A good guesstimate in Texas Hold'em: After Flop: OUTS*4 = P(completing your Hand) After Turn: OUTS*2 = P(completing your Hand) *OUTS = Number of Cards missing to complete any possible Hand
does the solution consider having a pair in the ' last 3 ' ? after the first pair , there is a choice of 48 cards for the 3rd card , 44 for the 4th and 40 for the 3rd .
It factors in the chance of there being a pair in the last three by using 12C3, which is the same as 12C1 x 11C1 x 10C1. The 3 for 12C3 basically means "choose three different values"
@@trolleymouse I have been constructing a card game for an assignment. I have removed aces from the deck for other reasons. I need to calculate the odds of drawing specifically one pair, when drawing 6 cards. 4C2 x 12C1 x 11C4 x (4C1)4 / 50C6 is the final solution i came up with. However this contradicts the video, where he cubed his value inside the brackets, rather than multiplying. Should i be multiplying by four or raising it to the fourth power?
I think there is a phrasing of the problem here, if we wanted to only have one pair in a hand the first 2 pick was perfectly fine, but for the third card there are 12 values and we can pick any four of them, for the 4th card we cannot pick the 12 values again if we do then we have pairs again so for the 4th pick it should be 11 out of 1 and 4 out of 1 and for the 5th card we pick 10 out of 1. this ensures that there are no other pairs of the card. the problem above ensures there is at least one pair not ONLY ONE PAIR.
Extremely well done, I'm still struggling with the extra cards, that's me, and I will get it after a few reviews. I don't believe most people know how to take the extra cards into account, making it one pair and not possible two pair or something else. Thanks, the Royal flush seems the simplest to figure, am I wrong wasn't this considered the dead man's hand, just asking.
1 thing I don't understand... is it the probability of picking the cards all at once? cause for me it seems like it. Then my question is, how much more complicated is it to calculate the probability of picking the cards one after another?
If you multiply it by 48 C 3 then you could get a card of the same rank as your pair, meaning you could have 3 or 4 of a kind. By multiplying it by 12C3×4×4×4 you guarantee that you wont pick a card of the same rank as your pair.
@Juan Sanchez - That is actually wrong. 48 choose 3 means that we are guaranteeing we will NOT choose any of the 3 remaining cards the same rank (value) as the pair we already chose. The problem with 48 choose 3 is that it is possible to choose another pair (thus making the hand have 2 pair) or even a triple (making the hand a full house). It would have to be 50 choose 3 to allow the same rank as the pair we already previously chose (the first 2 cards chosen from the deck).
His explanation of the 12c3 is confusing. He says you can't pick any of the remaining Kings (triples, quadruples). But you also can't pick any cards that match themselves. He left that part out, I think. That is why it is 12c3 (any three cards from a set of 12 different ranks) where card 3 is one of four suites 4c1, card 4 is one of four suites 4c1, and card 5 is one of four suites 4c1. Fun stuff can be a headache sometime.
for the 12c3 it works because if we have 12 values left and we need to fill up the remaining spots it'd be 12*11*10 and we don't need the remaining 9!. So we can write this as 12!/9!, but since the internal arrangement of 3! must be considered, (from the cards that have just been picked) we must divide this out as well giving us: 12!/(9!*3!). This can then be written as 12 choose 3, since 12!/(3!*(12-3)!)
You are right, he should have reduced the subsequent picks as well. His problem is at least one pair, this is the reason the probability was so high. the problem is to have AT LEAST ONE PAIR
@@davidjames1684 Ahh, but there's the problem: once you've phrased it well you've already solved half of the problem so there's just half a problem left. If you then phrase that half-problem well you have solved half of half-problem ie. a quarter-problem. There's still another quarter-problem to solve. This can be generalized: after n well-phrasings, you still have (1/2)^n of the problem to solve.
