A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.
1- Find the probability that the person who tosses first will win the game?
2- What are the odds against A's losing if she goes first?
difficult probablity !
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- ashish1354
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Let's assume A goes first. We can break it down:
Either:
A wins on first toss: 1/2
or A tails on first toss AND B wins on second toss: (1/2)(1/2) = 1/4
or A tails on first toss, B tails second toss, A wins on third: (1/2)(1/2)(1/2) = 1/8
or A tails, then B tails, then A tails, then B wins: 1/16
and so on.
So the probability A wins should be (since the game lasts at most 5 tosses each):
1/2 + 1/8 + 1/32 + 1/128 + 1/512 = 341/512
and the probability B wins should be:
1/4 + 1/16+ 1/64 + 1/256 + 1/1024 = 341/1024
and of course there's a 1/1024 chance that neither A nor B wins (if they both get tails, ten times).
Because there are three possible outcomes: A wins, B wins, or neither wins, one needs to be a bit careful calculating the odds against A losing. These odds should be equal to the ratio of:
(The probability A does not lose) to (the probability A does lose)
I presume the probability that A loses is equal to the probability that B wins- that is, 341/1024- so the odds against A losing should work out to
683 to 341
I wonder, however, if the question is really asking for the odds that A wins, assuming someone wins (that is assuming A and B do not both get tails every toss). Then the answer is much simpler:
682 to 341, or 2 to 1.
Either:
A wins on first toss: 1/2
or A tails on first toss AND B wins on second toss: (1/2)(1/2) = 1/4
or A tails on first toss, B tails second toss, A wins on third: (1/2)(1/2)(1/2) = 1/8
or A tails, then B tails, then A tails, then B wins: 1/16
and so on.
So the probability A wins should be (since the game lasts at most 5 tosses each):
1/2 + 1/8 + 1/32 + 1/128 + 1/512 = 341/512
and the probability B wins should be:
1/4 + 1/16+ 1/64 + 1/256 + 1/1024 = 341/1024
and of course there's a 1/1024 chance that neither A nor B wins (if they both get tails, ten times).
Because there are three possible outcomes: A wins, B wins, or neither wins, one needs to be a bit careful calculating the odds against A losing. These odds should be equal to the ratio of:
(The probability A does not lose) to (the probability A does lose)
I presume the probability that A loses is equal to the probability that B wins- that is, 341/1024- so the odds against A losing should work out to
683 to 341
I wonder, however, if the question is really asking for the odds that A wins, assuming someone wins (that is assuming A and B do not both get tails every toss). Then the answer is much simpler:
682 to 341, or 2 to 1.
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Isnt the odds against A = 682 : 342 --not assuming that A Loosing = B winning , in which case 2:1
Probably that was a typo.
Anyhow, the odds against A's loosing is 682:342 unless it is given in the question that A's Loosing = B's Winning . Hope I am correct!! [/b]
Probably that was a typo.
Anyhow, the odds against A's loosing is 682:342 unless it is given in the question that A's Loosing = B's Winning . Hope I am correct!! [/b]
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I'm not sure I understand your comment about typos, Rashmi - 341 is half of 682, and 342 is not, so 682:341 is a 2:1 ratio.Rashmi1804 wrote:Isnt the odds against A = 682 : 342 --not assuming that A Loosing = B winning , in which case 2:1
Probably that was a typo.
Anyhow, the odds against A's loosing is 682:342 unless it is given in the question that A's Loosing = B's Winning . Hope I am correct!! [/b]
In any case, the meaning of the question is ambiguous - the intended meaning of the second question is not clear - and it's not much like a real GMAT question. In particular, I've never seen a real GMAT question that required you to know what 'odds' are.
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Ian, I mean that odds against A = 682 : 342, considering the actual meaning of the problem,
However if we take A 's loosing is = B's winning. then it is 682:341 which is 2:1.
However if we take A 's loosing is = B's winning. then it is 682:341 which is 2:1.