A player plays a gam against the bank, where she tosses two coins. The player wins if two heads, whilst the bank wins if any other combination appears.
a) Is two heads equally likley to come up as any other single number?
b) if the coints are tossed 60 times, how many times woudl you expect wo heads to come up?
c) if you bet $1 on getting two heads, what would you have to win back so that in the long term you win as much as you lose.
I didn't understand the calculations or formula for this at all, so if you could explain in elementary terms, (make it long if you have to) that would be really helpful.
2006-10-29 10:51:49 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Okay, with two coins your sample space (what combinations you can get when you toss them) is {HH, HT, TH, TT} where H = heads and T = tails
Player wins with HH
Bank wins with HT, TH, TT
a.) Since there is 4 possible outcomes in our sample space and HH only appears 1 time. We know that the probability of two heads = P(HH) = 1/4. This is the same as two tails also P(TT) = 1/4 but the probability of one head and one tail P(TH or HT) =
P(HT) + P(TH) = 1/4 + 1/4 = 2/4. Therefore, NO two heads are not as equally likely as any other combination
b.) To solve this you would take the probability of two heads P(HH) = 1/4. Think of the 60 tosses as 60 marks on the paper ( ie. _ _ _ _ _ _ _ ... _ _ _ _) until the 60th one. the probability of getting two heads on the first _ is 1/4 and the 2nd _ is 1/4 and the third _ is 1/4... and the 60th _ is 1/4 (since each toss is independent of the previous or next toss) therefore the P(HH) for 60 tosses is (1/4) * (1/4) * (1/4) * (1/4)... * (1/4) up to the 60th one. therefore the answer is (1/4)^60
actually not sure if b is correct or not. HOpefully someone can explain better.
c.) just gonna stop here cause I'm not sure...Sorry
I hope what I gave was helpful.
2006-10-29 11:05:10 · answer #1 · answered by rachie_grl6 2 · 0⤊ 0⤋
a) One toss, 4 Possibilities: HH, HT, TH, TT, so
P(HH) = 1/4 = 0.25
b) because 2 coins are tossed each time, you would expect HH to come up only (1/4)*60 = 15 times.
c) With three losses for every win, the odds against HH are 3:1 (You can also calculate odds from probability by O = (1 - P)/P. With more complex problems this becomes easier than counting up all the possibilities)
In order to break even betting $1 on HH, you would need a $3 payout for winning because you are going to lose 3 times as many times as you win.
2006-10-29 19:15:36 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
To clarify the latest answer... you have to win $3 AND get your dollar back.
2006-10-29 20:17:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