LawLord 10 Posted June 25, 2009 Share Posted June 25, 2009 I assume this is the right board. Difficult to tell what probability maths comes under. I have a game. I start out with 26 cards. 13 of these are red. 13 of them are black. The cards are spread out face down on the floor randomly. Now I must flip over as many black cards while trying to limit the amount of red ones I flip over. Originally, I figured there is a 50 - 50 chance of getting either colour, so what I did was: Randomly select the first card to be flipped. If it was a black one, I would chose another card (only in my head this time, I would assume this other card would be red if I turned it over (seems as the last one was black). I would then ignore this card and flip the next card I chose. Usually this would be a black. etc etc. I became aware that there is a problem with this way of doing things. The number of cards (26) changes everytime I flip one over. Also, depending on what colour cards i've flipped over, the number of the coloured cards on the ground changes also. For example I could end up with 18 cards - 12 being red and only 6 being black. This means there is more than a 50 percent chance of turning over a red card. I have been trying to make up an algorithm to tell me how often I should flip a card in my head and move to the next one (flipping this one over for real). Is anyone able to give me some pointers? What particular part of probability am I working with here? I've come up with about a 5 different ideas, but all are flawed. The nearest I have so far: 1/2 - 1/2 1/4 - 3/4 1/8 - 7/8 1/16 - 15/16 1/32 - 31/32 1/64 - 63/64 1/128 - 127/128 1/256 - 255/256 1/512 - 511/512 1/1024 - 1023/1024 1/2048 - 2047/2048 1/4096 - 4095/4096 1/8192 - 8190/8191 The idea was that I go down the line. First flip: 50 - 50 Second flip: 25 - 75 chance it's the same as the last one, do I want it to be the same as the last card? If I do, flip a card in my head and then flip the next one for real. The problem, as mentioned above: Success causes the number of black cards in the spread to decrease. This totally changes the odds. I don't know how to to deal with this part. Link to post Share on other sites

Kyrisch 169 Posted June 25, 2009 Share Posted June 25, 2009 In fact, your whole idea is fundamentally flawed. No matter how many cards you "pick in your head", the probability of the next card you flip being red or black is based solely on the number of cards of each colour showing and face down. In fact, there are many outright false statements in the OP. First, of course, is the gambler's fallacy, that flipping over a black card first would cause the odds to shift drastically in favour of a red card immediately following. However, given 26 cards, 13 of each colour, the probability that you will choose a red card on the second flip given that one black card is face up already is only 13/(26-1) = 0.52, close enough to 50-50 that it would only make a difference an average of twice in one hundred turns. Second is this idea that because after 18 cards you ended up with 12 red and 6 black, the probability is somehow NOT 50-50; this is patently false. Just because an outcome does not match perfectly with its assigned probability does not mean that the probability is invalid. If you flip a coin four times, it is not very unlikely that you might produce three consecutive heads, and one tails (1 in 16 flips). In fact, this outcome is just as likely as heads, tails, heads, tails (1 in 16 flips) and does not indicate that the probability for heads is something like 3/4 and that of tails is 1/4. Probabilities are outcomes on average. Flip a coin 1,000 times and if the spread is something like 3:1, then you have an issue (or a weighted coin xP). Link to post Share on other sites

LawLord 10 Posted June 25, 2009 Author Share Posted June 25, 2009 In fact, your whole idea is fundamentally flawed. No matter how many cards you "pick in your head", the probability of the next card you flip being red or black is based solely on the number of cards of each colour showing and face down. In fact, there are many outright false statements in the OP. First, of course, is the gambler's fallacy, that flipping over a black card first would cause the odds to shift drastically in favour of a red card immediately following. However, given 26 cards, 13 of each colour, the probability that you will choose a red card on the second flip given that one black card is face up already is only 13/(26-1) = 0.52, close enough to 50-50 that it would only make a difference an average of twice in one hundred turns. Second is this idea that because after 18 cards you ended up with 12 red and 6 black, the probability is somehow NOT 50-50; this is patently false. Just because an outcome does not match perfectly with its assigned probability does not mean that the probability is invalid. If you flip a coin four times, it is not very unlikely that you might produce three consecutive heads, and one tails (1 in 16 flips). In fact, this outcome is just as likely as heads, tails, heads, tails (1 in 16 flips) and does not indicate that the probability for heads is something like 3/4 and that of tails is 1/4. Probabilities are outcomes on average. Flip a coin 1,000 times and if the spread is something like 3:1, then you have an issue (or a weighted coin xP). I'm a little confused. In particularly here: Second is this idea that because after 18 cards you ended up with 12 red and 6 black, the probability is somehow NOT 50-50; this is patently false. Perhaps what I said was not clear enough. how can 18 cards, 12 being red and 6 black, still give an equal opportunity of 50 - 50? To emphesis it a little more: if there where 400 hundred red and 6 black, would it still be 50 - 50? I think you might of answered this in the last paragraph. Just asking for clarification. Link to post Share on other sites

