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Tossing a coin 3 times

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Tossing a coin 3 times

by kukinessa » Fri May 13, 2011 1:50 pm
Please help me with this one:
the coin will be flipped 3 times. For each head you will be paid 200$. Assume that the coin comes up heads with probability 1/3. How to compute an expected value of the game? How much u will be willing to pai to play thi game? construct posibilities and probabilities.


I hope someone will know, cause in my opinion there is still 8 options but i really dont have a clue how to calculate it with 1/3 probability
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by djiddish98 » Fri May 13, 2011 3:51 pm
I think this is right.

Expected Value (EV) = Probability * payout. So the EV is (1/3)*200 for each flip * 3 flips.

As long as I'm paying less than 200 bucks, I'm in, since 1/3*200*3 = 200.

If you want to do it out long form, you can arrive at the same answer.

3 heads = (1/3)(1/3)(1/3) * 600 - you get 600 because you win 200 dollars 3 times
2 heads, 1 tail (1/3)*(1/3)*(2/3)*400*3 - 400 bucks due to 2 heads, times 3 since we can flip heads on the first two, the second two, or the first and the third flips.
1 head, 2 tails - (1/3)*(2/3)*(2/3)*200*3
no heads - (2/3)*(2/3)*(2/3)*0
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by kukinessa » Fri May 13, 2011 4:19 pm
but it is 8 possibilities for the outcome of the flip so doesnt that connected somehow in calculations? shouldn I multiply all the outcomes by 1/8 ?I cannot see the point of probability 1/3 in flipping a coin when it should be 1/2
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by djiddish98 » Fri May 13, 2011 4:25 pm
Pretend that it's a coin that's been modified or magic, so it's 2 times more likely to land on heads than tails.

You're right that a fair coin will land heads/tail 1/2 of the time, but that's not the coin that's been presented here.

Also, there are 8 options presented as you mentioned. all heads (1 option) + heads once (3 options; first second or third flip) + heads twice (3 options) + no heads (1 option). Notice that I'm multiplying the middle two results by 3 to account for how many ways they can occur.
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by kukinessa » Fri May 13, 2011 4:35 pm
Could u also colud help me with the last one? :) this is the the last question connected with that exercise as well: Consider the effect of a change in the game so that if tails comes up two times in a row, you get nothing. the same question but different situation
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by djiddish98 » Fri May 13, 2011 5:23 pm
kukinessa wrote:Could u also colud help me with the last one? :) this is the the last question connected with that exercise as well: Consider the effect of a change in the game so that if tails comes up two times in a row, you get nothing. the same question but different situation
This will impact our 1 heads options here (it doesn't impact the 3 tails scenario, since we already were getting zero in that instance). We'll have to split the category into two possibilities

Scenario 1: 1 heads, heads in the middle - (2/3)*(1/3)*(2/3)*200
Scenario 2: 1 heads, heads on first or last (1/3)*(2/3)(2/3)*0*2

so those 3 possibilities were split up in the following. I think that's the only impact of the new rule.

Previously, we were getting 200 dollars from the second scenario. The probability was 4/27 and it happened twice. So the expected value was 2*(4/27)*200. Now that amount is zero, so the difference in EV is negative of the previous amount or about 60 dollars.

So I wouldn't pay more than 140.
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