Two Card Dilemma
A standard deck with 52 cards—26 red and 26 black—is shuffled.
What is the probability that the top card is the same color as the bottom card?
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19 solutions
Moderator note:
Note that the fact the cards are checked in a "simultaneous" way doesn't matter in terms of checking their content, because they cannot be the same card. So we have to consider this a "probability without replacement" so that after considering top card which could be out of any of 52 cards, the bottom card must be out of 51 (since the top card can't be duplicated).
Also, be careful not to confuse the physical process of something happening with the calculation of its probability. Quite often events happen in a "joint" fashion but to analyze them mathematically we consider them one at a time. This is especially true when the probabilities are dependent (as they are in this problem).
I'm still unsure about this. With a 2 card deck, having cards of the same color is mutually exclusive, but that's not the case for more than 2 cards. Also, this doesn't seem to be a problem where a card is drawn without replacement, which is necessary for the chance to be < 50%. The way the problem is stated, it seems we are selecting both the top and bottom card at the same time as a pair, hence the choice is between BB, RR, RB or BR.
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Good point
I recommend trying listing larger card sets if you're still not sure. For example, with four cards, here is every possibility:
RRBB, RBRB, RBBR, BRRB, BRBR, BBRR
Two of them involve the same first as last color.
Four of them involve different first and last colors.
Picking "at the same time" is irrelevant for changing which lists are possible above - there are only two possible orders where the first and last are the same color.
Choosing the top and bottom card simultaneously is, necessarily, cards drawn "without replacement".
Sir,please reply on Ryan Flint’s post, as simultaneous choosing of two cards will contradict your reason.
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"Simultaneous choosing" doesn't matter here. In determining the probability, we still are choosing from a limited set.
If we have 26 red cards, just because we pick both cards "at the same time" doesn't mean we have 26 cards to pick from for both cards!
I do agree with Ryan : I think the problem is badly stated : the answer "less than 50%" is correct is we suppose we select one card and, then a second one somewhere among the 51 others; the probability they are the same color is less than 50%. When the 52 cards are shuffled and we look at them, there are only 4 situations for the first and the last one : they are either RR, RB, BR, BB which gives 1/2.
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There are four situations, but the four situations aren't equally likely.
Consider a four-card deck, with two red and two black. The colors fall out as follows: RBBR RBRB RRBB BRRB BRBR BBRR
It's easy to see that the probability of the cards matching is 2/6, or 1/3.
As we add cards to the deck, the probability of matching cards approaches 50%, but never reaches it.
Hmmm...This is wrong. I think your reasoning is valid only if you first see the top (or bottom) card, but the problem states "what is the probability that the top card is the same color as the bottom card." What I did was to take the permutations of a deck of cards. There were permutations [red (24 red, 26 blue in the middle) red], [red (25 red, 25 blue in the middle) blue], [blue (24 blue, 26 red in the middle) blue], and [blue (25 blue, 25 red in the middle) red]. The number of each of the "forms" of the permutations are equal to each other since we always have 50! possible permutations in the "middle" part. Restate the problem correctly.
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You seem to be having situations where the number of one color does not match the number of the other color.
I recommend trying with a smaller number of cards. For example, with four cards, here is every possibility that leaves the color counts the same:
RBBR RBRB RRBB BRRB BRBR BBRR
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A deck of cards not only has colors: it also has (A,2,3,4,5,6,7,8,9,10,J,Q,K). So following your 4- card deck, the RBBR would not be the only permutation since the 2 blues are different, distinct cards. The same for BRRB.
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@A Former Brilliant Member – Are you confusing suits with colors?
The answer is the same with suits, though.
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@Jason Dyer – I am just saying that with the 4-card deck, we have to consider suits because there are 2 ways (actually 4 since the reds could also shift but we only need the outter cards to be the same) to shuffle to a RBBR color combination: Lets put a specific example: any red card, A of clubs, K of spades, any other red card. But since in the problem says we shuffle the deck, we could also get: any red card, K of spades, A of clubs, any other red card.
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@A Former Brilliant Member – The fact we could have some face value and permute the face value contents isn't relevant because we can do the exact same thing with both colors, and get an equal number of choices.
Additionally the actual face values are unused in problem entirely and so are irrelevant to the probability. You can consider the cards, if you like, to be blank other than having red or black on their front.
