Anna and bob play a game in which Anna begins by rolling a fair dice, after which bob tosses a fair coin. They take turns until one of them wins. Anna wins when she rolls a 6. Bob wins when the coin lands on heads. What is the probability that Anna will win the game?
Well there are a number of ways for Anna to win like 1st it is 1/6 the second way can be 5/6*1/2*5/6*1/2*1/6 etc. Where do i go from here to find an exact number for the probability?
What happens if Anna rolls the 6 and Bob tosses a head? It it considered a tie, or does one or the other automatically win? This will change the probability of Anna winning the game.
So, if Ana rolls a 6, does the game end instantly, or does Bob get to flip a coin one last time? If he does, and gets a head, does the game continue or do they declare it a tie?
Does the question really ask for the probability that Ana wins the game, or does it ask for the probability that she wins the game within a certain number of rounds?
On Ana's first roll, she has a 1/6 chance of getting a 6.
In order to get to a second roll, she needs to roll something other than a 6 on her first roll, and Bob needs to get a tails. The probability of that happening is 5/6 * 1/2 = 5/12. Then she has a 1/6 chance of getting a 6 on her 2nd roll, so the probability that Ana wins on her second roll is 5/12*1/6 = 5/72.
probability that she wins on her third roll is 5/6*1/2*5/6*1/2*1/6=25/864
So far, the probability that she wins in three rolls or less is 1/6 + 5/72 + 25/864
This looks like a geometric series, with r=5/12.
Sum of an infinite geometric series is a1/(1-r) where a1= the first term = 1/6
a1/(1-r)=(1/6) / (7/12) = 2/7
To find the exact probability that Anna will win the game, you can use the concept of geometric series. Let's break down the different ways Anna can win the game.
1. Anna wins on her first turn (rolling a 6): The probability of this happening is 1/6.
2. Anna does not win on her first turn, but eventually wins after Bob's turn: The probability of Anna not winning on her first turn is 5/6. After Anna's turn, the probability that Bob does not win (coin lands tails) is 1/2. Then the probability of Anna not winning again is 5/6, and the probability of Bob not winning again is 1/2. This pattern continues until Anna wins.
Let's calculate the probability for this second case using a geometric series:
Probability (Anna wins on her second turn) = (5/6) * (1/2) * (5/6) * (1/2) * (1/6)
Probability (Anna wins on her third turn) = (5/6) * (1/2) * (5/6) * (1/2) * (5/6) * (1/2) * (1/6)
And so on...
To find the total probability, you can sum up all these probabilities:
P(Anna wins) = (1/6) + (5/6) * (1/2) * (5/6) * (1/2) * (1/6) + (5/6) * (1/2) * (5/6) * (1/2) * (5/6) * (1/2) * (1/6) + ...
This is an infinite geometric series with the first term being (1/6) and the common ratio being (5/6) * (1/2) * (5/6) * (1/2) * (1/6). To find the total probability, you can use the formula for the sum of an infinite geometric series:
Sum of infinite series = a / (1 - r)
where "a" is the first term and "r" is the common ratio.
So, the probability that Anna will win the game is:
P(Anna wins) = (1/6) / (1 - [(5/6) * (1/2) * (5/6) * (1/2) * (1/6)])
To find the exact probability that Anna will win the game, we can use the concept of geometric series.
Let's denote the probability that Anna wins in one turn as P(A) and the probability that Bob wins in one turn as P(B).
P(A) = 1/6 (because Anna wins when she rolls a 6 on the dice)
P(B) = 1/2 (because Bob wins when the coin lands on heads)
Now, let's consider the probability that Anna wins on her first turn (P1). This is simply P(A) = 1/6.
The probability that Anna wins on her second turn (P2) can be calculated as follows:
P2 = (5/6 * 1/2) * P(A)
Here's the breakdown of the calculation:
- Anna has a 5/6 probability of not rolling a 6 on her first turn, and Bob has a 1/2 probability of not flipping heads on his first turn.
- If neither Anna nor Bob wins on their first turns, the game essentially restarts. So the probability of Anna winning from this point is still P(A).
Similarly, the probability that Anna wins on her third turn (P3) can be calculated as:
P3 = (5/6 * 1/2) * (5/6 * 1/2) * P(A)
Notice that the pattern continues, where the probability of Anna winning on each subsequent turn is the product of (5/6 * 1/2) and the probability of winning from the previous turn.
To find the exact probability, we need to sum the infinite sequence of these probabilities, because the game can potentially go on indefinitely. This can be done using the formula for the sum of an infinite geometric series:
Sum of infinite geometric series = a / (1 - r)
In this case, a represents the first term (P1 = 1/6) and r represents the common ratio (5/6 * 1/2).
Now, let's substitute these values into the formula:
Probability of Anna winning = P(A) + P2 + P3 + ...
Probability of Anna winning = (1/6) / (1 - (5/6 * 1/2))
Simplifying the equation:
Probability of Anna winning = (1/6) / (1/3)
Probability of Anna winning = 1/2
Hence, the exact probability that Anna will win the game is 1/2 or 0.5.