8/31/2023 0 Comments How to do permutations![]() If I switch these around in my hand, I still have those three cards to play the game, right? THAT's what Sal means when he says 'the order does not matter'. ![]() Does this make sense? For example, say I have a Queen, a Jack and a Spade. You would still have the same cards in your hand even if you switch a few around. And just to be clear, imagine you are holding some cards in your hand. We want to know the probability of getting all four 1's in our hand of nine cards. ![]() I did not fudge on this one either! I got it right the first time! I am strongly convinced that Sal is correct. When you add each of the 4 possible hands above you get 94,143,280. (It is an excellent question that you asked.) So we use the same logic for all the possible hands: 4, 3, 2, 1, and 0 ones. Yes, this is exactly what Sal has proposed. Which is real exciting, because 4 choose 4 = 1. You then do permutations of 4 choose 4 times 32 choose 5. There are 5 slots to place them in A.K.A. So, then we have 32 cards left to freely choose from. That's Leibniz for you! And I should also remind you this was invented as a way to cheat at cards. neither does math evidently, because how can you choose 4 cards to fill 0 slots and not 9 and multiply 0 by something to equal 1? You should really remember how the formula works (that is all Liebniz did.)įor example, we can't 4 choose 9, because then you end up with (-5)! which is undefined, but you can do 4 choose 0 only because 0! = 1. Or counting how many ways to choose 4 cards for 4 slots, all at once.īecause 0! = 1 (and only just because) we wind up with 4!/4! which is equal to 1.Īnd your intuition should tell you, there is only one way to "4 choose 4." What Sal is doing when he says 1*1*1*1 for the four 1's in the hand, is 4 choose 4. They all mean the exact same thing and will make this MUCH easier to write. I am not trying to confuse anyone! All these previous statements (*) are equivalent and are all equal to 94,143,280. since we are in a sense choosing 27 cards NOT to be in the hand. This can be written as, "36 choose 9." Which is found by 36! divided by 27! and 9! Then, when all the possibilities are added together, they should have a 100% chance of success.(2)Įach of these events are mutually exclusive and cover the range of ALL possible hands:ģ6*35*34*33*32*31*30*29*28 / 9*8*7*6*5*4*3*2*1 If it is correct, then the odds of getting three 1s, two 1's, one 1's, and zero 1's should all be calculated in the exact same way.(1) The event he explained is getting all four 1's in a hand of 9, out of 36 cards numbered 1 through 9 and of four suits. Yikes! That explanation got kinda long! I hope it helped :-) Then, all he had to do was figure out how many 9-card hands he could have based on that assumption. This is essentially the same thing Sal did, but instead of assuming he had just one card in his hand at the outset, he assumed he had all four 1s. Then, just divide this by the total number of possible hands and you have your answer. This would tell you the total number of hands you could have (52 minus the four of hearts = 51). Using a standard deck of cards, what would the probability be of having the four of hearts in your 2-card hand? Well, you can start by assuming you have the four of hearts, then figure out how many options you would have for the other card in your hand. We can essentially ignore the four 1s because we already accounted for them by assuming we had them from the outset.Ī simpler example might examined a 2-card hand (perhaps we're playing blackjack or something). ![]() The question then became how to fill the other five "slots" in our hand, which turns out to be (32*31*30*29*28)/(5!). Sal took this idea (the idea that there is only one way way to have all of the ones if our hand only had four cards), and he built from there. Unfortunately, we have 5 other cards in our hand as well, which make this more complicated (and interesting). If the size of our hand were just four cards, there would be exactly one way to do this. We know our hand will have four 1s in it. ![]()
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