@A Former Brilliant Member – If you're looking at RBBR, those could be HSCD, HCSD, DSCH, or DCSH, certainly, but all four of those permutations are exactly the same for the purposes of this problem. Looking at color only means that each specific iteration of the problem requires 1/4 of the potential calculation.
The same goes for considering the value of the cards. K and Q are, in this case, equivalent cards, so it doesn't matter what order the deck is in, it only matters whether the cards in question are Red or Black. For that matter, RBBR and BRRB are equivalent, since we're only looking at the odds that the cards are matching in color on top and bottom.
If we phrase it another way, if the first card is Red, what are the odds that the last card is also Red? The only possible configurations with a Red top card are BBR, BRB, and RBB, and it's easily seen that the odds are 1 in 3. If the first card is Black, we have the same question, with the same result. The next three cards are either RRB, RBR, or BRR, and there's a 1 in 3 chance that the bottom card is Black.
Regardless of the color of the top card, the odds that the bottom card matches its color are less than 50%, and the suit and value of that card doesn't affect those odds.
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@Brian Egedy – They are the same color combination but there are more than one way to shuffle to thar color combination.
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@A Former Brilliant Member – But those variations don't change the odds of the corresponding colors occurring. If you're looking at RBBR, there are four different suit combinations that will result in that color combination, but if the question is whether a four-card combination matches Red Black Black Red, the odds of any RBBR meeting that condition is 100%. The question of whether it's 1 of 1 (RBBR) or 4 of 4 (HCSD, HSCD, DCSH, DSCH) doesn't change the probability of RBBR matching RBBR. HCSD = DSCH in this type of problem.
@Jason Dyer – I said with the 4-card deck, but I meant with any deck.
you are wrong as the two blacks are unique and your RBBR has four solutions R1B1B2R2 can also be R1B2B1R2 ... ... so lets not go there
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@Sean Houston – I agree, let's not go there. The values of the cards, or even the suits, have no bearing on the odds of pulling a particular color from a deck of 52. For the purpose of this stated problem, the Ace of Spades is exactly the same card as the 5 of Clubs. This problem only recognizes two values: Red and Black.
Read through some of the other comments that have already come up if you're concerned about the uniqueness of individual cards as far as this problem is concerned.
'The number of each of the "forms" of the permutations are equal to each other since we always have 50! possible permutations in the "middle" part.' This statement is incorrect. There are fewer permutations of (24 red, 26 blue) than there are of (25 red 25 blue), even though there are 50 total cards in each case. If there are 50 distinguishable cards, there are 50! permutations. But introducing identical colors reduces those permutations a lot.
With 24 red cards and 26 blue cards, we have 50 total 'slots' and we have to pick 24 of them to be red. So there are P 1 = C ( 5 0 , 2 4 ) = 2 4 ! ⋅ 2 6 ! 5 0 ! permutations.
With 25 red cards and 25 blue cards, we still have 50 total slots, but we now have to pick 25 of them to be red. So there are P 2 = C ( 5 0 , 2 5 ) = 2 5 ! ⋅ 2 5 ! 5 0 ! permutations.
Comparing the two expressions shows that P 1 is less than P 2 . P 2 P 1 = 2 4 ! ⋅ 2 6 ! 5 0 ! ⋅ 5 0 ! 2 5 ! ⋅ 2 5 ! = 2 4 ! 2 5 ! ⋅ 2 6 ! 2 5 ! = 1 2 5 ⋅ 2 6 1 < 1
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Agreed. But the problem states a standard deck of cards, so every card is distinct.
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@A Former Brilliant Member – The permutation computation already accounts for the uniqueness of every card in the deck.
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@Brian Egedy – Wrote it out and saw that it was. Sorry for the arguing :v
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@A Former Brilliant Member – Not a problem. Have a good one.
My solution would be : It doesn't matter what colour the top card is. The probability that the bottom card is the same colour would be 24/50. Where is the mistake in my logic ?
there are 26 red and 26 black the only combinations of cards top and bottom are RR RB BB BR this is the universal set and there can be none other the answer is 50%. The problem was never "given one card is red" what is the probability of drawing the same colour. You have imposed a condition that was NOT in the original problem. The deck is shuffled and put aside.The answer is 50
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The broader question, "what is the probability that the top and bottom cards match" has two cases. The top card must either be red or black. If the top card is red, the probability of the bottom card being red is the same as the probability of the bottom card being black if the top were black. Those two identical cases, together, are the complete case of the top and bottom cards matching.