i think you need to subtract the number of ways for the full house before multiplying by the pair - the three other cards must not be of the same value
Another way of solving the problem is thinking this way: you have 52 out of 52 cards to choose from (52/52), then, after that, you have only 3 cards out of 51 to choose from (3/51). And the position of these 2 cards can be swapped 2 times between them, so you must divide by 2! So, the first 2 slots equation will be (52/52 * 3/51) / 2!. For the 3 remaining positions, in the third slot, we have 50 cards at our disposal, but we can only pick 48, because we have already chosen 2 cards of the same value and there are another 2 cards of the same value yet, so we have 48/50. At the fourth slot, from 49 cards remaining, we must subtract 2 cards from the first pair and 3 cards from the third slot, so 44/49. For the fifth slot, we will have 48 cards at our disposal, but we will only can pick 40, because we must subtract 2 cards from the first pair, 3 from the third slot and 3 from the fourth slot, so 40/48. And the 3 remaining slots must be divided by 3!, the number of ways the 3 cards can be swapped between them. Finally, we must multiply all this by 5!, the number of ways the 5 cards selected can swap positions between them. In summary, the equation will be: (52/52 * 3/51 * 48/50 * 44/49 * 40/58 ) * (5! / 2! * 3!).
Didn´t understand, if i pick one king, three others left, so i should just combine C3,1 to form a pair not a C4,2. Sorry for bad english, am not native
I agree...this is fantastic.... but I disagree that for a Pair we need consideration of the fact that once one card is picked the second card as condition probobility and not funny consderation of the suit.
Can you help me calculate how many hands are expected to get 3 cards for a royal, I heard is every 92 hands approx but I dont find a correct way of calcualted, I know there are 40 distinc hand but with the 2 remaining spots should I multiply by (38c2), so I exclude the 5 for the royal and also any 3 of a kind, @eddie woo
8:26 The more cards you get, the more likely a pair will be in there somewhere is incorrect. If you already have a pair say in the first 2 cards, then getting 3 more cards will NOT increase your chances of getting a pair somewhere in there as you stated. Actually in a 5 card poker hand dealt from a 52 card deck, it is MORE likely that you do NOT get just one pair. 42.2569% chance of getting exactly one pair means 57.7431% or getting something else, so once you get a pair (such as dealing the cards one at a time), then no, there is NOT a higher probability of getting a pair once you already get it and draw more cards. For example, K, K, A, 7, K.
@@karolisgudziunas9738 Not at least one pair. Specifically one pair. Thats why when u pick the other three cards u only have 12 to chose from, not 13. Kings would make it a better hand (trips or quads).
@@nicolasseminara9307 I don't think within the other three cards could be another pair, since you pick the value 12C3. So the three cards can never be the same.
Wait, is this percentage ONLY for the two cards in my hand of getting a pair, or your best hand of 5 cards which means that percentage includes a pair on the flop as part of your best 5 cards
@Paul MácRéamóinn - It could be that you get all 5 cards or you get 2 cards and there are 3 "community" cards (a 3 card flop). The probabilities are the same for both situations. This is assuming a standard full 52 card deck.
@@davidjames1684 it's actually the probability of only have a one pair. that's why he factored in 12c3, to choose a 3 sized combination of the remaining 12 card values. i believe.. probability for AT LEAST one pair would be 13c1 x 4c2 x 52c3
In most probability questions, including this one, they're talking about 5 card per person poker, not holdem. They aren't even considering discards and draws. It's simply if you're given 5 cards at the beginning, and that's it.
This is at least one pair, not exacly one pair, 12C3 would be just for the first card, then the card after it can't be 12 again, in his example, Kings AND Aces are forbidden for card number 4
12C3 makes sure that the 3 chosen cards can't have two cards of more with the same value. By expanding 12C3 = 12*11*10 / 3! we see that the third card (when the pair is in the 1st and 2nd cards) can be one of 12 values, the fourth card can be one of 11 values and the fifth card can be one of 10 values.
You thinking that really just means you understand what he's talking about. He talks that much to explain the subject to people that don't. Be considerate of people who need more explaining than you do to figure out the topic, and be glad you can understand it quickly enough for his explanation to become boring.
