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Suppose you play a game where you role a single dice and [#permalink]
 03 Nov 2009, 21:23
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Suppose you play a game where you role a single dice and whatever you roll you get that dollar amount. For example, if you roll a 1 you get $1. If you roll a 6 you get $6. If you are unhappy with the first roll, you can roll again. However, if you get lower the second time, you cannot take the first roll. If you are unhappy with the second roll you can roll a third and final time. Again, if you get lower on the third roll, you have to keep this roll and cannot take the first or second roll.
What is the expected value of this game?
For example, if you could only roll one time, the expected value would be 1/6*(1+2+3+4+5+6) = 3.5. However, with three opportunities to roll higher, this value should increase.
Last edited by baileyf16 on 03 Nov 2009, 21:58, edited 1 time in total.
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Re: Tough probability question [#permalink]
03 Nov 2009, 21:58
The value of the game will not change. Every time a dice is thrown, the value of the game will be the latest result having no resemblance on the earlier result.
The probability of scores rising is similar to the probability of scores declining and hence the value of the game will remain constant in all the 3 attempts.
Consider this addition info:
One needs to consider the cost of each throw. The opportunities given to the player are three-fold, hence we can say, player has 3 options in which he can probably win upto $6. If there is cost assigned to each throw, same needs to be deducted from the result, which is a probable as compared to the cost which is certain.
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Re: Tough probability question [#permalink]
03 Nov 2009, 22:13
baileyf16 wrote: Suppose you play a gain where you role a single dice and whatever you roll you get that dollar amount. For example, if you roll a 1 you get $1. If you roll a 6 you get $6. If you are unhappy with the first roll, you can roll again. However, if you get lower the second time, you cannot take the first roll. If you are unhappy with the second roll you can roll a third and final time. Again, if you get lower on the third roll, you have to keep this roll and cannot take the first or second roll.
What is the expected value of this game?
For example, if you could only roll one time, the expected value would be 1/6*(1+2+3+4+5+6) = 3.5. However, with three opportunities to roll higher, this value should increase. Notice that you won't see anything like this on the real test.Expected value for one roll is 1/6*(1+2+3+4+5+6)=3.5$. This part is easy. Let's proceed: if it's 3.5$, when will I agree to continue the game and roll the second time? Well obviously if the first roll will give me the value less than 3.5 or 1, 2, 3. In other cases if I get 4, 5 or 6, it won't be clever to continue as I already have from my first roll the value higher than the expected value of the second. So what is the expected value of two rolls: 1/6*6+1/6*5+1/6*4+1/2*3.5=2.5+1.75=4.25 The same for the three rolls: if I know that expected value for two rolls is 4.25$, so I'll continue to roll if the first roll will give me the value less than 4.25 or 1,2,3,4. So for three rolls expected value must be: 1/6*6+1/6*5+ 2/3*4.25=1.833+2.833=~4.67$
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Re: Tough probability question [#permalink]
05 Nov 2009, 19:18
Good question, +1. I believe the expected value of the game to be $4.67. Here's my thought process:
The expected value of any given roll is 3.5. However, when deciding whether or not to forsake the first roll, you need to determine the expected value of playing the game with two rolls (since you have a 2nd and 3rd roll option) NOT just one roll. So let's isolate the game to two rolls.
Expected Value of Game with 1 roll: 3.5.
Expected Value of Game with 2 rolls: (1/2)(5) + (1/2)(3.5) = 4.25 (you keep the first roll if its 4, 5, or 6)
Now let's consider a game with three rolls:
You only keep the first roll if its 5 or 6. Otherwise, you dump it, since the expected value of a game isolated to two rolls is 4.25.
Therefore, EV = (1/3)(5.5) + (2/3)(4.25) = 14/3 = 4.67.
Edit: Just realized Bunuel used almost this exact same logic, lol. Good answer.
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Re: Tough probability question [#permalink]
03 Dec 2009, 07:29
Hi! I have two questions in caps as below. Please explain.
Let's proceed: if it's 3.5$, when will I agree to continue the game and roll the second time? Well obviously if the first roll will give me the value less than 3.5 or 1, 2, 3. In other cases if I get 4, 5 or 6, it won't be clever to continue as I already have from my first roll the value higher than the expected value of the second. So what is the expected value of two rolls:
1/6*6+1/6*5+1/6*4+1/2*3.5=2.5+1.75=4.25
PLEASE EXPLAIN HOW AND WHY DID WE MULTIPLY 1/2X3.5
The same for the three rolls: if I know that expected value for two rolls is 4.25$, so I'll continue to roll if the first roll will give me the value less than 4.25 or 1,2,3,4. So for three rolls expected value must be:
1/6*6+1/6*5+2/3*4.25=1.833+2.833=~4.67$
SIMILARLY HERE WHY 2/3*4.25. I AM JUST UANBLE TO GET IT. PLEASE HELP
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Re: Tough probability question [#permalink]
04 May 2011, 09:07
As we all know this forum is intended solely for GMAT aspirants. Unless I am totally ignorant, these sort of questions usually are not presented in Official GMAT exam. Please do not dilute the utility and value of this forum by posting such questions. I leave it for the moderator and others to decide.
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Re: Tough probability question
[#permalink]
04 May 2011, 09:07
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