A coin is flipped repeatedly. One man bets that the sequence TT will come up first, the other bets on HT instead. Is there a difference in odds?
2007-07-15 09:13:16 · 8 answers · asked by Scythian1950 7 in Science & Mathematics ➔ Mathematics
A coin is tossed repeatedly in an endless sequence until either TT or HT comes up first.
2007-07-15 09:19:17 · update #1
Consider the four equally likely outcomes of the first two tosses.
HH, HT,TH,TT
If HT occurs one man wins, if TT occurs the other man wins. Now let's look at the cases in which TH occurs or HH occurs. In either case, the man who bet on TT cannot win, because the man who bet on HT will win as soon as a tail is tossed, and the man who bet on TT needs two tails to win. So the probability that the man who bet on TT wins is 1/4, while the HT man will win with probability 3/4.
2007-07-15 10:20:06 · answer #1 · answered by Anonymous · 0⤊ 0⤋
From the first posting of this problem:
Nice problem. There is a difference. Just examine the sample spaces for small numbers of coins. Say Ted bets TT, Hattie bets HT.
HH
HT Hattie wins
TH
TT Ted wins
In two flips, each has a 1/4 chance of winning.
HHH
HHT Hattie wins
HTH (Hattie already won)
HTT (Hattie already won)
THH
THT Hattie wins
TTH (Ted already won)
TTT (Ted already won)
With three coins, Hattie gains a 2/8 probability of winning, while there is no new chance of a win for Ted (he's already won with TTT, and HTT does no good). So at this point, Hattie has a 1/2 probability of winning, Ted still 1/4.
With four coins, Hattie has new wins on HHHT and THHT, giving probability of 5/8 now, while Ted is still at 1/4.
In fact, Ted's probability of winning will never increase. If he doesn't win in two tosses with TT, that means there's an H in the sequence. Before there can be another TT, there has to be an HT, and Hattie will win.
Just to be thorough, with n tosses (n >= 2), only H...H and TH...H are tosses where no one has won. P(Ted wins) = 1/4, P(Hattie wins) = 3/4 - 2/2^n, leaving probability 2/2^n of needing to flip again.
2007-07-15 10:16:06 · answer #2 · answered by brashion 5 · 1⤊ 0⤋
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2017-01-27 01:12:15 · answer #3 · answered by Roshan 1 · 0⤊ 0⤋
Hi. No. They both have a 1 in 4 chance. (TT, TH, HT, HH.)
2007-07-15 09:17:34 · answer #4 · answered by Cirric 7 · 0⤊ 1⤋
there is no difference in odds because its a coin with two side unless or if are counting out the heads and tails. the chance will be fifty-fifty.
2007-07-15 09:27:27 · answer #5 · answered by Anonymous · 0⤊ 0⤋
P(T T ) =1/4
P(h t in that order ) = 1/4
P(H Tin any order ) = 2/4 = 1/2
2007-07-15 09:19:19 · answer #6 · answered by harry m 6 · 0⤊ 0⤋
well you have a better chance getting HT.. i think, then TT.
2007-07-15 09:17:28 · answer #7 · answered by Kassidee 3 · 1⤊ 1⤋
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2014-04-14 16:47:29 · answer #8 · answered by Anonymous · 0⤊ 0⤋