Question

We are in casino playing roulette, American version. We decided to play only red/black, i.e. to bet on red only.

We will use the Martingale strategy. A game starts with the first bet of 1 chip. If you win the game is over. You won a game. If you loose, you double your bet. You proceed to double your bet till you finally win or if you loose 6 in a row. Either way that concludes one game. You start a new game by betting 1 chip again.

Fill up the following table

Game ended as we	Bet (no. of chips)	Casino pays	Profit	Probability of this happening
Won in the 1st round
Won in the 2nd round
Won in the 3rd round
Won in the 4th round
Won in the 5th round
Won in the 6th round
Lost 6 in a row

From the table compute the probability of winning in a game. A game has two outcomes therefore it is a

______________________ experiment.

We made a PDF table for the game. Include the gain function.

Sample Space	Win	Loose
PDF
Gain

Now we decide to play the game 12 times. Fill the table below.

Number of games won (out of 12)	Probability	Gain
0
1
2
3
4
5
6
7
8
9
10
11
12

For this table we used __________________ distribution

What is the probability of winning(making a profit) after 12 games?

What is the expectation of the gain after 12 games?

Should we Gamble?

Yes or No

Answer 1

American Roulette: There are 18 Reds and the total number of choices is 38. Thus the probability of winning a bet on Red is 18/38. The probability of NOT getting red P(Not red) = 20/38

The traditional martingale betting strategy calls for the better to double the wager after each loss until finally winning. A game starts with the first bet of 1 chip. If you win the game is over. You won a game. If you lose, you double your bet. You proceed to double your bet till you finally win or if you lose 6 in a row. Either way that concludes one game. You start a new game by betting 1 chip again. The payoff for this bet is 1-1, so if you bet 1 chip on red and the ball lands on a red number, you would win your original 1 chip back, plus an additional 1 chip in winnings.

game ended as we	Bet	Casino pays	Net gain	Profit	Probability of this happening	Probability of this happening
	(no. of chips)
Won in the 1st round	1	1	2	1	18/38	0.473684211
Won in the 2nd round	3	2	4	1	20/38*18/38	0.249307479
Won in the 3rd round	7	4	8	1	20/38*20/38*18/38	0.131214463
Won in the 4th round	15	8	16	1	20/38*20/38*20/38*18/38	0.069060244
Won in the 5th round	31	16	32	1	20/38*20/38*20/38*20/38*18/38	0.036347497
Won in the 6th round	63	32	64	1	20/38*20/38*20/38*20/38*20/38*18/38	0.019130261
Lost 6 in a row	63	0	0	-63	20/38*20/38*20/38*20/38*20/38*20/38	0.021255846

Won in second round implies lost in first (1chip) and doubled the bet to 2 chips, thus total bet = 3 chips. The casino pays 2 chips and you 2 chips are returned so net gain 4 and profit is the net gain - bet = 4-3 = 1. The probability of losing first and winning second is given by 20/38*18/38. Similarly, we can calculate for the rest.

A game has two outcomes, therefore, it is a binomial experiment.