Kyrisch 169 Posted June 25, 2009 Share Posted June 25, 2009 Ah, I misunderstood you. I thought you were concluding that after turning over 18 cards, because 12 of them were red and 6 of them were black, the initial probability was not 50-50. I see now that you're saying that given that spread of 18 cards face up, there will be more black cards face down, and the new (called conditional) probability is different. This is true -- however, the turns (unlike coin-flipping, I'm regretting using that analogy), are not independent of each other, as you realize. This means that turning over a black card will increase the probability of turning over a red card, BUT in the end, as long as you choose randomly, you will still get an approximately even number of black and red cards face up, no matter what "system" you use. Link to post Share on other sites

DJBruce 143 Posted June 25, 2009 Share Posted June 25, 2009 I'm a little confused. In particularly here: Second is this idea that because after 18 cards you ended up with 12 red and 6 black, the probability is somehow NOT 50-50; this is patently false. Perhaps what I said was not clear enough. how can 18 cards, 12 being red and 6 black, still give an equal opportunity of 50 - 50? To emphesis it a little more: if there where 400 hundred red and 6 black, would it still be 50 - 50? I think you might of answered this in the last paragraph. Just asking for clarification. You are right that if you start with a deck that has 18 card, of which 12 are red and 6 are black the probability would not be 50-50. The probability of drawing a red card would be 2/3, while the probability of drawing a black card would be 1/3. In the case of the 406 card deck the odd are again not 50-50. Instead the probability of drawing red is roughly 98.5%, while the probability of drawing black is approximately 1.5%. As for your algorithm I cannot help you that much, but my one guess is use something similar to card counting. Everything Kyrisch said is accurate it just appears, at least to me, that there was a little confusion. Link to post Share on other sites

LawLord 10 Posted June 28, 2009 Author Share Posted June 28, 2009 Would it make a difference if there was a third choice? For example black, red, or face cards (King, Queen, Jack, Ace, Joker)? In this case, there would be 24 cards - 7 of each. I remember now, about the gamblers fallacy. It is the same as someone thinking 777777 is less likely to come up in a lottery than 187294. However, today I came across a web page: http://numb3rs.wolfram.com/508/ Scroll down and you will see a mention of Roshambo (known to me as "Paper, scissors, rock"). The web page (as well as the Numbers episode) suggest that mathematics can be used to give someone an edge in Roshambo. Unfortunately it doesn't seem to explain how. I am wondering, does anyone know how to use math/probability to gain an advantage in Roshambo and can this theory be applied to my current situation? @DJBruce: I have a look at card counting. It's interesting, but it doesn't seem to give someone much of an advantage. Also, I'm not sure how to apply it to my problem yet- I am working on that though. Link to post Share on other sites

Sisyphus 993 Posted June 28, 2009 Share Posted June 28, 2009 7x3=21 What difference do you have in mind? The probability you'll pick up a certain kind of card is always [number of that type of card]/[total number of cards]. That's it. There's no more to it. Link to post Share on other sites

LawLord 10 Posted June 28, 2009 Author Share Posted June 28, 2009 silly me, yes 24/3 is 8 not 7. What difference do you have in mind? The probability you'll pick up a certain kind of card is always [number of that type of card]/[total number of cards]. That's it. There's no more to it. Yes, that right. If we apply this to "Roshambo/Paper, scissors, rock" we get [Variable (Scissor,paper, rock (1)]/[total number of possibilities (3)]. Which means there is a 1/3 chance of getting either rocks, paper, scissor. However, the web page indicated that there are math based strategies to help you win a game. I cannot even comprehend how mathematics can be used to help win a game of Roshanbo. So what i'm asking is: how does it help? If math can help with roshambo (I cannot comprehend how this would work due to ignorance) perhaps I can apply the same idea to my situation (I don't see how - due to ignorance of this area of maths). What i'm trying to say, I know my request seems like a long shot, but if math can be used in roshambo which at first glance doesn't seem possible, perhaps it can be used in my situation which as also doesn't seem to be possible at first glance. Am I making sense? My head hurts a little. Link to post Share on other sites

Klaynos 741 Posted June 28, 2009 Share Posted June 28, 2009 Rock paper scissors is more complicated as you are playing against other people... People work in set ways and our brains are weighted to select certain choices more often at different stages of the game... or so it seems. Link to post Share on other sites

Sisyphus 993 Posted June 29, 2009 Share Posted June 29, 2009 Yeah, I think that's being misleading. What you'd really use is psychology, perhaps with math-aided analysis (though I don't know what that would be). However, if you were playing against a computer randomly picking rock, paper, or scissors, there is nothing you could possibly do that would ever change your chances from 1/3 win, 1/3 lose, 1/3 tie. Link to post Share on other sites

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