Yes there are restrictions that you didn't state. You need to state that after you pick a card, you do NOT replace it into the deck. Otherwise if you don't state that, then I, or anyone else, is free to replace the card, and then your probability would be wrong. You also need to state a well shuffled deck. Since you didn't, then I can put the cards in rank/value order, then your probability goes "out the window". You will get quads everytime that way.
That's not what he meant by "restrictions in the sample space". There are assumptions, yes. The example is considering a poker hand and in poker you don't replace drawn cards back into the deck. Also I'm quite sure that ordering the deck isn't allowed in poker.
This problem was difficult for me at the beginning. After many attempts I found this processes. **************************** Using a tree diagram and the idea given in the problem of the socks + permutation with repetition. Problem of the socks (ua-cam.com/video/Rd1VHhwoBJ4/v-deo.htmlsi=20iNcil9s1OtCeFN) permutation with repetition (ua-cam.com/video/1Uy2E2ncazg/v-deo.htmlsi=0ECKpBeej48rt6_K). Using P(SS~S~S~S), this is the probability of a card (S), the same card again (S), a different card (~S), a different card (~S), a different card (~S). This gives me : (52/52)(3/51)(48/50)(44/49)(40/48). Then P(SS~S~S~S) could have also happened as P(~S ~S~S S S) or P(S ~S~S~S S) just like a permutation with repetition in the form of 5! / (2!3!). The final answer is obtained by multiplying (52/52)(3/51)(48/50)(44/49)(40/48) * 5! / (2!*3!) ************************ The other way is a mixture of Combination and Permutation with repetition. You have an array of 5 places For the 1st card, you have 4C1 * 13 (You have 13 options and you must select 1 card) For the 2nd, you have 3C1 * 1 ( You only have one option to create a pair, so you select 1 card out of the 3) For the 3rd, you have 4C1 * 12 (You have 12 different options Suppose you have a pair of Kings, now you have 12 remaining options A, 2,3,4,5...). For the 4th you have 4C1 *11 For the 5th you have 4C1 * 10. So far, you have created the sequence Pair, Pair, not Pair , not Pair, Not Pair (PPNNN), but this could also have appeared as PNNNP or PNNPN ... creating a permutation with repetition so you are overcounting by (2! * 3!) The numerator is obtained as (13) (4C1) * (1)(3C1) * 12(4C1) * 11(4C1)* 10(4C1) / (2! *3!) Final answer = numerator / (52C5) These methods are not meant to be easier. They only offer a different approach .
Need to calculate some more probabilities (Three of a kind, two pair, straight, flush, royal flush). & Then I am on my way to Vegas.
A good guesstimate in Texas Hold'em:
After Flop: OUTS*4 = P(completing your Hand)
After Turn: OUTS*2 = P(completing your Hand)
*OUTS = Number of Cards missing to complete any possible Hand
This helps for holdem, he calculated that in holdem we will have a paired board 42% of the time. That's schwifty
does the solution consider having a pair in the ' last 3 ' ? after the first pair , there is a choice of 48 cards for the 3rd card , 44 for the 4th and 40 for the 3rd .
It factors in the chance of there being a pair in the last three by using 12C3, which is the same as 12C1 x 11C1 x 10C1.
The 3 for 12C3 basically means "choose three different values"
@@trolleymouse I have been constructing a card game for an assignment. I have removed aces from the deck for other reasons. I need to calculate the odds of drawing specifically one pair, when drawing 6 cards. 4C2 x 12C1 x 11C4 x (4C1)4 / 50C6 is the final solution i came up with. However this contradicts the video, where he cubed his value inside the brackets, rather than multiplying. Should i be multiplying by four or raising it to the fourth power?
