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Server Time: 12/3/2008 3:05:04 PM PACIFIC |
 Maths question, calculators out.., chasepoker, 17. Jun 2003 17:36 | ||
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| Quick maths question, I play a game called 10 card brag with a standard 52 card deck, you get 10 cards dealt to you and i for the purposes of the game i would like to know a) what are the odds that you will get at least 4 pairs dealt to you and b) what are the odds of getting dealt four of he same number ( quads ie K-K-K-K ). I know there are some maths legends out there, anyone know the answer ( a simpletons explanation of how you got there would be useful as well ) I do actually have a few other statistics questions if anyone can be bothered to work this out i might ask them. Cheers Chase | ||
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| Re: Maths question, calculators out.., jumpthru, 18. Jun 2003 01:26 | ||
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| Well...this is my stab at it. I can't guarantee I did the math right, but I spent about a half hour on it, so hopefully... Sincerely, Nate To find probability, it is the chance of hitting what you want, over all possibilities. ---------------------------------------------------------------------- Q: Chance of hitting four of a kind (after 10 cards) A: 0.000000172566% of the time. ----------------------------------------------------------------------- The probability of hitting four of a kind after getting 10 cards, is 13*C(4,4)*C(10,4) and the total number of possibilities after 10 cards is C(52,10). 13*C(10,4) ---------- c(52,10) or 2730 ----------- 15820024220 or 3 -------- 17384642 or 0.000000172566% of the time. ---------------------------------------------------------------------------- Q: Chance of hitting at least four pair (after 10 cards) Including double pairs... AKA: 44 55 66 44 9K or 33 22 33 22 9K A: .069386% of the time. ---------------------------------------------------------------------------- The probability of hitting four pair after getting 10 cards, is ((13*6)*(12*6+1)*(11*6+2)*(10*6+3))*C(10,8) and the total number of possibilities after 10 cards is C(52,10). ((13*6)*(12*6+1)*(11*6+2)*(10*6+3))*C(10,8) --------------------------------------------------------------- C(52,10) or 1097689320 ------------------- 15820024220 or 35478 ---------- 511313 or .069386% of the time. ---------------------------------------------------------------------------- Q: Chance of hitting at least four pair (after 10 cards) Without double pairs... AKA: 22 33 44 55 98, not 22 33 44 33 98 A: .06326% of the time. ---------------------------------------------------------------------------- The probability of hitting four pair after getting 10 cards, is (((13*6)*(12*6)*(11*6)*(10*6))*C(10,8)) and the total number of possibilities after 10 cards is C(52,10). ((13*6)*(12*6)*(11*6)*(10*6))*C(10,8) --------------------------------------------------------------- C(52,10) or 1000771200 ------------------- 15820024220 or 349920 ---------- 5531477 or .06326% of the time. | ||
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| Re: Maths question, calculators out.., Paul Stine, 18. Jun 2003 06:11 | ||
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| Nate, Without doing any math I suspect that your first number is way off; orders of magnitude off. Have you every player Chinese (13 card) poker? Quads are dealt regularly, certainly not in the 100,000ths of a percent range. It is only slightly unusual to be dealt two quads in 13 cards. A quicky, spreadsheet cacluation of the probability of being dealt quads in four cards (like in Omaha) is 0.0000480192. This based on the calculation P(any quads) =(3/51)*(2/50)*(1/49). 20824:1 against Your approach *looks* right, but I think you have left something out. Somewhere there is a publish stat for making quads in 7-card stud. This will give a closer approximation. OK here it is. If you are dealt trips in 7-stud you the probability of being dealt a four in the next 4 cards is 8.17% or 0.0817. We know the odds of being dealt trips are 424:1 against. so this can be expresssed as 1/425 or 0.002352941. The product of these probabilities is 0.000192235 or 0.019% or 5200:1 against. So, by adding three more chances we have shortened the odds considerably. Don't have time to take this further, but I'm sure you can find the hiccup. Paul Stine College Station, TX on 18. Jun 2003 01:26 jumpthru wrote: > Well...this is my stab at it. I can't guarantee I did the math right, but I spent > about a half hour on it, so hopefully... > > Sincerely, > Nate > > To find probability, it is the chance of hitting what you want, over all > possibilities. > > ---------------------------------------------------------------------- > Q: Chance of hitting four of a kind (after 