Focusing on the half of the problem in which the top card is red allows you to simplify the broader problem by focusing on a specific instance of it.
"...the only combinations of cards top and bottom are RR RB BB BR..." This is true, but misleading. If the card colors were based on coin flips, then it would, truly, be 50%. Because the cards are drawn from a limited set, the odds of each color go down as that color continues to be chosen. If you draw two cards from the deck, the probability that the first card is a particular color is 26/52. The probability that the SECOND card is the same color is 25/52. This means that the chances of drawing two consecutive cards of the same color, regardless of the color, is slightly less than 50%.
RB and BR are slightly more likely than RR and BB, because you're drawing from the deck.
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because you're drawing from the deck. is the error. There is no draw its shuffle and stop somewhat have we got ...
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@Sean Houston – see also : I do agree with Ryan : I think the problem is badly stated : the answer "less than 50%" is correct is we suppose we select one card and, then a second one somewhere among the 51 others; the probability they are the same color is less than 50%. When the 52 cards are shuffled and we look at them, there are only 4 situations for the first and the last one : they are either RR, RB, BR, BB which gives 1/2.
G B - 1 day, 4 hours ago
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@Sean Houston – Can you clarify what you think is badly stated about the problem?
Shuffle a normal deck of 52 cards, and compare the top and bottom cards. What are the odds that they're the same color?
The odds would be exactly the same if we compared the first and second cards, the 51st and 52nd cards, or any card in any position to any other card in any other position.
@Sean Houston – Incidentally, did you see the response I made to GB when he originally posted that?
@Sean Houston – Drawing the top card from the deck and looking at it is exactly the same mechanism as turning the deck over, fanning out the cards, and looking at all of them. I'm not changing the position of the top card, nor am I changing its identity, by removing it from the deck to look at it.
But there are 52 cards.
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You start with 52 cards, and look at the top one. Then, you take the bottom card from the remaining 51.
The main confusion here is that there is nothing to say that the 2 cards aren't revealed simultaneously. Unclear question. They could be revealed simultaneously and then it's 50%
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The identity of the cards is set when the shuffle is completed. The timing of revealing them, either top/bottom or bottom/top, or simultaneously, doesn't change the identity of the cards themselves.
We can't physically have the cards be the same card, even if we pick them "simultaneously" - so with the calculation of probability, one of them has to be out of 26 and the other one has to be out of 25.
Agree. Top and bottom positions are irrelevant, the problem as stated is just pick 2 cards, are they same or different colours
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The positions are irrelevant, on that we agree. The disagreement is about whether "same color" and "different color" have the same odds in a finite deck. They don't.
Opposing colors each have 26 cards to choose from.
The math of opposing colors is the same as drawing one card from 26 and the second card from 26.
Matching colors have 26 cards from which to choose {both}.
The math of matching colors is the same as choosing one card from 26, and then one card from the 25 remaining.
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@Brian Egedy – This is not quantum physics, looking does not change the result of selecting 2 cards from 52.
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@Philip Sabine – I didn't say anything about looking changing the results. I said, "the math of matching colors is the same as choosing one card from 26, and then one card from the 25 remaining."
It's also the same math as writing a list of all the possible permutations of the deck that have the same color for the top and bottom cards, and a list of all the possible permutations that have opposite colors. There are more versions of the deck that have opposite colors.
Here's an example, posted elsewhere in this discussion by me and other people, with a smaller deck. The model is the same.
Picture a four-card deck, with two cards of each color. All of the possible permutations of that deck are as follows: BBRR, BRBR, BRRB, RRBB, RBRB, RBBR.
With the small deck, it's easy to see that there are two instances in which the top and bottom (first and last) cards match, and four in which they don't, resulting in a 33% chance that the cards match for a given shuffle.
Another way of looking at this, is that if the first card is Black, then there is one configuration that results in a Black bottom card, and there are two configurations that result in a Red bottom card. Same discussion, same result, but a different approach to it.
Another way, and this is the one that establishes my general case, is to say that there are two chances for the top card to be Black, since there are two black cards available, and IF the top card is Black, then there is only ONE chance for the bottom card to be black. That's the total number of black cards to start, and then the total number of black cards, minus one, for the bottom card.
