Alfred and Bob toss a coin alternatively, and whoever gets a head first will win. They play many times and whoever loses in the previous play will toss first at the next play. If Eric tosses first at the first play, then what is the probability that Eric will win at the 6th play?
1/2
http://drdavespsychologypage.homestead.com/Odds__Figuring.pdf
Yeah........ 1/2 that's an easy one.
To find the probability that Eric will win at the 6th play, we need to consider the possible outcomes at each play leading up to the 6th play.
Since Eric tosses first at the first play, there are two possible outcomes: H (Heads) or T (Tails). If he gets a Head (H) on the first play, he wins immediately. However, if he gets a Tail (T), then Bob will toss next.
On the second play, Bob has to get a Head (H) in order to win, otherwise, Eric will toss again. If Bob gets a T, then it goes back to Eric for the third play.
This pattern continues until the 6th play. At each play, one of the players needs to get a Head (H) to win, otherwise, it goes to the next player.
To calculate the probability, we can break it down into two cases:
Case 1: Eric wins on the first play
The probability of this happening is simply the probability of Eric getting a Head (H) on the first toss, which is 1/2.
Case 2: Eric does not win on the first play
For each subsequent play, the players need to get a Tail (T) before it goes back to Eric again. Since the probability of getting a Tail is also 1/2, the probability of this happening on each play remains constant.
Therefore, the probability of Eric not winning on the first play and then winning on the second play, and so on until the 6th play can be calculated as:
(1/2 * 1/2 * 1/2 * 1/2 * 1/2) = 1/32
Since there are 2 cases, we need to add the probabilities of both cases to get the final probability:
Probability of Eric winning at the 6th play = Probability of Case 1 + Probability of Case 2
= 1/2 + 1/32
= 16/32 + 1/32
= 17/32
Therefore, the probability that Eric will win at the 6th play is 17/32.