5/30/2023 0 Comments Coin flip probability![]() ![]() (ii) P(getting two heads) Number of times two heads appeared P(getting three heads) Number of times three heads appeared ![]() Let E 1, E 2, E 3 and E 4 be the events of getting three heads, two heads, one head and zero head respectively. Number of times three heads appeared = 21. (i) three heads, (ii) two heads, (iii) one head, (iv) 0 head. So, by definition, P(F) = \(\frac\).Ģ. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. = 3 (as HH, HT, TH are having at least one head). = Number of outcomes having at least one head (i) Number of favourable outcomes for event E The possible outcomes are HH, HT, TH, TT. = P(T) = total number of possible outcomesġ. Therefore, P(getting a tail) Number of favorable outcomes (ii) If the favourable outcome is tail (T). = P(H) = total number of possible outcomes Therefore, P(getting a head) Number of favorable outcomes These turn out to be the only ways to win with this strategy, leading to a (4 3 3)/16 = 62.5% > 50% chance of winning.(i) If the favourable outcome is head (H). Again, this accounts for 3/16 new win cases (again the "A:01 and B:01" being the case that was already counted). Moreover, since A and B's strategies are the same, if they get the same sequence of flips, they'll win. If B has "01", they win for the same reason, accounting for another 3/16 victory cases (since the "A:01 and B:01" case was counted above). When examining the possible first 4 flips of the warden, the case above accounts for 4/16. If A has "01" as their first two flips, and B has anything, then they'll win since A will pick the first flip from B, and B will pick "0" from A if B has a 0 as their first flip, and "1" if B has a 1 as their first flip. ![]() You can draw out a table of possibilities to determine there is a 62.5% chance of winning with this strategy, but I'll describe the victory cases below anyways: Otherwise, if it's heads (represented by a 1), they pick the second flip they can't see. If it's tails (represented by a 0), they pick the first flip they aren't allowed to see. ![]() This is exactly the sum I had before.Įach person looks at the first flip they are allowed to see. Therefore, the overall probability of winning is If $i\neq j$, then the conditional probability of winning is exactly $\frac12$. In this case, they are guaranteed to win. Player $B$ finds the smallest $j$ for which $B_j$ is heads, and points to $A_j$. Player $A$ finds the smallest $i$ for which $A_i$ is heads, and points to $B_i$. Say that player $A$ sees the flips $A_1,A_2,A_3,\dots,$ and player $B$ sees the flips $B_1,B_2,B_3,\dots$. Let me use a different numbering system for convenience. Although I don't fully understand why.Įdit: a 70% chance of winning has been found on my Puzzling StackExchange duplicate post! After running this on a computer simulation I get a 60% winrate. A sees a tail on coin flip 2 and 4 so he picks 3, B does the same. I recently tried this problem in university and I am interested to see other people's solutions, and perhaps the optimal solution.Įdit: My strategy is that both players agree that as soon as they see 2 tails in a row for themselves, they pick the number in the middle, i.e. The Warden told them what they were going to do and let them agree upon a common strategy in advance but after that they can't communicate.įind a strategy such that the chance of winning is higher than $0.5$. If they are different, they will spend life in jail. If the outcomes of the trials A and B picked are the same, the prison warden will release them. Now A and B are separately told to pick a trial whose outcome they don't know, i.e., A is supposed to pick an even number, and B is supposed to pick an odd number. They flip a fair coin an infinite number of times and told outcomes of odd trials to A and even trials to B. Known for being exceedingly smart, the prison warden set a test for them. They were then locked in two separate rooms. 2 criminals A and B, were recently captured and brought to prison. ![]()
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