So, 2 then 1. IF the top card is black, which had a 2/4 chance, THEN there is a 1/3 chance that the bottom card is also black, and a 2/3 chance that it's red. IF the top card is red, THEN there is a 1/3 chance that the bottom card is also black. Since these are the only two cases, then overall there is a 2/3 chance that the cards don't match, and a 1/3 chance that they do.
It parallels the full-deck solution, where there are 26 chances for the top card to be black, and IF it is black, then there are 25 chances for the bottom card to be black.
I'm not saying that looking at the top card changes the odds for the bottom card. I'm establishing that, in the instances where the top card IS black, there is a 25/51 chance for the bottom card to also be black, and a 26/51 chance that it's red. Overall, there is a 25/51 chance that the top and bottom cards match, regardless of whether the top card is red or black.
I'm seeing a lot of people trying to justify it being 50% rather than trying to learn why it isn't exactly 50%. I quickly guessed 50%, got it wrong, and came and LEARNED WHY and its crystal clear to me now why it isn't.
Remember, this is in the BASIC section lol.
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Can you think of a better way to explain it? I feel like I'm not getting through.
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@Brian Egedy Look, I found it in the way of considering all possible permutations of the deck, 52! as there are 52 different cards. Then, for the number of ways the top and bottom cards are equal using the product rule, We can have 52 possible choices for the top or bottom one, but then for the other one to be equal you have only 25 (26 - 1) options. Then for the rest of cards you have 50 × 49 × 48 ... = 50!, as they are different cards. Then the probability of the top and bottom cards being equal is
5 2 ! 5 2 × 2 5 × 5 0 ! = 5 1 2 5 = 0 . 4 9 0 1 9 . . . ≈ 4 9 %
Which is less than 50%. I hope this helps you get it Brian.
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@Javier Álvarez – I appreciate it; I got it a while ago and have been trying to explain it to other people. The factorial method for breaking out the overall probability seems to be the best method for explaining it when it doesn't make sense intuitively.
Thanks for the suggestion!
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@Brian Egedy – You're welcome! Also thank you for your effort on trying to make people understand! I've seen you in many of these replies
$(ax^2 + bx + c = 0)$
It doesn't say the top card has been pulled off the deck so there would be 26 of each color so the chances are 50/50. If the top card had have been drawn off the deck then it would have been 25/26. Why can't you just peek at the top card?
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I'm unclear why peeking matters? We have two different cards. They can't be the same card. So if we're working out their probability of color-matching, we consider two cases (one when the top card is red, the other when the top card is black) and when looking at the possibilities for the other card we can't pick the same card as the top one.
The top card doesn't have to be physically removed from the deck. The problem setup removes the top card from the group of 26 that you can include in the potential cards for the bottom card.
If you look at it as separate groups of cards, i.e., {the top card} + {all remaining cards}, you can see that there are two possible makeups of the deck:
{one red card} + {25 Red cards + 26 Black cards} or {one black card} + {25 Black cards + 26 Red cards}
In either case, the bottom card is part of a group of 51 cards, not 52, so the odds of it being a matching card are 25/51. Again, we don't have to know what the top card is, in order to calculate the odds of the bottom card matching it.
Another way to look at the problem, more intuitively, is to isolate three groups: {one red card} + {25 red cards and 25 black cards} + {one black card} = not matching {one red card} + {24 red cards and 26 black cards} + {one red card} = matching {one black card} + {25 red cards and 25 black cards} + {one red card} = not matching {one black card} + {26 red cards and 24 black cards} + {one black card} = matching
It should make sense that the odds of having a 24/26 split in either direction are slightly lower than the odds of having a 25/25 split.
reducing the card deck to 4 cards: 1) R1 R2 B1 B2 2) R1 R2 B2 B1 3) R1 B1 R2 B2 4) R1 B2 R2 B1 5) R1 B1 B2 R2 6) R1 B2 B1 R2 Analizing the cases where R1, R2, B1, and B2 are on the top of the deck, we have 6 x 4 = 24 possible ways on which the deck can be arranged (or shuffled). In each group of 6 cases on wich a specific card is on top, we have 2 cases on which the bottom card is of the same colour. Considering the 24 possible ways of arranging the deck, we would have 8 cases on which the top card and the bottom card are of the same colour. The odds of having in a 4 card deck the bottom and the top card of the same colour would be 8/24 = 0,333333
Thank you moderator for the comment.
There are 4 possible outcomes for the top and bottom cards, regardless of the "number of cards left in the pack after looking at the first": RR, BB, RB, BR. The odds are 50:50 for the top and bottom cards to be the same.
