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Cineikcn 30lnpelgamblor plays only clle qame MOSL one nour Cuch avenino Tho chance novana prolii Iha ed of lta Ir[1 plave rQulello and blacrJC 3096 and 2080 respect...

Question

Cineikcn 30lnpelgamblor plays only clle qame MOSL one nour Cuch avenino Tho chance novana prolii Iha ed of lta Ir[1 plave rQulello and blacrJC 3096 and 2080 respectively He chooses blackjack on 609 of Ihe ovenings endtlbi61 remainno Oveninos Sudoocahe shons ptolt alter playing Oriu parlicular evoning Whal is the probabilly 47a' ihet& playud blackiack?0.890.62Rescl Scltdion

Cineikcn 30ln pel gamblor plays only clle qame MOSL one nour Cuch avenino Tho chance novana prolii Iha ed of lta Ir[1 plave rQulello and blacrJC 3096 and 2080 respectively He chooses blackjack on 609 of Ihe ovenings endtlbi61 remainno Oveninos Sudoocahe shons ptolt alter playing Oriu parlicular evoning Whal is the probabilly 47a' ihet& playud blackiack? 0.89 0.62 Rescl Scltdion



Answers

Express all probabilities as fractions.
In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning “blackjack” hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order.Find the probability of being dealt a blackjack hand. What approximate percentage of hands arewinning blackjack hands?

And problem number 49 were asked to analyze a standard deck of cards and were asked to find the probability um for different scenarios. So the first scenario is that the probability that both cards are aces or 10 point cards. And in the prompt, it told you that there were 16 10 point cards And then there are four aces. And if you didn't know there were 52 cards total in a deck. Okay, So keep that in the back of your mind. All right. We're gonna need to know that. So there's a couple ways to go about this. There's a formula way and then there's kind of a logical way. So I'll show you the formula way. Um So here there are 20 you can think of like favorable. There are 20 that you want or the probability that both cards or races or or 10. So 20. And it says that we're selecting to to both cards. So we're choosing to And then there are 32 non 10 point or aces. That's the 23456789 For all four suits. There are 32 of them and we're choosing zero of those. And then that's divided by 52 choose to. Now. A way to double check to make sure you do the formula right? 20 plus 32 is 52 then two plus zero is too Okay. So then you can use the formula or I like to use a calculator personally. So if you go to 20 and then if you have a. T. I. T. For you go to math and then you go to probability and it's that N. C. R. So 20 choose to and then times um 32 math choose zero. And then that's divided by 52 Math choose to. Okay and that gives us about .1433. Let's say it's a .14 33 and that is the correct answer. Now, also, if you think of it logically we're assuming this is without replacement. So you can kind of say like or Um there are 20 possible choices out of the 52 for your first selection. And then whenever you draw that 10 or ace or king or queen, you know of any suit. Then there are 19 left over out of a total of 21. Whenever you do that, you take 20 times 19 divided by 52 times. I'm sorry, not 21. That should say 51. 51. That gives you the same thing. So basically, you know exactly the same way. Just kind of a different way to think about it. So then the second part of this it says what's uh find the probability that both cards are aces. All right. So there are four aces. Were having four total and we're choosing two of them. And that means there are 48 that are not aces. And we should choose zero those. And then that's still divided by the 52 choose to. And we saw with the exact same way exactly as Um I did earlier. I'm not going to show that again, but you should get .0045. Okay, Similarly, if you think of it logically, that's the same thing as saying, well, therefore Out of 52 aces and then you choose one, you don't put it back. That means there are three aces with 51 left over. And that still gives you the .0045. Okay, So either way is fine. All right. And then this 3rd 1 With the hyper geometric it says probability that both cards will have a point value of 10. It's the same thing. There are 16 of those totals to 16 and we're choosing two of them. All right. So that's the four tens, the four jacks, the four queens, the four kings. That's the 16. All right. And then there are 36 left over. Sometimes 36 choose zero. Now there's only one way to do that. So, it's really just like multiplying by one. But for formula sake, I went ahead and wrote it like that And then that's divided by the 50 to choose to again. And that should give you .09 05 Okay? So again, you could do that as 16/52 And then once you choose one you don't put it back. So then 15 Over 51 and that's also .0905. So either one of those works. Alright. So now we're finding the probability of the last one, Find the probability of a blackjack. And that's a special case. That's where you have an ace with Um a 10 value. So 10 jack queen king. And basically what you're going to use is the the first three answers here. So this total here is the probability that we get aces or um or 10 point card. So that's that 10.0.1433 Okay. But in those scenarios we have um not non blackjacks. So in those scenarios we could have you know like two tens or two aces or two jacks or something and we don't want to count those. So we need to subtract those out. So we're gonna subtract The .0045, you know the few times that you can get two aces. And then we're also going to subtract The .0905 a few times where you can get to 10 point Cards. And that should give you whenever you subtract all that .0483. So you have about a 5% chance of making a blackjack with the standard deck of cards.

