can only happen if the two tails are selected for the second throw, and then they both come up heads. This will happen with probability 1/3*1/4=1/12. Next let's consider the probability of reaching the end state of 0 heads on the second throw. This can only happen if one of the two tails are not selected for the second throw, and then the selected coins both come up tails. This will happen with probability 2/3*1/4=1/6. So if the original throw has produced 1 head, the probability of reaching an end state of 0 or 3 heads on the second throw is 1/12+1/6=1/4.

Now assume the original throw has produced 2 heads. Then let's consider the probability of reaching an end state of 0 heads on the second throw. This can only happen if the two heads are selected for the second throw, and then they both come up tails. This will happen with probability 1/3*1/4=1/12. Next let's consider the probability of reaching an end state of 3 heads on the second throw. This can only happen if one of the two heads are not selected for the second throw, and then the selected coins both come up heads. This will happen with probability 2/3*1/4=1/6. So if the original throw has produced 2 heads, the probability of reaching an end state of 0 or 3 heads on the second throw is 1/12+1/6=1/4.

So if the original throw produces 1 or 2 heads, the probability that an end state of 0 or 3 heads will be reached on the second throw is 1/4+1/4=1/2. The probability of doing the second throw at all is 1/2, so if the original throw produces 1 or 2 heads, there a 1/2*1/2=1/4 probability of reaching an end state after the second throw of 0 or 3 heads. The probability of the original throw producing 1 or 2 heads is 3/4. Multiplying and adding to the probability that the original throw produced 0 or 3 heads, we get an overall answer of 1/4+3/4*1/4=7/16.

Answer 3

the consequences of the test are pairs (s, t), the place s is between the six rankings, and t is one in each and every of the two consequences (T, H) of tossing a coin. consequently, the pattern area on your test has 6*2 = 12 outcomes. considering the fact that the two the die and the coin are honest, it fairly is shown that each and all the 12 outcomes has the comparable danger, which consequently equals a million/12. the form "An H seems on the coin" incorporates right here six outcomes: {(a million,H), (2,H), ..., (6,H)}. including up their guy or woman possibilities, we acquire the danger of the form: 6/12 = a million/2. the different possibilities could be computed analogously, by ability of explicitly writing down the consequences that characterize the form and summing up their possibilities.