Team 13: Cián Mc Leod, Eoghan O Neill, Ruaidhri O Dowd, Luke Mulcahy

Transcription

1 Team 13: Cián Mc Leod, Eoghan O Neill, Ruaidhri O Dowd, Luke Mulcahy Our project concerns a simple variation of the game of blackjack (21s). A single player draws cards from a deck with or without replacement. The game only involves one player. The player must take a card on each draw regardless of what sum he has. Each card has a fixed value within a game. Question 1 [With replacement] The images here are taken from the worksheet Simulation without replacement (1). This worksheet simulates a simple game of blackjack with one player who draws cards from a deck with replacement. The game has been simplified in that the Ace card has a fixed value and there is only one player. This player must draw a card on each pass. [Ai] Using the images here, explain how the simulation works. Illustrate your answer with reference to the selected cell. There are four of each of the 13 cards in the deck of cards. As the game is played with replacement, there is a probability of of selecting any of the cards on any of the draws. From this the cumulative probabilities are calculated. A stream of random numbers are drawn and looked up in the cumulative probabilities. Twenty once calls to the random number generator are used as this is the maximum amount of draws a player can have to draw 21. Using the example, lies in the interval (7,8) and so is assigned card 8. The value of this card is then found and added to the sum of the previous sum to give a running sum. [Aii] Consider the alternative process of simulating the game here after the process of using random numbers to generate random values and mapping them to a specific card in the deck, explain the algorithm to get the frequency of blackjacks. After each whole random number has been generated it s ranked to an already pre-determined value in the pack e.g. A 10 results in 4 being picked the probability of selecting any card is the same (1/13) for each draw as the card is replaced after being picked, although as picture cards (bar the ace) are valued as being 10 hence there a bigger chance of drawing a card of value 10 (4/13) as there are 16 rather than 4 of these valued cards. The running score is found by summing all the card values in a game after each respective draw, as can be seen the screenshot below.

2 A countif algorithm is then used to search through the data of running scores to determine the number of blackjacks (21s) obtained after each draw for each game, this can be altered to also determine the frequency of any score from 1 to 210. (B) A thousand of these games are simulated with the table below summarising the results. The average of the first draw is Confirm this using theory developed in class. Use this result to derive expectations for the sum of the second and hence x n draw, where x i refers to the sum of card values after the i th draw. [ ] [ ] If we were to do an infinite number of simulations, we would expect our simulation to equal this theoretical value. As the game is played with replacement, the probabilities for choosing a card are constant for any particular draw. As a result; E[1 st draw] = E[2 nd draw] = E[n th draw] Therefore; [ ] [ ] [ ] [ ] [ ] [ ]

3 Average Score obtained Average Score after each draw Draw Number Average card value drawn Sum after draw n (C) Being in play means that the sum of cards is strictly less than 21. From the above table, we can see that in the simulation, 547 of our games are still in play at the end of the third draw. Use theory to confirm this result. Consider the implications of changing the card values and or number of cards. P(in play at end of third draw) = P(x 3 <21), where x i is the sum of cards at the end of the i th draw and v j is the card value on the j th draw. P(x 3 <21) = P(x 3 = 1,2,3,..,20) = = Sum P(Draw 3) Sum Carrying out the calculations (as per theoretical probabilities without replacement sheet in excel file) will give you the results in the table beside. The probability that a game is still in play at the end of draw three is So in a 1000 games, we would expect 539 to be in play at the end of the third draw. Considering we only achieved 1,000 simulations, a figure of 547 is remarkably close to the theoretical probability calculated here. The question of changing card values and or number of cards is quite an interesting one. Of course, we can engineer the probabilities so that the probability of being in play at the end of draw 3 is zero (set the lowest card value to be 7) or the probability to be 1(set the highest card value to be 6). However, let s consider some interesting blackjack related situations. Commonly, one can choose ace to be high (i.e. 11 rather than 1). Inputting this change to the simulation gives a much less number of games in play at the end of draw 3, with only 365 such games. Removing the picture cards (Jack, Queen & King) gives a deck of cards with each value 1-10 having equal probabilities of being drawn, 0.1. When this is implemented in the simulation we get a probability of 0.78 of being in play at the end of draw three.

4 Frequency Number of 21s obtained after draw n Draw (D) Derive the probabilities of getting 21 at each draw and compare these with those obtained by simulation. We can proceed as we did before in part c. P(x i =21) = Proceeding with these calculations, we get the probabilities for achieving 21 on each draw. The image on the left summarises the probabilities for the first few draws. The highest probability of achieving 21 occurs on the third draw (as one would expect due to the high number of cards valued at 10 in the deck). This shows that if one has not achieved 21 by the third draw that it is unlikely that he will be successful on the fourth draw. Graphing the probabilities for obtaining 21 on each draw for the theory and simulation gives the following image which shows our simulation is a very good substitute for the theoretical calculations which might be quite surprising as we only achieved 1,000 games.