10 cards) > > A: 0.000000172566% of the time. > ----------------------------------------------------------------------- > > The probability of hitting four of a kind after getting 10 cards, is > 13*C(4,4)*C(10,4) and the total number of possibilities after 10 cards is C(52,10). > > 13*C(10,4) > ---------- > c(52,10) > > or > > 2730 > ----------- > 15820024220 > > or > > 3 > -------- > 17384642 > > or > > 0.000000172566% of the time. > > ---------------------------------------------------------------------------- > Q: Chance of hitting at least four pair (after 10 cards) > > Including double pairs... > AKA: 44 55 66 44 9K or 33 22 33 22 9K > > A: .069386% of the time. > ---------------------------------------------------------------------------- > > The probability of hitting four pair after getting 10 cards, is > ((13*6)*(12*6+1)*(11*6+2)*(10*6+3))*C(10,8) and the total number of possibilities > after 10 cards is C(52,10). > > ((13*6)*(12*6+1)*(11*6+2)*(10*6+3))*C(10,8) > --------------------------------------------------------------- > C(52,10) > > or > > 1097689320 > ------------------- > 15820024220 > > or > > 35478 > ---------- > 511313 > > or > > .069386% of the time. > > ---------------------------------------------------------------------------- > Q: Chance of hitting at least four pair (after 10 cards) > > Without double pairs... > AKA: 22 33 44 55 98, not 22 33 44 33 98 > > A: .06326% of the time. > ---------------------------------------------------------------------------- > > The probability of hitting four pair after getting 10 cards, is > (((13*6)*(12*6)*(11*6)*(10*6))*C(10,8)) and the total number of possibilities after > 10 cards is C(52,10). > > ((13*6)*(12*6)*(11*6)*(10*6))*C(10,8) > --------------------------------------------------------------- > C(52,10) > > or > > 1000771200 > ------------------- > 15820024220 > > or > > 349920 > ---------- > 5531477 > > or > > .06326% of the time. | ||
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| Re: Maths question, calculators out.., timmypitt, 18. Jun 2003 07:34 | ||
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| without doing any stats, the answer to the first question must be way off because the chances of getting 4 pair when allowing quads was .069386 and the chance of getting 4 pair without quads was .06326. This implies that the chance of getting quads is at least .069386-0.6326, or .006126. However, there will be times when quads come up and there are not four pair. So I would estimate that quads should come up about 1/100 times when dealt 10 cards??? Hope that makes sense. | ||
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| Re: Maths question, calculators out.., stdioh, 18. Jun 2003 08:23 | ||
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| I think it is time to address an issue in enumeration and probability. What you're doing here is trying to find the probability that the cards come up in the desired way. Your method is right, but there are a lot of ways to make mistakes and it is arduous. Often with cards, a much better approach is to think of the total possible number of configurations in the universe (in this case a 10 card hand) and then to think of the total number of configurations that fit into your cases (4 pair+ or quads). When I have a minute today I'll do the calculation that way and post it here. | ||
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| Re: Maths question, calculators out.., stdioh, 18. Jun 2003 08:31 | ||
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| Oh, and for the sake of it, calculating the universe is easy. You're looking at 10 cards from 52 and the order is irrelevant, so you've got 52 choose 10. That is 52!/(10!)(42!) which is equivalent to the number of possible ways to draw 10 cards in order (52!/42!) divided by the number of ways we can arrange those 10 things (10!). The result is 57,407,703,889,536,000 / 3,628,800 and the result of that division is 15,820,024,220. Now all we need to do is for each case find the number of ways we can build a set of 10 cards into a valid configuration, divide by the universe and we're there. | ||
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| I got impatient., stdioh, 18. Jun 2003 08:59 | ||
