Can anyone solve these Brain Teasers?
1) Counterintuitive Cards -- You have 7 cards: two are red on both sides, two are black on both sides, and three are red/black. You choose a card at random, and one of its sides is red. What is the probability that the other side is also red? (Hint: The answer is not 2/5, though that's what everyone thinks.)
2) Filling a Bag -- You place some red and some blue marbles in a bag. The probability of drawing two marbles of the same color is 1/2. How many marbles of each color are in the bag? (Hint: The correct answer does NOT have the same number of red and blue marbles.) Note that there are many possible answers. Try to find one. Then try to see if you can find others and identify a pattern.
3) You will be paid $1 times the number you roll on a six-sided die. You can re-roll the first time or the second time if you want, but must keep the third roll if you do so. How much can you make on average playing this game and what is your re-rolling strategy?
I don't quite understand this logic. I'm just thinking that if I place all of the cards with red faces up, I will only have five of those (discounting the 2 red faces that are downwards, since they are the outcomes). If I flip them all over, two of the five will have red faces on the opposing sides.
Therefore, 2/5 right?
I'm not very good with probability though.
It's not about the cards, but the faces -- a card with both sides red may have either side facing up and still be red side up, while the mixed cards must have a particular face up.
For #1, I think it's 2/5.
If you KNOW that one side is red, then you know that it is one of five cards (2 double red or 3 red/black). Out of the five cards, there are only two cards that can possibly be red on the other side.
You can rephrase this in a simpler way:
There are five cards. Two are red and three are black. You pick one at random without looking. What is the probability of picking a red.
1 : 2/7, since you made a random choice before having any info and there are 2 cards that are red on both sides.
OP, I think you may have worded #1 wrong? I can't see any other way around it being 2/5.
Since one face is red, it can't be Black/Black= 7-2; so it's out of 5.
Since you're looking for the other side being red, that leaves only two possibilities. 2/5
I get 4/7 also. Tried it using Bayes at first but that seems to result in a circular thing where you need the probability of the top being red given that the bottom is red, which is symmetrical to the original problem.
Each of seven cards can be selected, and either side A or side B can be on top. There are 14 distinct possibilities for the top card -- 7 cards * 2 orientations (A side up and B side up). The rest is as SirTradesaLot said -- 7 of those orientations are red, and 4 of them are paired with a red back.
One roll left: 50% threshold is obviously 3.5. Re-roll if 3 or under. Two rolls left: p-value of 4 or less is 4/6 * 4/6 = 4/9, which is less than half. p-value of 5 or less is 5/6 * 5/6 = 25/36, which is greater than half. The 50% threshold is somewhere in between, so re-roll if you have 4 or less with two rolls left.
Glad you took a crack at this. I'm too lazy ... will attempt later.
either I have misunderstood the question, or this is just a simple http://en.wikipedia.org/wiki/Monty_Hall_problem
1: 2/7, you have a 5/7 chance of picking a card with a red side, then given that the first side is red, you have a 2/5 chance of the card having a second red side, 5/7*2/5=2/7
This is conditional on the fact that you already drew one card and one of its side is red.
1 Def of conditional prob: P(othersidered | first side red) = P(othersidered and firstsidered)/P(first side red) =
(2/7) / [(2/7) + .5*(3/7)] = 2/(2+1.5) = 4/7
2 Suppose there are R red, B blue, T = R+B. Then 1/2 = ([R(R-1)] + [B(B-1)])/T*(T-1) . Substituting T = R+B will give a relation between R and B.
Solving we get the relation R^2+R(-1-2B) + (B^2-B) = 0, thus by the quadratic formula we require 8B+1 to be a perfect square. One can do more number theory but fuck that.
To get one particular solution, let us set B = 3. Then R = 1. Notice that the probability is 0 + 32 / 43 = 1/2 as desired. Other solutions are B = 6, R = 10 or B = 6, R = 3.
3 The expected value of a game with one reroll option is .5 * 3.5 + (4+5+6)/6 = 4.25
Therefore in a game with two reroll options, if you get 1-4 you should roll again. The value of the game is then (4/6) * 4.25 + (5+6)/6 = 4 and 2/3.
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Not sure there is a single answer to #3 unless more information is given on risk appetite. An assumption one can make to simply the problem is to be risk neutral (i.e. maximize expected payoff). Without this information, there is not necessarily a "right" answer.
In any case, you should work from the end towards the front. Step A: If you decide to do your third roll, your expected payoff is $3.5. Step B: Once you roll the second time, if you roll a $4 or higher, you keep it If you roll a 1,2,3, you take the third roll.
Step C: Your expected payoff prior to your second roll is: (1/6)6+(1/6)5+(1/6)4+(1/2)3.5 = 4.25 (check my math there) Step D: After your first roll, if you have rolled a 5 or 6, you take your payoff. Anything else, you roll again. Therefore, you're expected payoff from the game is: (1/6)6+(1/6)5+(4/6)4.25 = 4.667 (check my math again)
Therefore, for a risk neutral player, the expected payoff from the game is $4.667. The strategy for taking your payoff or rolling again is described above.