I think there is a phrasing of the problem here, if we wanted to only have one pair in a hand the first 2 pick was perfectly fine, but for the third card there are 12 values and we can pick any four of them, for the 4th card we cannot pick the 12 values again if we do then we have pairs again so for the 4th pick it should be 11 out of 1 and 4 out of 1 and for the 5th card we pick 10 out of 1. this ensures that there are no other pairs of the card.
the problem above ensures there is at least one pair not ONLY ONE PAIR.
Hi, the thing is that when he does 12C3 he is saying that is choosing 3 diferent VALUES of the remaining 12 VALUES. I don't know if that helps.
This is what I was wondering about. Thank you.
I have found a wonderful way of looking at counting probabilities. Thank you, Eddie.
Extremely well done, I'm still struggling with the extra cards, that's me, and I will get it after a few reviews. I don't believe most people know how to take the extra cards into account, making it one pair and not possible two pair or something else. Thanks, the Royal flush seems the simplest to figure, am I wrong wasn't this considered the dead man's hand, just asking.
Dead man's hand is A8 with a black suit as a starting hand iirc. Cheers mate!
@@jamalblack8995 Thanks
@@jamalblack8995 thank you
1 thing I don't understand... is it the probability of picking the cards all at once? cause for me it seems like it. Then my question is, how much more complicated is it to calculate the probability of picking the cards one after another?
Why after choosing the pair (13* 4C2) cant you multiply by 48C3 ? What makes the answers so different?
If you multiply it by 48 C 3 then you could get a card of the same rank as your pair, meaning you could have 3 or 4 of a kind. By multiplying it by 12C3×4×4×4 you guarantee that you wont pick a card of the same rank as your pair.
@@juanmse Thank you! That makes sense.
@Juan Sanchez - That is actually wrong. 48 choose 3 means that we are guaranteeing we will NOT choose any of the 3 remaining cards the same rank (value) as the pair we already chose. The problem with 48 choose 3 is that it is possible to choose another pair (thus making the hand have 2 pair) or even a triple (making the hand a full house). It would have to be 50 choose 3 to allow the same rank as the pair we already previously chose (the first 2 cards chosen from the deck).
His explanation of the 12c3 is confusing. He says you can't pick any of the remaining Kings (triples, quadruples). But you also can't pick any cards that match themselves. He left that part out, I think. That is why it is 12c3 (any three cards from a set of 12 different ranks) where card 3 is one of four suites 4c1, card 4 is one of four suites 4c1, and card 5 is one of four suites 4c1. Fun stuff can be a headache sometime.
for the 12c3 it works because if we have 12 values left and we need to fill up the remaining spots it'd be 12*11*10 and we don't need the remaining 9!. So we can write this as 12!/9!, but since the internal arrangement of 3! must be considered, (from the cards that have just been picked) we must divide this out as well giving us: 12!/(9!*3!). This can then be written as 12 choose 3, since 12!/(3!*(12-3)!)
You are right, he should have reduced the subsequent picks as well. His problem is at least one pair, this is the reason the probability was so high. the problem is to have AT LEAST ONE PAIR
This helped so much! Thank You!
Why can we not do 4C3 to find the remaining cards? I feel like 12C3 x 4C3 should be the same for the second part but it’s not.
it's 4C1^3 instead of 4C3 because the remaining cards can be the same suit, just not the same number
“A problem well phrased is half solved.”
Then just phrase it well twice and the problem should solve itself.
@@davidjames1684 Ahh, but there's the problem: once you've phrased it well you've already solved half of the problem so there's just half a problem left. If you then phrase that half-problem well you have solved half of half-problem ie. a quarter-problem. There's still another quarter-problem to solve.
This can be generalized: after n well-phrasings, you still have (1/2)^n of the problem to solve.
i think you need to subtract the number of ways for the full house before multiplying by the pair - the three other cards must not be of the same value
never mind. I was wrong with my statement above. 12C3 means the three cards will never be of the same value
What about subtracting out two pair if you can only get one pair?
A model class in permutations: Chances and uncertainty. Thank you🌹.
brilliantly explained
You are gifted !
Thanks a lot, very helpful!