5 P r o b a b i l i t y Distribution of 21s Draw number Theory Simulation (E) Before the game, the dealer allows you to alter the cards in the deck. He gives you a standard deck and also a pack of 100 Jack cards and allows you to add as many Jacks as you would like to the deck. Treating the value of cards as before, how many jacks would you place in this deck to maximise the chance of obtaining 21 at the earliest point. Take x i to be the sum of cards after the i th draw and v i be the value of the card drawn on the i th draw. If we treat the values of cards as before (Ace being one, picture cards 10, and the number cards taking their respective number), then to maximise the chance of obtaining 21 at the earliest point is to maximise the probability that we reach 21 on the third draw. i.e. max{p(x 3 =21)} P(x 3 =21) = Let x be the number of jacks in the deck, where Then: While P(v i = 1) = P(v i = 2) = P(v i = 3) = = P(v i = 9) = P(v i = 10)= Using P(x 3 =21) =, we can derive the probability of x 3 =21 in terms of x; [ ] [ ( ) ) ] Simplifying this expression gives: P(x 3 =21) = Differentiating this to get its maximum: [ ]

6 Probability As [ ], we know that (48+x) 4 >0 for all x. Therefore: =0 and ( ) Therefore x is approximately Looking at our theoretical probabilities with replacement sheet we can implement values for the number of jacks. So having 19 Jack cards in the pack gives us a probability of of obtaining a 21 on the third draw compared with and for 18 and 20 cards respectively. This shows that 19 is indeed a local maximum. So, if given this option we would add a further 15 Jacks to the standard deck. We can actually graph P(x 3 =21) against the number of jacks to see that the maximum does occur at approximately 19 jacks Number of Jacks (F) Let s say that you are playing with this augmented deck (19 Jacks and 4 of each other card) and cards take on their usual values. Given that you achieved 21 in your third draw, what is the probability that you drew a card of value ten (ten, jack, queen or king) in your first draw? Compare this with the simulations We seek P(v 1 = 10 x 3 = 21). Using Baye s theorem, we have ( ) ( ) ( ) ( ) P(x 3 =21) = as above and P(v 1 =10) = ( )

7 Looking at the clipping above taken from our simulation with replacement sheet (1) we can see that out of the 1,000 simulated games, 75 of them were successful in the third draw ( #(x 3 =21)=75). Of these 75 successful games, 33 of them had drawn a card of value ten at draw 1, giving us a probability of 0.44, hence validating the theoretical probability of Question 2 [Without replacement] [Ai] It could be claimed that the 1000 games simulated for looking at sums of cards after drawing from a deck without replacement is too small. What could be meant by this? In general, increasing the number of simulations will make the distribution of the generated counts more representative of the theoretical probability distribution. However, for this particular example, perhaps the generated distribution could be much more accurate. If one were to look at the counts table, it can be observed that each column (representing each draw) should add up to 1000 (as there are 1000 simulations). Considering that for example, for draw 6, the sums can be between 6 and 60 (and larger ranges for further draws), this is a large spread for only 1000 counts. Also a draw affects the probability of future results therefore the data for later draws can become quite skewed. A few clicks of the F9 button may show that the count table shows quite a variable distribution of counts.

8 [Aii] Even so, it can still be claimed that the simulation is very useful. Explain Given the drawback noted in the previous question, the simulation matches the theory for draws 1 to 6 remarkably well. It can be observed that the number of different methods used in deriving the theoretical probabilities were rather elaborate and time consuming in comparison with the rather straightforward simulation. In addition, producing the theory for higher numbers of draws requires ever larger numbers of calculations. Therefore a simulated estimated probability distribution may even be preferable (particularly if it was extended to more simulations, which is a simple task). Final probabilities for draws 1 to 6 from theory compare with a simulation probs sum draw 1 draw 2 draw 3 draw 4 draw 5 draw 6 draw 1 draw 2 draw 3 draw 4 draw 5 draw E E E E E E E E E E E [Bi] For the calculations involving the theoretical probabilities of the sums after 3 draws and the sums after 4 draws, is the step take account of ordering necessary? No, all possibilities could be listed. However, for any given combination, a different ordering has the same probability; therefore some simple rules involving factorials can reduce the number of possibilities that need to be listed. For example, for 3 draws, to incorporate the combinations involving drawing two 2s and one 3, we can consider the following 3 combinations: Draw 1 Draw 2 Draw And then consider the probability of each of these combinations individually. However, these three probabilities are equal three.. Therefore we can list one and multiply the associated probability by In general, we can find the number of orderings by using the fact that the number of ways of ordering n objects where p of type 1 are the same, q of type 2 are the same, r of type 3 are the same.etc. is given by. The only issue now is how to make a list that does not repeat the same combinations of numbers (in different orderings). This can be done in a straightforward (rather algorithmic) manner.