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It's not the "number of cards left in the pack after looking at the 1st card", it's the number of cards left in the pack after the deck is defined by the shuffle.
The odds of any one card in the deck being a certain color are 50:50. The odds of any one card in the deck matching any other card in the deck are slightly less.
Consider the odds of the 1st 26 cards in the deck being all the same color. Then the first 25, then 24. It should be clear that those odds start very low and then raise as you compare fewer cards, but that it doesn't get to 50:50 if you're still matching any 2 cards in the deck.
I don't agree.
The problem is: A standard deck with 52 cards—26 red and 26 black—is shuffled. What is the probability that the top card is the same color as the bottom card?
The problem does not state that the top is red or black. So the odds are 50:50 the top is the same color as the bottom.
If we assume the first card is either a red or black, then yes, the odds the bottom card will be the same color as the top card are < 50%.
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You seem to be saying that knowing the top card changes what the bottom card is. There's no sense in that statement.
The odds are 50:50 that the top card will be red or black. The odds are also 50:50 that the bottom card will be red or black. There are slightly fewer permutations of the deck that allow the top and bottom cards to match, so it's slightly below 50:50 for them to be matching.
Declaring a particular color for the top card is done to simplify the analysis. It doesn't change the odds at all.
49:51 matching, not 50:50.
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Knowing what the top card is ABSOLUTELY changes the probability of what the bottom card is. If one color is postulated as the top card, is it inevitable that there are 49 cards left of which 50 are not the color of the top card and 49 are.
The initial question cannot be confused with conditional probability of which clearly the odds are less than 50% that the bottom card matches the top card.
Focus on what the INITIAL problem is stating. It's clear after that. You are answering the wrong question.
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@Peter Kuchnicki – Knowing the top card changes your odds of subsequently guessing what color the bottom card is (similar to the Monty Hall problem). It doesn't change the probability of the bottom card being a particular color (50%), and it doesn't change the probability that the top and bottom cards match (49%).
The initial problem asks what the probability is that the top and bottom cards match. Those odds are the same, regardless of what that matching color is. Knowing the color doesn't change the odds.
The deck represents 26 possible events. Each event is two cards. This is a non replacement probabillity yes but the non replacement is applied to two cards not one and this question is only about the first event. So the possible outcomes are potentially 26 R R or 26 B B or 26 R B or 26 B R . There are 104 possible arrangements of which 2 X 26 are matched for colour 52/104 is 50%
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The deck represents 52 non-replacement events, since the identity of a single card in the deck is a unique value. There are 52! unique configurations for the total deck. If you define an event as a comparison of two cards at a time, then there are 1326 unique pairs of cards, and 52 x 51 = 2652 unique configurations for each pair.
Of those, there are 676 B R, 676 R B, 650 B B, and 650 R R. 1300 matching chances per 2652 total configurations = 49% probability that the any two cards in the deck are matching in color.
Thinking further about my motivations for posting earlier I want to lift the lid on this as many of the other respondents may have had a similar reaction. If the question had been "You have drawn a card from a deck what is the probabillity of the next one matching colour?" then the question would have been seen by many of us as having been correct; however I would be willing to bet that many of us would still have answered 50% and been wrong. The logic of Jason's solution is solid despite the arguments that the solution belongs to a different question. In fact there has been much to learn in the solution and the counter arguments.
So why do I feel a need to dispute a question that I likely would have gotten wrong anyway it was written? Perhaps there is left over trauma from exams where we feel robbed for a mark or two because of how the question was written or how we thought it was written; here at last is an opportunity to "get 'em back"; save face for incorrect thinking because there is opportunity to dispute how the question was asked. So yes there is a lot to learn from questions like this; not just maths but also about myself. Thank you Jason for an excellent thought provoking question! More please :-)
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Wow, this comment is so positive! Thank you Chris for the positivity!
As the problem says, the deck is shuffled. So until you check, it is 50%
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Your odds of correctly guessing either the top or bottom card are 50%, but the odds of those cards matching each other, regardless of whether you've checked yet, are less than 50%, due to the finite nature of the deck.
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Just made a comparison with Schrödinger's cat, as a joke. I understand, that the moment you saw the first card its less than 50% to see it again.
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@Mher Kosakian – In retrospect, that's hilarious.
How about, until you check, all the cards are grey, so they're 100% matching in color?