Another way of solving the problem is thinking this way: you have 52 out of 52 cards to choose from (52/52), then, after that, you have only 3 cards out of 51 to choose from (3/51). And the position of these 2 cards can be swapped 2 times between them, so you must divide by 2! So, the first 2 slots equation will be (52/52 * 3/51) / 2!. For the 3 remaining positions, in the third slot, we have 50 cards at our disposal, but we can only pick 48, because we have already chosen 2 cards of the same value and there are another 2 cards of the same value yet, so we have 48/50. At the fourth slot, from 49 cards remaining, we must subtract 2 cards from the first pair and 3 cards from the third slot, so 44/49. For the fifth slot, we will have 48 cards at our disposal, but we will only can pick 40, because we must subtract 2 cards from the first pair, 3 from the third slot and 3 from the fourth slot, so 40/48. And the 3 remaining slots must be divided by 3!, the number of ways the 3 cards can be swapped between them. Finally, we must multiply all this by 5!, the number of ways the 5 cards selected can swap positions between them. In summary, the equation will be: (52/52 * 3/51 * 48/50 * 44/49 * 40/58 ) * (5! / 2! * 3!).
Didn´t understand, if i pick one king, three others left, so i should just combine C3,1 to form a pair not a C4,2. Sorry for bad english, am not native
4c1 x 3c1 = 4c2
Have you got videos on how to get a 2 pair, royal flush etc... great video sir 👍
Superb.
I agree...this is fantastic.... but I disagree that for a Pair we need consideration of the fact that once one card is picked the second card as condition probobility and not funny consderation of the suit.
Clearly explained,Thank you
Excellent
Thank you Joshua Lee
Thank you for this video
Very helpfull !
1: A8
2: JJ
AA8J4
Who won this hand,, please explain
A8- it’s a full house.
A8 won. They both have a full house, but the higher trips are the deciding factor. So AAA88 vs JJJAA. If trips are the same the higher pair decides.
Not enough information given. The person with A8 may have folded for example.
Best Explanation Ever... Helped a lot :)
How can we represent n choose k without using a calculator?
n choose k is shorthand for n! / [ (n-k)! * k! ].
Can you help me calculate how many hands are expected to get 3 cards for a royal, I heard is every 92 hands approx but I dont find a correct way of calcualted, I know there are 40 distinc hand but with the 2 remaining spots should I multiply by (38c2), so I exclude the 5 for the royal and also any 3 of a kind, @eddie woo
it is not good but GREAT
8:26 The more cards you get, the more likely a pair will be in there somewhere is incorrect. If you already have a pair say in the first 2 cards, then getting 3 more cards will NOT increase your chances of getting a pair somewhere in there as you stated. Actually in a 5 card poker hand dealt from a 52 card deck, it is MORE likely that you do NOT get just one pair. 42.2569% chance of getting exactly one pair means 57.7431% or getting something else, so once you get a pair (such as dealing the cards one at a time), then no, there is NOT a higher probability of getting a pair once you already get it and draw more cards. For example, K, K, A, 7, K.
Is this the probability of ONLY ONE pair, or AT LEAST ONE pair of cards?
only one
At least one pair
@@karolisgudziunas9738 Not at least one pair. Specifically one pair. Thats why when u pick the other three cards u only have 12 to chose from, not 13. Kings would make it a better hand (trips or quads).
@@jamalblack8995 but the other 3 can include a pair, that's what they are asking. So you got the 2 Ks but the other 3 could be two 7s and a 9.
@@nicolasseminara9307 I don't think within the other three cards could be another pair, since you pick the value 12C3. So the three cards can never be the same.
Wait, is this percentage ONLY for the two cards in my hand of getting a pair, or your best hand of 5 cards which means that percentage includes a pair on the flop as part of your best 5 cards
@Paul MácRéamóinn - It could be that you get all 5 cards or you get 2 cards and there are 3 "community" cards (a 3 card flop). The probabilities are the same for both situations. This is assuming a standard full 52 card deck.