9 e.g [Bii] Compare the probabilities of winning the game for the game played with and without replacement. Game played with replacement Game played without replacement Theoretical probabilities Simulation with probabilities Theoretical probabilities Simulation probabilities Draw Draw Draw Draw Remarkably, the probability of achieving 21 is higher for the games played without replacement than it is with replacement. This is due to higher valued cards being removed. [C] Calculate the expected value from the values and variances for sums of drawn cards from one draw to six draws from the theoretical without replacement slides. Also calculate estimates for the expected values and variances for one to six draws from the without replacement file. Increase the number of simulations and then compare the theoretical expected value and variance with the estimate. Calculations as per question 1b. The example below involves 1000 simulations. expected values estimates of expected values (using a simlation) draw 1 draw 2 draw 3 draw 4 draw 5 draw 6 draw 1 draw 2 draw 3 draw 4 draw 5 draw 6 E[x] E[x] E[x^2] E[x^2] Var[x] Var[x] [D] There were a number of methods used for calculating the theoretical probability distributions for sums of cards without replacement. Create a method for devising the theoretical probability distribution for the sums of drawn cards for 4 draws from a simplified deck that consists of two aces (valued at 1), two 2s, two 3s and four 4s. You may use one of the methods used in the theoretical probability distribution for a 52 card deck or a combination of methods, you may even devise your own method (or alternatively use a simulation, if it is sufficiently accurate). You must then justify your choice of method, discussing

10 issues such as clarity, simplicity, information given and usefulness (e.g. does it give information on conditional probabilities? Can it be used as an easy method for other problems/decks/distributions?, how much time does it take to c Below is an example of a possible method of creating the probability distribution of sums of cards up to 4 draws. This involves calculations for 2 and 3 draws, but note that sums need only be taken account of after 4 draws. Also note that due to the simplification that there are only types of cards, there are relatively few possible combinations and therefore there is not much point in taking account of ordering. Note that there are there are 4 matrices for three draws and our probabilities for 4 draws could also be listed as 16 matrices. The third draw matrices corresponded to different first draws, then they gave probabilities for those first draws and certain 2 nd and 3 rd draws(with the numbers drawn on the second draw represented on the columns and the numbers drawn on the third draw on the rows, however, it could also be displayed the other way around- i.e the transpose of this.) If we were to display the fourth draw probabilities as matrices, we would have 16 matrices corresponding to the 16 possible combinations of first and second draws (and the columns and rows corresponding to the 3 rd and 4 th draws).

11 In principle this method can be extended to n draws with 4 n-2 matrices. However, this may be time consuming and the method displayed below is also effective for the 4 th draw. Note that when finally calculating the probabilities of various sums, it is first necessary to look at the possible events and what sums they correspond to and then take account of their corresponding probabilities (hence for the fourth draw he sums for all possible events are listed- and a sumif function is then used to get the overall probabilities). [E] Below there is a probability distribution associated with a simplified deck and two draws. Does the below distribution show probabilities conditional upon the first draw? Explain you answer. The probabilities are not conditinal in the sense that they give the probability of drawing a card of value x in the second draw given that a card of value y was drawn in the first draw. Rather the probabilities are for all possible events involving two draws, i.e the probability of drawing an x on the first draw and drawing a y on the second draw. This should be apparent form the sums given at the bottom of the table. [F] Amend the table given in the above question so that the probabilities for the second draw conditional upon he first draw are given. Would it be easier to do this by theory or simulation? This simply involves the same values as in the table for the previous question, however it is no longer necessary to multiply by the probability for the corresponding first draw. This can be generalised to further draws. This is easier to do by theory as it just involves lists (and then using the countif function) than by simulation (as this would involve taking account of the number of counts for the first draw and then the number of counts for the second draw. Therefore an attempt to answer this question by

12 simulation would involve simulating random orderings of the 10 cards (by using random numbers and a rank function), then the probability of drawing card x given that card y is drawn is estimated by: This appears to be more difficult/tedious than the theoretical approach. However in principle this same general method can be used to give estimates of a wide variety of conditional probabilities for any number of draws.