So in the situation we How axe? Because who? Minus one if we don't lean. And we are how actually coastal truly revealing. Because are you pay off this street? Other are $1 Buy it. If the numbers like this one of those Children So we have the probability of winning is point who? Five Because of probability the summits one. So we got us oneness. One Manus Going to five were tree coastal born in silence And for B we can see the expected value dress equal to my nose. One time point, Someone sigh And then last two ready Zags Palms P the 0.25 princey does echoes through my nose $0.25.

For 43. We're trying to figure out the probability of a blackjack occurring. So remember when you go blackjack or it doesn't matter if you get the ace and the face card or the face card than the ace card, as long as you're getting it. So we gotta calculate each probability. So there is a four out of 52 chance that I'm going to get an ace, then a 12 out of 51 because you're not going to replace that 52nd card option. There. Also, there's the 12 out of 52 possibility uh times the four out of 51 possibility. So you go and you do the math, you get 48 out of 26, And 48 out of 26, So when you do that, you know, 48 out of 26, That's about 1.8% plus another 1.8% Gives you approximately 3.6% probably.

In this question for started the first part. They have said that whole game there are 38 tickets. The tickets okay, that are marked plus $2 and, uh, 26 tickets. Meta marked minus $1 in the books. Such chance off drawing plus $2 over Shores has to be 12. Jan's will be 12 out of 38 and we'll hear the Johns. It's going to be 26 out of 28 right now. Some of the numbers in the box We're going to calculate them so we can say this will be equal to 12 in tow. $2. And, uh, plus, we have 26 into minus $1. Right? So this actually gives us the average or the summer's negative toe, also, to find the average off numbers in the box now, so that is going to be minus $2 that were formed upon total, which is 38 so for simplified against it as minus $0.5. Moving further, they have said that the number of roller places 100 so I would write down the number off Roll it please is given as 100 waas, so the expected value for the net gain is a number off draws. Therefore, the expected value for the next game Well here is given by the number off the laws right into the average off the books. So number of draws enormous 100 average of the books if guys late as it is minus $0.5. So it actually comes to be minus $5 or here. So we therefore say that we are expecting a loss off $5 and moving for the and what do now, huh? Talk about the standard division of the books so we can see that the standard deviation off the box when the tickets in the boss show only two different numbers. Big and small. This is going to be equal to the big number, minus D small number. Okay. On. Until we dick squared off, the fraction meant the big number, okay. And into the fraction with this morning number. Right now, the big number we know is $2. Okay. And minus the smaller number we know is $1. It's actually minus $1 and in tow, squared off. We have the faction for the first big numbers 12 by 38 into for the others, 26 by 38. And, uh, so if you simplify, this gives you $3 in tow. This part if you saw, we get 0.47 So approximate value I'm getting here is $1.41. As a standard illusion Nobiz on this, I'm going to find out the standard editor also. So the standard error is given by square root off number of drawers into the standard deviation off the box we just found. Right. So this is square root off. The number of draws is 100 into this is $1.41. So any multiply, we get this approximately as $14. This is an approx value. Okay, Now, therefore, I would like to conclude that in 100 please the gamblers next game, we'll be around minus $5 give our big will be approximately $14. No going on to the next part for the Barbie. We're supposed to now count the number of friends in the game, so I'm going to use a box, uh, like this over here. I'm going it. Someone to have 12 tickets, which will be off one option and 26 tickets, which will be off auction. Zero. All right now, this game is in the average off. The books will be equal to begin to 12 into one, unless we have 26 in true zero on the whole thing, divided by 28. So they sold us their living this value approximately as 0.31 So therefore, expected value for the some off 100 such draws. So there's a vehicle 200 into 0.31 which approximately gives me 31 now for the after. Some careful it the standard division. So the standard deviation the formalize again one minus zero the values on in tow. The factions off the provided He's just well by 38 26 by 28. So this is going to be a zero point for seven. Next time. Finding is the standard editor just squared off. Number off draws into the standard deviation of the box. So this is against squared off 100 Indo 0.47 spaces, approximately five like so well. Therefore say that in 100 please. The gambler should win 21 times, give or take five or so. Right? Okay. Moving on to the last part, which is the part. See off this question. So we see that in single normal play, you can expect Oh knows around, uh, see, $25 in 100 place. Okay, So as a number off place increases to 1000 we say that as the number off plays increases true, 1000. We should expect the laws to be, uh, around $50 like they've been 25 in tow. That 25 4 100 flights so to 54,000 place. And that is more replay. The more we lose. So we conclude that more we play more me to lose here. Okay. So therefore, if you consider the column back, there's much instrument again. So if you consider the column bed, then that is more chance. Go win this game.


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