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| ok...I had to do it right away. We've already shown that the universe has 52!/10!42! configurations. Now for quads, how many ways can we make this? There are 13 ways to take quads out of the deck and we don't have to care about the other 6 cards at all. Thus there are 13x(48 choose 6) ways to make quads. That's (13*48!) / (6!*42!) ... now we divide that by (52! / 10!*42!) and we get 13*10!*42!*48! / 6!*42!*52! = 13*10! / 6! * 52*51*50*49 = 13*10*9*8*7 / 52*51*50*49 ... which is just 65,520 / 6,497,400 which is a 1/ 99 and 1/6 chance. ... so for all intents and purposes you should get quads 1 time in 100. As for the other one, we do the same calculation here again. When you say that you get 4 pair, I assume that means no trips or quads allowed, but 5 pair is allowed. The number of configurations for one pair is 13*6 since there are 13 ranks and 6 ways to pick 2 cards. After you make a pair out of the deck there will only be 12*6 ways to pick a different pair, etc. Thus we get 4 pair as (13*6)*(12*6)*(11*6)*(10*6) Now we have to take our last 2 cards. There are 36 cards left in the deck that do not make us trips and we take any two of those (it is ok if they pair eachother) which is 36 choose 2 which is 36!/2*34! which is 36*35/2 which is nicely 630. Thus we have 13*12*11*10*6*6*6*6*630 configurations which is just 14,010,796,800 configurations. This is rather astonishingly high, as 14,010,796,800 divided by 15,820,024,220 is 0.08856. Thus there is a 1/11.29 chance that you will be dealt 4 or 5 pairs. I hope that helps anybody who reads it in terms of the thought process of calculating odds from a deck of cards. Most of the time you want to find the total number of possible configurations, the total number of "winning" configurations, and then divide. | ||
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| Re: I got impatient., Mark, 19. Jun 2003 11:15 | ||
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| I just looked at your answer for quads, it seems way too high, 1 in 100. Take out a deck of cards and try it, you won't need to do it 100 times. My answer may be a bit low due to my rounding off. Also, logic should tell you that 1 in 100 is too high. You are taking 1/5 of the total deck. Getting quads in in the same 1/5 of the deck is not that rare. mark | ||
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| Re: I got impatient., stdioh, 19. Jun 2003 12:03 | ||
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| Well, it did match the results from that other poster's simulation so that's worth something. I think I did make an error in the 4 pair calculation however. I missed divinging one piece by 4!. Thus it should be 3.69% chance of having 4 or more pair. And that's pretty close to he 3.44% from simulation. As for justification of the quads, think of it this way. How many ways can you make quads with 4 cards? ->13. How many ways can you make an additional 6 cards? 48 choose 6. Thus quads can be had by multiplying 13 times 48 choose 6 and dividing by the universe. That gets you about 1 in 99 which means that the simulation showing 1.01% is very close to theoretical numbers. | ||
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| Re: Maths question, calculators out.., Mark, 18. Jun 2003 08:49 | ||
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| The actual math behind the accurate answer is very long, and since i don't remember how to shorten it, i will show how to get an approximate answer. To show the math, 2 pair in 4 cards. You have to start by fliping one card (X). You have 3 chances to pair X, and the odds of pairing on any of the three chances is 3/51, 3/50, and 3/49. To figure out the chances of at least a pair you need to add these fractions toghether 3/51 + 3/50 + 3/49 to do this you must get a common denominator 51 * 50 * 49. the short way to show this is with a factorial (51-49)! You get an answer of 22497/124950 or 4.5:1 Now to this you must multiple the odds of pairing which ever other card falls when X is not paired. This cards is card Y. Card Y only has 1 chance to pair. 3 cards in 49 = approx 15:1 So the odds of these 2 events happening together is 15 * 4.5 = 67:1 Quads your first card is card X. You have 9 chances to pair it, 3/51, 3/50, 3/49, 3/48, 3/47 and so on. this produces a large equation with a large factorial (51-42)!, which is a pain to calculate. So we will round off by using 3/51 *9 (chances) and get 0.8 : 1 to get 3 of a kind, you have 2/50 with 8 chances (1 of the original 9 must be used up with the pair) giving 2.1 : 1 and to get 4 of a kind, you have 1/49 with 7 chances giving 6:1 now, you multiply all these chances together and get 10:1 This is only a rough approximation because i was not going to do the lengthy calculations by hand. Also, my sample sized (1 in 49 for quads, are not accurate, but again, I wasn't going to do those factorials by hand) mark | ||
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| Re: Maths question, calculators out.., stdioh, 18. Jun 2003 09:01 | ||
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| I think you're a bit off here Mark - maybe because of approximation - I'm really not sure what math you are using in this one. See my solution for a quick answer an an easy algorithm. | ||
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| Re: Maths question, calculators out.., chasepoker, 18. Jun 2003 10:43 | ||