4/7 is the answer for sure. Think about it logically. There were 7 possible equally like red sides you could have selected. 4 of them have red on the other side, 3 have black on the other side.
For #3 you get 3 rolls: The EV for 1 die is 3.5. For 2 roles it's (11+23+35+47+596611)/36= 161/36 which is between 4 and 5. So after one roll you'll roll again if it's under 5, after 2 rolls you'll roll again if it's under 4. So long as your EV > your current # you'll roll again, because in the long run you'll make more money, specifically it will converge to EV#trials
to i.short.you.you You can't calculate the strategy because it assumes independence of rolls. This isn't true in this case since you are given the option to roll or not.
Further, your expected payoff will not converge to EV*#trials where EV=3.5. This is because if you go towards and infinite number of trials, your expected payoff actually converges to 6, because with an infinite number of rolls you would just keep rolling until you roll a 6 and then stop.
You're right that my EV is off. If you do it out: ((5+6)/2)(1/3)+(2/3)(1/2)((4+5+6)/3)+(1/3)((1+2+3+4+5+6)/6)=(11/6)+(15/9)+(21/18)=(33+30+21)/18=84/18=168/36, not the 161/36 I had before. The re-rolling strategy is the same. If not 5 or higher roll the die a 2nd time, if not 4 or higher roll a 3rd time.
As for what I said about the expected payoff, I was referring to an infinite number of trials of the OP's description, not rolls. I guess it doesn't converge in the mathematical sense, but if you play the OP's game enough and divide it by the number tirals the game was played you should maximize your EV to 168/36.
got the marble one, I believe.
R is # of red marbles. B is # of blue. Call X the # of total marbles (R+B) P(R1) is probability of red marble on first pick. P(R2) is probability of red marble on second pick. P(B1) is probability of red marble on first pick. P(B2) is probability of red marble on second pick.
P(R1)=R/X P(R2)=R-1/X-1 P(B1)=B/X P(R2)=B-1/X-1
P(R1)P(R2)+P(B1)P(B2)=(1/2) Simplifies to R^2+B^2-2RB-R-B=0 Then to (R-B)^2-R-B=0 Then sub in to get X^2+X-2R=0 AND X^2-X-2B=0 B=(X^2/2)-(X/2) R=(X^2/2)+(X/2)
Any combination of that where R and B are both 2 or greater work I think.
Also, for the card one: People are arguing based off a misunderstanding. Many people are answering the question as if they picked a random card, and the side they saw was red - that may be the intention of the question. But the actual phrasing is: "one of its sides is red", as opposed to "the side you see is red".
Based off the actual question, it looks like 2/5 to me.
The word 'also' implies that you are seeing red now. Refer to my earlier reply for longer description (too lazy to retype).
A less ambiguously worded question would be something like this:
You have 5 cards in a stack. 2 of them are red on both sides and 3 of them are red on one side and black on the other side. The cards have been shuffled. If you pull a card at random and the top is red, what are the odds that the other side is also red?
Careful on the first one. The important consideration is that the selected card has at least one red side, meaning the side facing up can be black or red. This suggests two cases, with the total probability being their sum (by the total law of.probability).
Say the side facing up is black. Then, the probability that the other side is red is 1, and since there are 3 black faces out of 10 for cards with at least one red side, I get (1)*(3/10)=3/10
Now, say the side facing up is red. The other side can be red only if one of the two R/R cards was selected with 2/5 probability, and with 7/10 faces being red, I get (1)(7/10)(2/5) = 7/25
Thus, 3/10 + 7/25 = 29/50
i.short.you.you...
right on!
1) There's seven cards, two of which when found to be red side up will also be red side down and three of which when found to be red side up will be black side down. So 3/7 cards fit the definition of not being a match for what we're looking for. Mutually exclusive outcomes means the portion of cards that fit our definition is 4/7. Did anyone confirm an answer for this?
Idk, maybe I'm just stupid. However this is just one of my questions:
If you only have five cards, this assumes that your odds will be out of 5 draws (given that the cards only have two sides). Therefore the number of cards (5), dictates the number of outcomes possible (5) which is the numerator of a probability ratio.
Am I missing something? Still feel like I'm the only one stuck on 2/5.
An alternative way to do #3:
Create a function where f(x,y) is equal to the average payoff and x and y represent the minimum number you would take on the first and second turn respectively.
When simplified, the formula reduces to: f(x,y) = (y^2-6x^2-xy^2+8xy+41x-8y+217)/72
As mentioned by others earlier, the best move would be to roll a second time if the first roll yields less than 5 and to roll a third time if the second roll yields less than 4.
f(5,4) = 336/72 = 14/3 = 4.6667
The reason the answer is not 2/5ths for the first question is because, picking a card at random and finding the upturned side to be red, it is more likely you picked a card with two red sides than with one red side - in particular, it is twice as likely you picked a card with two red sides.
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