Question 1 of 2

Player A puts up $20
Player B puts up $5

Player A gets 4 Rolls
Player B gets 1 Roll

Tie is push. Highest Roll Wins.

Player A's EV?

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Question 2 of 2

One extra variable, if player A rolls two six's he auto wins.

Player A's EV?

Thanks Guys

Odds for rolling dice are:

2 - 1/36
3 - 2/36
4 - 3/36
5 - 4/36
6 - 5/36
7 - 6/36
8 - 5/36
9 - 4/36
10 - 3/36
11 - 2/36
12 - 1/36

If player B rolls a 2, the only way you couldn't win would be if you rolled four 2's in a row. The odds of that are (1/36)^4. So your odds of winning would be 1-(1/36^4). That comes out to .9999994, your odds of losing would be 0. Your odds of tying would then be 1-0-.9999994=.0000006

If player B rolls a 3, the only way you couldn't win would be if you rolled a 2 or a 3 four times in a row. The odds of rolling a 2 or a 3 are (1/36+2/36)=3/36. So your odds of winning would be 1-(3/36^4)=.9999518. The only way you could lose would be if you rolled four 2's in a row. That = .0000006. So the odds of tying would be 1-.9999518-.0000006=.0000476

Using this method for every possible outcome you get (In Win/Loss/Tie form):

Player B rolls a 2: 0.999999404625819 / 0 / 5.95374180765129E-07
3: 0.999951775 / 5.95374180765129E-07 / 0.0000476299344612486
4: 0.999228395 / 0.0000482253086419753 / 0.000723379629629578
5: 0.994046258 / 0.000771604938271605 / 0.00518213686937969
6: 0.969859182 / 0.00595374180765127 / 0.0241870760935833
7: 0.884211034 / 0.0301408179012346 / 0.0856481481481484
8: 0.727928288 / 0.115788966049383 / 0.156282745579942
9: 0.517746914 / 0.272071711629325 / 0.210181374790428
10: 0.293933256 / 0.482253086419753 / 0.223813657407407
11: 0.10656662 / 0.70606674382716 / 0.187366636183509
12: 0 / 0.893433380010669 / 0.106566619989331

We now multiply each answer by how likely the occurrence is. Player B will roll a 2 or a 12 1/36 times, so we multiply the win, tie, and loss percentage in the 2 and 12 row individually by 1/36. 3's and 11's are multiplied by 2/36 and so on. The numbers are:

2: 0.0277777612396061 / 0 / 1.65381716879202E-08
3: 0.0555528763717421 / 3.30763433758405E-08 / 2.64610747006937E-06
4: 0.0832690329218107 / 4.01877572016461E-06 / 0.0000602816358024648
5: 0.110449584243594 / 0.0000857338820301783 / 0.000575792985486632
6: 0.134702664180384 / 0.00082690858439601 / 0.00335931612410879
7: 0.147368505658436 / 0.00502346965020577 / 0.0142746913580247
8: 0.101101151162594 / 0.0160818008401921 / 0.0217059368861031
9: 0.0575274348422497 / 0.0302301901810361 / 0.0233534860878253
10: 0.0244944380144033 / 0.0401877572016461 / 0.0186511381172839
11: 0.00592036777718506 / 0.03922593021262 / 0.0104092575657505
12: 0 / 0.0248175938891852 / 0.00296018388859253

Add all the wins together, all the losses together, and all the ties together and you get:

Win - 0.748163816412005
Loss - 0.156483436293375
Tie - 0.0953527472946197

Multiply 0.748163816412005 by $5 = 3.74081908206003

Multiply 0.156483436293375 by -$20 = -3.1296687258675

3.74081908206003 + -3.1296687258675 = 0.61115035619253

So Player A will expect to win $0.61115035619253 or $0.61/go or $61.12/100 gos or however you want to look at it.

Player A is going to win every time he rolls a 12 unless Player B also rolls a 12. So the auto-win when player A rolls a 12 only changes the results when Player B rolls a 12. This happens so infrequently that Player A's win rate will only increase by about $0.01/go.

4 rolls of how many dice at a time?

I forgot to include the ties a minute ago.

61 cents.

We can calculate this exactly in R with

> b = 1/36*c(0,1,2,3,4,5,6,5,4,3,2,1)
> a = 1-(cumsum(b))^4
> pwin = sum(b*a)
> ptie = sum(b[2:12]*(a[1:11] - a[2:12]))
> plose = 1 - pwin - ptie
> ev = pwin*5 - plose*20
ev
[1] 0.6111504
> pwin
[1] 0.7481638
> plose
[1] 0.1564834
> ptie
[1] 0.09535275

b is the probability that B gets each of the numbers. a is the probability that A beats each of those which is the 1 minus the cumulative sum of those raised to the 4th power. We multiply those 2 sets of numbers and sum to get the probability that A wins. To get the probability of a tie, we multiply the probability that B gets each number times the probability that A ties it, which is the difference between the probability that A beats the next lower number and the probability that he beats that number, and we sum those products. Then we use the probabilities of a win and tie to get the probability of a loss, and we use these to compute the ev. If you want to see the probabilities that A beats each number, that would be

> a
[1] 1.0000000 0.9999994 0.9999518 0.9992284 0.9940463 0.9698592 0.8842110
[8] 0.7279283 0.5177469 0.2939333 0.1065666 0.0000000


62.6 cents. A would win anyway unless B rolled a 12. So we add the 1/36 probability that B would have rolled a 12, times the probability that A rolled a 12 which is 1 minus the probability that he didn't roll a 12, or 1 - (35/36)^4.

> pwin
[1] 0.751124
> ev = pwin*5 - plose*20
> ev
[1] 0.6259513

So A wins about 75% of the time either way.

One, but weather we rolled 4 at a time or all individual what does it matter? what am I missing?

One die? We assumed 2 dice. So "two sixes" means 2 out of 4 rolls of 1 die? Or is it at least 2?

Thanks a lot guys! very interesting calculations.

but...

Sorry about the confusion, All done with a single Die.

Question 1:

EV = $1.02

Note that p(tie) = 1/6, so p(lose) = 5/6 - p(win).

> b = 1/6*c(1,1,1,1,1,1)
> a = 1-(cumsum(b))^4
> pwin = sum(b*a)
> plose = 5/6 - pwin
> ev = pwin*5 - plose*20
> ev
[1] 1.019162
> pwin
[1] 0.7074331
> plose
[1] 0.1259002
> ptie
[1] 0.1666667


Or if you prefer:

> plose = 1/6 * ((1/6)^4 + (2/6)^4 + (3/6)^4 + (4/6)^4 + (5/6)^4)
> pwin = 5/6 - plose
> ev = pwin*5 - plose*20
> ev
[1] 1.019162
> pwin
[1] 0.7074331
> plose
[1] 0.1259002


Question 2:

EV = $1.13

Assuming 2,3,or 4 sixes wins automatically, we multiply the 1/6 probability that B would have gotten a 6, times the probability that A gets 2,3,or 4 sixes, which is 1 minus the probability that A gets 1 or less, which comes from the binomial distribution.

> pwin = pwin + 1/6 * (1 - pbinom(1,4,1/6))
> ev = pwin*5 - plose*20
> ev
[1] 1.129115
> pwin
[1] 0.7294239
> plose
[1] 0.1259002

I simplified that last post considerably with the obvious realization that p(tie) will be 1/6, so p(lose) = 5/6 - p(win). My initial approach was more general since it was originally for 2 dice.

Thanks BruceZ, That is a solid answer, much appreciated!