@@davidjames1684 it's actually the probability of only have a one pair. that's why he factored in 12c3, to choose a 3 sized combination of the remaining 12 card values. i believe.. probability for AT LEAST one pair would be 13c1 x 4c2 x 52c3
In most probability questions, including this one, they're talking about 5 card per person poker, not holdem. They aren't even considering discards and draws. It's simply if you're given 5 cards at the beginning, and that's it.
🃏🃏
What’s the answer for 2 pairs
(13C2)(4C2)^2(11C1)(4C1)
This is at least one pair, not exacly one pair, 12C3 would be just for the first card, then the card after it can't be 12 again, in his example, Kings AND Aces are forbidden for card number 4
12C3 makes sure that the 3 chosen cards can't have two cards of more with the same value.
By expanding 12C3 = 12*11*10 / 3! we see that the third card (when the pair is in the 1st and 2nd cards) can be one of 12 values, the fourth card can be one of 11 values and the fifth card can be one of 10 values.
dudes talk too much, just go straight to the point.
You thinking that really just means you understand what he's talking about. He talks that much to explain the subject to people that don't. Be considerate of people who need more explaining than you do to figure out the topic, and be glad you can understand it quickly enough for his explanation to become boring.
you still can get a flush
Yes there are restrictions that you didn't state. You need to state that after you pick a card, you do NOT replace it into the deck. Otherwise if you don't state that, then I, or anyone else, is free to replace the card, and then your probability would be wrong. You also need to state a well shuffled deck. Since you didn't, then I can put the cards in rank/value order, then your probability goes "out the window". You will get quads everytime that way.
That's not what he meant by "restrictions in the sample space".
There are assumptions, yes. The example is considering a poker hand and in poker you don't replace drawn cards back into the deck. Also I'm quite sure that ordering the deck isn't allowed in poker.
Its wrong
When he takes 12 to mantain his hand . He might have another pair . Bloody bullshit
6:13 "twelve choose thirteen". What a screwup. Downvoted for inaccuracy.
This problem was difficult for me at the beginning. After many attempts I found this processes.
****************************
Using a tree diagram and the idea given in the problem of the socks + permutation with repetition.
Problem of the socks (ua-cam.com/video/Rd1VHhwoBJ4/v-deo.htmlsi=20iNcil9s1OtCeFN)
permutation with repetition (ua-cam.com/video/1Uy2E2ncazg/v-deo.htmlsi=0ECKpBeej48rt6_K).
Using P(SS~S~S~S), this is the probability of a card (S), the same card again (S), a different card (~S), a different card (~S), a different card (~S). This gives me : (52/52)(3/51)(48/50)(44/49)(40/48). Then P(SS~S~S~S) could have also happened as P(~S ~S~S S S) or P(S ~S~S~S S) just like a permutation with repetition in the form of 5! / (2!3!).
The final answer is obtained by multiplying (52/52)(3/51)(48/50)(44/49)(40/48) * 5! / (2!*3!)
************************
The other way is a mixture of Combination and Permutation with repetition.
You have an array of 5 places
For the 1st card, you have 4C1 * 13 (You have 13 options and you must select 1 card)
For the 2nd, you have 3C1 * 1 ( You only have one option to create a pair, so you select 1 card out of the 3)
For the 3rd, you have 4C1 * 12 (You have 12 different options Suppose you have a pair of Kings, now you have 12 remaining options A, 2,3,4,5...).
For the 4th you have 4C1 *11
For the 5th you have 4C1 * 10.
So far, you have created the sequence Pair, Pair, not Pair , not Pair, Not Pair (PPNNN), but this could also have appeared as PNNNP or PNNPN ... creating a permutation with repetition so you are overcounting by (2! * 3!)
The numerator is obtained as (13) (4C1) * (1)(3C1) * 12(4C1) * 11(4C1)* 10(4C1) / (2! *3!)
Final answer = numerator / (52C5)
These methods are not meant to be easier. They only offer a different approach .