How would this coin flip simulation work ?

[Godzilla_Hat]

This is what I have so far.

Data roll;
MyMoney=100;
	Die1 = ceil(6*ranuni(-1));
	Die2 = ceil(6*ranuni(-1));
sum = Die1+Die2;
if sum > 9 then MyMoney + 5;
else if sum < 9 then MyMoney - 2.50;


Run;

proc print data=roll;
run;

[PaigeMiller]

Are you asking for help?

[Godzilla_Hat]

Yes please

[ChrisHemedinger · Accepted solution]

How about something like this?

data RandInt;
length die1 8 die2 8 winnings 8 balance 8;
/* streaminit for reproducible sequence, comment out for different results */
call streaminit(123); 
retain balance 100;
do i = 1 to 100;
   die1 = rand("Integer", 1, 6);  /* requires SAS 9.4M5 or later */
   die2 = rand("Integer", 1, 6);  
   winnings = ifn(die1+die2>=9,5,-2.50);
   balance+winnings;
   output;
end;
drop i;
run;

Edit:

At first I had payout when the total > 9 only, and that's a sucker's game. But corrected to >= 9 and it seems better.

[ballardw]

Still a suckers game or at least not not for me.

Expected winnings per play, assuming fair dice, is -$0.41666666...

If you end up with more than 58.3333.... at the end of 100 plays you beat the house (a bit).

[ballardw]

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@Godzilla_Hat wrote:

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Why did you call this a "coin flip"? There are more than two outcomes per die and coins, barring the "lands on edge" event typically have only 2 outcomes.