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| I have no idea how you calcualte it but experience tells me that it must be about 50-1 to get 4 pairs and about 500+-1 to get quads ( THAT IS JUST AM APROXIMATION TO SEE IF YOU ARE IN THE RIGHT AREA ! ) Chasepoker | ||
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| Re: Maths question, calculators out.., stdioh, 18. Jun 2003 12:12 | ||
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| Nope...way off. My calculation has it as an 1:10 ish shot for getting 4+ pairs and a 1:100 ish shot for getting quads. Unless somebody can find a hole in my math (which I think is fairly rigorous), that's how it is. | ||
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| Re: Maths question, calculators out.., flintsword, 18. Jun 2003 23:13 | ||
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| I love a challenge. I will include all gruesome steps so even Members of Parliment and Congressmen can follow the logic, ... or worse ... someone can spot my mistake and pounce ... :( The easiest way to do this is to work out the odds of NOT getting ANY pair AT ALL when dealt ten cards: (52/52)*(48/51)*(44/50)*(40/49)*(36/48)*(32/47)*(28/46)*(24/45)*(20/44)*(16/43) = (20,927,899,238,400/1,103,994,305,568,000) = 0.019 or 1.9% So the odds of getting ten cards with NO pairs is 1.9% Therefore the odds of getting AT LEAST one pair is 1 - 0.019 = 0.981 or 98.1% By definition, getting four pairs exactly means six non-paired cards plus four pairing cards only. So the first six cards are: (52/52)*(48/51)*(44/50)*(40/49)*(36/48)*(32/47) The seventh card pairing "one" of the first six cards is: (3+3+3+3+3+3)/(46) = (18/48) = (3/8) The eigth card pairing "one" of five remaining unpaired cards: (3+3+3+3+3)/(45) = (15/45) = (1/3) The ninth card pairing "one" of four remaining unpaired cards is: (3+3+3+3)/(44) = (12/44) = (3/11) The tenth card pairing "one" of three remaining unpaired cards is: (3+3+3)/(43) = (9/43) Therefore the odds of having exactly four pairs in ten dealt cards is: (52/52)*(48/51)*(44/50)*(40/49)*(36/48)*(32/47)*(3/8)*(1/3)*(3/11)*(9/43) = 7,882,997,760/3,199,983,494,400 = 0.0025 or 0.25% That is (I hope) the answer to the first question. flintsword | ||
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| Re: Maths question, calculators out.., Peter, 19. Jun 2003 03:13 | ||
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| I thought I'd give a shot at solving this problem, but rather than working out all the math by hand, I wrote a little java applet that deals out cards then looks to see how many pairs, quads, etc... are in the hands. I ran the program for a while (about 16 million hands) and here are it's findings for dealing 10 cards per hand. Chance of getting 4 or more pair --- 3.44% Chance of getting 1 or more quads --- 1.01% If anyone's interested to run the program or check out the source code, here are some urls for you. 10-card app - http://www.bluesmith.com/10card/pokerstats.html 10-card source - http://www.bluesmith.com/10card/PokerStats.java I also put together a 5 card version to double check my logic and make sure the program came up with the right numbers. 5-card app - http://www.bluesmith.com/5card/PokerStats.java 5-card source - http://www.bluesmith.com/5card/PokerStats.java -- pete -- | ||
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| Re: Maths question, calculators out.., stdioh, 19. Jun 2003 09:29 | ||
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| Your quads data is in line with the math for my quads calculation, but I'm a bit suspicious of your 4 pair data. Are you sure that you're calculating 4 or more pair and not 4 pair exactly? | ||
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| Re: 4 of a kind in a 10-card hand, johnph77, 28. Jun 2003 20:41 | ||
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| Formula for 10 cards out of a 52-card deck: (52!-42!)/(10!) = 15,820,024,220 possibilities. What happens next is that four specific cards - your 4 of a kind - are selected, leaving 6 cards to be dealt to your 10-card hand out of the remaining 48. The formula for possibilities is: (48!)/(6!) = 12,271,512 SInce there are 13 such specific possibilities, the numer of hands in which this will occur is 159,529,656 Taking this number against the total number of possible hands leaves the odds of collecting 4 of a kind in a 10-card hands is 1::91.666667 or a probability of 0.010084. I'll work on the 4 pair possibilities later. | ||
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| Re: 4 of a kind in a 10-card hand, johnph77, 28. Jun 2003 20:43 | ||
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| Formula for hand possibilities is wrong - sorry - should read: (48!-42!)/(6!) - my mistake. | ||
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