Probability Game quesiton regarding Biassed coin
We're given an unfair coin with the following probability:
H-2/5
T-3/5
Two players play a game where each of them chooses a sequence of 3 tosses of this unfair coin,then the coin is tossed again and again until one of the sequences emerges.
Player A chooses the sequence TTT,is there a sequence Player B can choose such that his probability to win will be greater than half?
TTT seems tempting but H must have accurred before such a sequence completes , but i have no idea how to write it mathematically.
probability
asked Nov 20 at 17:58
user610402
194
add a comment |
We're given an unfair coin with the following probability:
H-2/5
T-3/5
Two players play a game where each of them chooses a sequence of 3 tosses of this unfair coin,then the coin is tossed again and again until one of the sequences emerges.
Player A chooses the sequence TTT,is there a sequence Player B can choose such that his probability to win will be greater than half?
TTT seems tempting but H must have accurred before such a sequence completes , but i have no idea how to write it mathematically.
probability
asked Nov 20 at 17:58
user610402
194
2
Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38
add a comment |
We're given an unfair coin with the following probability:
H-2/5
T-3/5
Two players play a game where each of them chooses a sequence of 3 tosses of this unfair coin,then the coin is tossed again and again until one of the sequences emerges.
Player A chooses the sequence TTT,is there a sequence Player B can choose such that his probability to win will be greater than half?
TTT seems tempting but H must have accurred before such a sequence completes , but i have no idea how to write it mathematically.
probability
asked Nov 20 at 17:58
user610402
194
We're given an unfair coin with the following probability:
H-2/5
T-3/5
Two players play a game where each of them chooses a sequence of 3 tosses of this unfair coin,then the coin is tossed again and again until one of the sequences emerges.
Player A chooses the sequence TTT,is there a sequence Player B can choose such that his probability to win will be greater than half?
TTT seems tempting but H must have accurred before such a sequence completes , but i have no idea how to write it mathematically.
probability
probability
asked Nov 20 at 17:58
user610402
194
asked Nov 20 at 17:58
user610402
194
edited Nov 20 at 19:18
asked Nov 20 at 17:58
user610402
194
asked Nov 20 at 17:58
user610402
194
asked Nov 20 at 17:58
user610402
194
194
2
Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38
add a comment |
2
Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38
2
2
Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38
add a comment |
1 Answer
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Claim: If any of the first three rolls is H, then HTT beats TTT. Hence in the HTT vs. TTT game, TTT wins with probability $(3/5)^3=27/125$ and HTT wins with probability $98/125$.
proof: Suppose at least one of the first three rolls is H, and for contradiction, suppose that TTT appears before HTT. Choose the first occurrence of TTT. By assumption these are not the first three flips of the game. The flip immediately preceding this occurrence of TTT can't be T by choice of "first occurrence" nor can it be H since then HTT would have won. Contradiction.
So TTT only wins if the first three rolls are all T's, which happens 27/125 of the time.
I'm not sure, but it may well be that B can always pick a string to beat A (on average) if she is allowed to choose after A and knows what A chose.
answered Nov 22 at 13:14
Ned
2,003910
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Claim: If any of the first three rolls is H, then HTT beats TTT. Hence in the HTT vs. TTT game, TTT wins with probability $(3/5)^3=27/125$ and HTT wins with probability $98/125$.
proof: Suppose at least one of the first three rolls is H, and for contradiction, suppose that TTT appears before HTT. Choose the first occurrence of TTT. By assumption these are not the first three flips of the game. The flip immediately preceding this occurrence of TTT can't be T by choice of "first occurrence" nor can it be H since then HTT would have won. Contradiction.
So TTT only wins if the first three rolls are all T's, which happens 27/125 of the time.
I'm not sure, but it may well be that B can always pick a string to beat A (on average) if she is allowed to choose after A and knows what A chose.
answered Nov 22 at 13:14
Ned
2,003910
add a comment |
Claim: If any of the first three rolls is H, then HTT beats TTT. Hence in the HTT vs. TTT game, TTT wins with probability $(3/5)^3=27/125$ and HTT wins with probability $98/125$.
proof: Suppose at least one of the first three rolls is H, and for contradiction, suppose that TTT appears before HTT. Choose the first occurrence of TTT. By assumption these are not the first three flips of the game. The flip immediately preceding this occurrence of TTT can't be T by choice of "first occurrence" nor can it be H since then HTT would have won. Contradiction.
So TTT only wins if the first three rolls are all T's, which happens 27/125 of the time.
I'm not sure, but it may well be that B can always pick a string to beat A (on average) if she is allowed to choose after A and knows what A chose.
answered Nov 22 at 13:14
Ned
2,003910
add a comment |
Claim: If any of the first three rolls is H, then HTT beats TTT. Hence in the HTT vs. TTT game, TTT wins with probability $(3/5)^3=27/125$ and HTT wins with probability $98/125$.
proof: Suppose at least one of the first three rolls is H, and for contradiction, suppose that TTT appears before HTT. Choose the first occurrence of TTT. By assumption these are not the first three flips of the game. The flip immediately preceding this occurrence of TTT can't be T by choice of "first occurrence" nor can it be H since then HTT would have won. Contradiction.
So TTT only wins if the first three rolls are all T's, which happens 27/125 of the time.
I'm not sure, but it may well be that B can always pick a string to beat A (on average) if she is allowed to choose after A and knows what A chose.
answered Nov 22 at 13:14
Ned
2,003910
Claim: If any of the first three rolls is H, then HTT beats TTT. Hence in the HTT vs. TTT game, TTT wins with probability $(3/5)^3=27/125$ and HTT wins with probability $98/125$.
proof: Suppose at least one of the first three rolls is H, and for contradiction, suppose that TTT appears before HTT. Choose the first occurrence of TTT. By assumption these are not the first three flips of the game. The flip immediately preceding this occurrence of TTT can't be T by choice of "first occurrence" nor can it be H since then HTT would have won. Contradiction.
So TTT only wins if the first three rolls are all T's, which happens 27/125 of the time.
I'm not sure, but it may well be that B can always pick a string to beat A (on average) if she is allowed to choose after A and knows what A chose.
answered Nov 22 at 13:14
Ned
2,003910
answered Nov 22 at 13:14
Ned
2,003910
answered Nov 22 at 13:14
Ned
2,003910
answered Nov 22 at 13:14
Ned
2,003910
2,003910
add a comment |
add a comment |
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Well, if B chooses HTT after A makes the TTT choice, then A will only win if the FIRST 3 tosses are TTT, which has only 27/125 probability. B will win every other time, 98/125. Convince yourself by writing out some strings of tosses.
– Ned
Nov 20 at 19:19
@Ned - that's an answer, not a comment.
– Paul Sinclair
Nov 21 at 1:38