Sorry, but I believe this staff solution is wrong. The correct answer should be exactly 50 percent. To illustrate, assume the "deck" only had 4 cards, 2 black and 2 red. By Jason's argument, if you look at the top card and it's black, then you have only a 1/3 chance that the second card is black - way less than 50. But there are only 4 choices when you pick 2 cards: BB, RB, BR, and RR, and half of them match. The correct answer must be exactly 50 percent.
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Try this experiment. Make up a 4 card deck, with two red cards and two black cards. Shuffle them. Deal two cards, and record if the match or not. Shuffle again, then deal, compare, mark, and repeat 19 more times. Then see if the results are closer to 10 matches ( if 50/50, or 6.6 ( if 1 in 3). Have some friends do it to, and compare.
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I tried this, and used a method that would reduce my ability to track cards as much as possible.
I made 13 piles of 4 cards, each with 2 red and 2 black. I shuffled each pile, and then randomly stacked them back into a deck.
Then I dealt 4 cards from the recombined deck, shuffled them, and revealed 2 cards.
After revealing the cards, I recorded the result, shuffled the 4 cards again, and put them on the bottom of the deck, and repeated 100 times.
My result was 38 matching colours, and 62 non matching colours.
Fun experiment!
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@Jonathan Quarrie – Thanks for trying the experiment and telling the results!
I was wrong. The staff abswer is right. I learned this with the experiment that I devised above. I shuffled the 4 card deck that contained 2 red and two black cards, checked the bottom card, and compared it to the top card. If it matched, I put tallied in a column called matched, and if it was different, I put it in a column called not matched. If it's really 50-50, as I thought, then with enough trials, the two columns should be roughly equal. If the staff answer is correct, the not match tally should contain roughly twice the count of match count. The first time I tried it, I got exactly 3 matches, and 9 not matches. Being stubborn, I tried it more times. Each time, the not match count was roughly twice the match count. Thanks for making me think!
Take into account the following, the ways of getting BB or RR if you pick two cards would be 4(for the first card)×1(as there is only one option left for it to be the same colour as the first)×2×1/4!=8/24=1/3, the same as if we were using Jason's argument. The problem is you are assuming that the probability of getting either BB, RR, BR, RB are equal, when they aren't, they are 1/6, 1/6, 1/3 and 1/3 respectively.
Consider walking outside with a coffee mug, and calculating the odds of a gold coin falling into your mug. The two options are either {coin in mug} or {no coin in mug}, but those options don't have an equal probability.
The fact that there are only four options BB/BR/RB/RR doesn't mean that each of the options has an equal likelihood of occurring.
Wait, so sample space is{RR, BB, RB, BR} which means no of outcomes is 4 out of which favourable outcomes are 2 i.e. {RR, BB}. Shouldn't the probability be 2/4 = 50%?
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Counting the outcomes simply at the B/R level is not the same as weighing their probability.
| R 1 | R 2 | B 1 | B 2 | No Match; 1 |
| R 1 | R 2 | B 2 | B 1 | No Match; 2 |
| R 1 | B 1 | R 2 | B 2 | No Match; 3 |
| R 1 | B 2 | R 2 | B 1 | No Match; 4 |
| R 1 | B 1 | B 2 | R 2 | Match; 1 |
| R 1 | B 2 | B 1 | R 2 | Match; 2 |
| R 2 | R 1 | B 1 | B 2 | No Match; 5 |
| R 2 | R 1 | B 2 | B 1 | No Match; 6 |
| R 2 | B 1 | R 1 | B 2 | No Match; 7 |
| R 2 | B 2 | R 1 | B 1 | No Match; 8 |
| R 2 | B 1 | B 2 | R 1 | Match; 3 |
| R 2 | B 2 | B 1 | R 1 | Match; 4 |
| B 1 | B 2 | R 1 | R 2 | No Match; 9 |
| B 1 | B 2 | R 2 | R 1 | No Match; 10 |
| B 1 | R 1 | B 2 | R 2 | No Match; 11 |
| B 1 | R 2 | B 2 | R 1 | No Match; 12 |
| B 1 | R 1 | R 2 | B 2 | Match; 5 |
| B 1 | R 2 | R 1 | B 2 | Match; 6 |
| B 2 | B 1 | R 1 | R 2 | No Match; 13 |
| B 2 | B 1 | R 2 | R 1 | No Match; 14 |
| B 2 | R 1 | B 1 | R 2 | No Match; 15 |
| B 2 | R 2 | B 1 | R 1 | No Match; 16 |
| B 2 | R 1 | R 2 | B 1 | Match; 7 |
| B 2 | R 2 | R 1 | B 1 | Match; 8 |
OK I have finally come out the other side. Many of us are a bit caught up over wording of the question, simultaneity and sequence. But as the moderator states it is about the content. Also Jason Dyer with his "2 card deck" is extremely helpful for me as it is a practical example of 'Proof by Induction' an important concept that can be hard to grasp but has been put here so well.
