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A Side Bet


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Mr. Gamble and Mr. Prude went to the Brain Den Casino, which offered the following game:

A customer bets on a number on a die (1 through 6). The dealer rolls a single die. If the customer’s number comes up, he is paid 4 times his bet, if not - customer loses his bet.

Without delay, Mr. G makes his mind to play 5 rounds betting $1 on number “1” each time. Mr. P, believing the game is in favor of the house offers a side bet, whereby if after 5 rolls Mr. G ends up behind he must pay $5 to Mr. P, otherwise Mr. P pays $5 to Mr. G.

After 5 rounds, what is the average payoff to:

1) BrainDen Casino?

2) Mr. Gamble?

3) Mr. Prude?

Results may be rounded to the nearest cent. Calculators are allowed.

Answers obtained by way of computer simulations may not claim the “best answer” spot.

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Gamble vs Casino for 0 1 2 3 4 5 "1"s showing is now better for Gamble:
-5 0 5 10 15 20.
Which means Gamble trails Casino only if no "1"s show at all.
This hurts Prude, who gets $5 from Gamble only if no "1"s show at all.

Revised numbers and expected payoffs.

Probabilities are still calculated as given in past post.


# "1"s 0 1 2 3 4 5

Winings

Casino 5 0 -5 -10 -15 -20
Gamble -10 5 10 15 20 25
Prude 5 -5 -5 -5 -5 -5
Total 0 0 0 0 0 0

Probabilities

.4019 .4019 .1608 .0322 .0032 .0001

Wins x Probabilities

Casino 2.0094 .0000 -.8038 -.3215 -.0482 -.0026
Gamble -4.0188 2.0094 1.6075 .4823 .0643 .0032
Prude 2.0094 -2.0094 -.8038 -.1608 -.0161 -.0006

Expectations to nearest penny.

Casino $.83
Gamble $.15
Prude -$.98

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Mr. Gamble and Mr. Prude went to the Brain Den Casino, which offered the following game:

A customer bets on a number on a die (1 through 6). The dealer rolls a single die. If the customers number comes up, he is paid 4 times his bet, if not - customer loses his bet.

Without delay, Mr. G makes his mind to play 5 rounds betting $1 on number 1 each time. Mr. P, believing the game is in favor of the house offers a side bet, whereby if after 5 rolls Mr. G ends up behind he must pay $5 to Mr. P, otherwise Mr. P pays $5 to Mr. G.

After 5 rounds, what is the average payoff to:

1) BrainDen Casino?

2) Mr. Gamble?

3) Mr. Prude?

Results may be rounded to the nearest cent. Calculators are allowed.

Answers obtained by way of computer simulations may not claim the best answer spot.

I think Mr. P has the right idea. He and the casino will win. I'll give part of the analysis.

Here are Mr. G's payoffs for each of the 5 rolls.

1. A 1 shows: his stake increases by $3. Casino loses $3.

2. A 1 does not show: his stake decreases by $1. Casino wins $1.

After 5 rolls, the payoffs to all parties can be any of 6 cases,

described by how many individual payoffs Mr G. received: 0 1 2 3 4 or 5.

Let these cases have respective probabilities p0 p1 p2 p3 p4 and p5.

First we note that Mr. G's winnings with respect to the Casino are -5 -1 3 7 11 and 15.

He comes out ahead if two or more "1"s show.

Casino's winnings are the opposite: 5 1 -3 -7 -11 and -15.

For the side bet,

Mr. P's winnings are 5 5 -5 -5 -5 and -5.

Mr. G's winnings are adjusted by -5 -5 5 5 5 and 5.

# "1" Mr. P Mr. G Casino

0 5 -10 5

1 5 -6 1

2 -5 8 -3

3 -5 12 -7

4 -5 16 -11

5 -5 20 -15

OK now multiply the rows by p0 p1 p2 p3 p4 and p5 then add the columns.

Oh yah, first compute p0 et al. :)

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The six probabilities are

.4018776 .4018776 .1607510 .0321502 .0032150 .0001286

p0 = 1 x (1/6)0 x (5/6)5

p1 = 5 x (1/6)1 x (5/6)4

p2 = 10 x (1/6)2 x (5/6)3

p3 = 10 x (1/6)3 x (5/6)2

p4 = 5 x (1/6)4 x (5/6)1

p5 = 1 x (1/6)5 x (5/6)0

The weighted table from my previous post is

Mr. G. Casino Mr. P

-4.01878 2.00939 2.00939
-2.41127 .40188 2.00939
1.28601 -.48225 -.80376
.38580 -.22505 -.16075
.05144 -.03537 -.01608
.00257 -.00193 -.00064

And the column sums are

Mr. G. Casino Mr. P

-$4.7042 $1.6667 $3.0376

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Gamble vs Casino for 0 1 2 3 4 5 "1"s showing is now better for Gamble:

-5 0 5 10 15 20.

Which means Gamble trails Casino only if no "1"s show at all.

This hurts Prude, who gets $5 from Gamble only if no "1"s show at all.

Revised numbers and expected payoffs.

Probabilities are still calculated as given in past post.

# "1"s 0 1 2 3 4 5

Winings

Casino 5 0 -5 -10 -15 -20

Gamble -10 5 10 15 20 25

Prude 5 -5 -5 -5 -5 -5

Total 0 0 0 0 0 0

Probabilities

.4019 .4019 .1608 .0322 .0032 .0001

Wins x Probabilities

Casino 2.0094 .0000 -.8038 -.3215 -.0482 -.0026

Gamble -4.0188 2.0094 1.6075 .4823 .0643 .0032

Prude 2.0094 -2.0094 -.8038 -.1608 -.0161 -.0006

Expectations to nearest penny.

Casino $.83

Gamble $.15

Prude -$.98

That is a good solve! Exact and exhaustive proof.

Complexity could be reduced just a bit, though.

A number is rolled 1 time out of 6 on average. If Mr. G played 6 rounds, his expectation would be to win once: +$4 and lose 5 times: -$1*5 = -$5, for a total of $4 - $5 = -$1. Since Mr. G plays only 5 rounds, his expected loss to casino must be reduced accordingly: -$1*5/6 ~ -$0.83.

As for the bet with Mr. P, since it is enough for Mr. G to win just one out of 5 rounds to break even, the only time he ends up behind is when he loses all 5 rolls. The probability of that is (5/6)5. Thus this bet’s the payoff to Mr. G is $5*(1 - 2*(5/6)5) ~ $0.98. The payoff for this bet is greater than Mr. G’s loss to the casino.

Overall:

Casino: +$0.83; Mr. G: +$0.15; Mr. P: -$0.98.

Thanks to Mr. Prude, his friend is not expected to lose money in this game.

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The dice is rolled 5 times. Mr G will be behind only if #1 never shows, so probability of Mr P winning the side bet is

    p = (5/6)^5

Therefore his average payoff is

    +5*p -5*(1-p) = 10p - 5 = 10*(5/6)^5 - 5 = -0.98

Casino wins on average

    +1*(5/6) -4*(1/6) = 1/6

with each roll, so expected casino's payoff after 5 rolls is

    5/6 = +0.83

The average payoff for Mr G is

    5 - 10*(5/6)^5 - 5/6 = +0.15

Edited by witzar
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