We would all agree that as we drew cards and had results 'in the past' that the future probability is of course changing; this is because we have information prior to the next draw. But we can get too caught up in our time-line I think. Is there information prior to our first draw? Effectively many of us are saying no; We are in effect having the draws on the table but not looking; this isn't what the question asked. We do have info before the first draw, the top card will be a color, this is known. The bottom card will either match or not, this is unknown. So in the first instance we know that irrespective of the color on the top that it is 'a color' and there remains less than 50% of matching color (the first card is removed) For me this has been a powerful cognitive shift on probability. Brilliant!
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It just occurred to me that some of the confusion may have been people envisioning that the first card is drawn, returned to the deck, and the deck re-shuffled prior to revealing the bottom card, which would be 50:50, since both cards have a 26/52 chance of being a particular color.
Welcome to the other side!
The 51 non-top cards contain 25 cards that have the same color as the top card and 26 cards that have the opposite color. Each of them is equally likely to be the bottom card. Therefore P ( bottom card = top card ) = 5 1 2 5 < 5 0 % .
Excellent and succinct solution.
Very nice!
Simple,
Ways it can happen = 2 6 C h o o s e 2 (both black) + 2 6 C h o o s e 2 (both red).
The total number of ways in which 2 cards can be drawn from the deck = 5 2 C h o o s e 2 .
Answer: 5 2 C h o o s e 2 2 6 C h o o s e 2 + 2 6 C h o o s e 2 = 5 1 2 5 < 0 . 5
:)
Before you look at the top card, there are still 26 red and 26 black. What are the odds that the top and bottom cards are the same before looking at either? Wouldn't that be 50/50? Would that not be the answer to the original question?
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Are you sure that the query is posted for my solution?
If yes: No, it wouldn't be the same. It would have been the case with replacement, where you have exactly the same number of cards in each case (one when you draw a card to place at the top, and other when you draw for the bottom).
Consider, in without replacement, you want the same color card, you shuffle the deck to draw a card - let it be red. Now, your deck is not the same as it was before; it is BIASED towards black colored cards.
In without replacement, you will have no such bias. :)
Are you sure that the query is posted for my solution?
If yes: No, it wouldn't be the same. It would have been the case with replacement, where you have exactly the same number of cards in each case (one when you draw a card to place at the top, and other when you draw for the bottom).
Consider, in without replacement, you want the same color card, you shuffle the deck to draw a card - let it be red. Now, your deck is not the same as it was before; it is BIASED towards black colored cards.
In without replacement, you will have no such bias. :)
Let's say the top card is red. There are 25 remaining red cards 26 remaining black cards. Therefore, it is less likely that the bottom card is red. The same logic applies if the top card is black.
Exactly thge way I solved it too.
Let us suppose that the total number of cards is 4 instead of 52 among which two of them are black and two of them are red. The possible arrangements of these cards are B R B R, B B R R, R R B B, R B R B, B R R B and R B B R. ONLY two of the arrangements have the top card having the same colour as the bottom card. The total percentage is 2/6 that is 33.33% less than 50%
(26/52) (25/51)+(26/52) (25/51)=25/51<0.50
假设最上方的牌是红色的，那么其下的牌有25张红和26张黑，最下面那张牌是红的概率是25/51，自然小于50％。同理，最上方牌是黑色，结果也是一样的。 思考一个极端情形，这幅牌只有2张，那么最上方和最下方的牌一定是不同色的，此时概率是0，小于50%。类比一下，52张牌的情形，概率也应该小于50%。
P(BB or RR)= 2(26x25)/(52x51)=25/51 which is <50%
Simple. It doesn't matter what color will be put in the top. There will always be 25 remaining cards with the same color. And since there remains 51 cards, then the answer will be
P ( top color = bottom color ) = 5 1 2 5 < 5 0 % .
Here the thing matters is choosing 2 cards whether from top and bottom ones or any other pair.Now choosing 2 cards we have 2 possibilities either a pair same color or different color.So (26C1 26C1)=676 ways for getting a pair of different colors and (26C2 2)=650 ways for getting a pair of similar color.So required probability is (650/(676+650))=0.490196..< 0.50. In here the redundancy about the same color is not considered beforehand, as it will eventually going to get eliminated.
As one of any type card occupied top place in shuffled deck. It means there will be 1 card less then that of other hence probability will be <50% because both are having same no. At beginig.
Say the top card is red, there are now 25 red cards that could be on the bottom compared to 26 black ones. Therefore it is less likely to be red. Less than 50%
Top card say it be a red card, we have 26 choices of red card. Next card has to be red but the first one is already red so 25 choices. And for the rest cards cards can be of any color.
So prob=(26X25X50!)/52!
50!=50X49X48.... (50 is the choices of color for third card and it keeps decreasing till we reach the last card with only 1 choice of color)
Same will be the case for black color.
Final p=probX2
=> p=25/51
roughly less than 50 %
I used methods of probability and guessing first of all I have divided the cards of 52 in 50 that is 25 and 25 and here is given that we need to find the probability for 26 cards so as 26 is more than 25 and as far as I concerned while doing probability that the finding number that is 26 is more therefore the answer must be than less than 50 rather than more than 50 so I did the same.
Disregarding math:
- Intuitively, the chance might seem to be equal for "two different cards" and "two equal cards". The options are RR, BB, RB, BR, right? Two out of four pairs are equal!
- They would certainly be, if you drew a random card, looked at it, put it back, and drew another random card.
- But because you are NOT putting it back, you are "using up" a card.
- If the top one is red, the REST of the deck contains still the same about of black cards, but a little bit less red cards. One less red card, to be precise!
The "putting it back" part is key!
If you draw a card, write it down and put it back, the chances stay the same, 50% for each draw. If you don't put the card back into the set of cards you are going to look at next, the chances get a tiny little bit skewed. The new deck isn't equally divided between red & black anymore!
We realize that this is just the probability of them both being black plus the probability of them both being red. The probability that they're both black is just 26/52 * 25/51 = 0.245 , which we know from dependent probability. Likewise the probability that they are both red is 26/52 * 25/51 = 0.245 . The sum of the probabilities being 0.49 , or 49%.
The position of the cards doesn't matter in this problem. The two cards can be top, bottom or any other position in the deck. The problem can be restated as what are the odds of drawing two cards of the same color?
You start with 52 cards, 26 are red and 26 are black. What is the probability they you will draw a red card for example? Well it is exactly 50% because the number of favorable choices is 26 and the total number of choices is 52. So 26/52=0.5=50%.
Now there are 51 cards in the deck: 25 are red (because you drew one) and 26 are black. What is the probability that the second card you draw will also be red? Well, the number of favorable choices is 25 and the total number of choices is 51. So 25/51=0.49<50%.
Also, you can calculate it like this:
Odds of drawing two red cards are 5 2 2 6 5 1 2 5 and the odds of drawing two black cards are the same. So the odds of drawing two cards of the same color are: 2 5 2 2 6 5 1 2 5 = 5 1 2 5 ≈ 4 9 %
Define the color of the top card as T. Define the opposite color as N. The bottom card can be any of the 51 cards that are not the top card. Of those 51, 25 are T, and 26 are N.
The odds of the very last card being T, then, are 25 / 51.
The deck is of 52 and we have 26 red and black so the top has chance of 50% and bottom top has chance of 50% .If both have same chance then 1/2+1/2 should give a1/4 that is lesa than 50% oh 25%
Suppose the top card is red. Then of the 51 cards remaining, 25 are red, so there is a 25/51 or roughly 49% chance the bottom card is also red.
Suppose the top card is black. Then of the 51 cards remaining, 25 are black, so there is a 25/51 or roughly 49% chance the bottom card is also black.
Since red and black might be the top card with equal probability, the overall probability of the top and bottom colors matching is roughly 49%.
The main confusion here can be isolated if we fix the condition that the number of red and black cards are equal, and reduce the problem to a 2 card deck. We might think BB, RR, RB, and BR are all of equal probability (and hence both cards being the same is as likely as them being different) but neither BB or RR are possible because the deck has an equal number of red and black cards!