Expected Value of a hand

Hello. I have been a fan of Wizard of Odds for a long time, and I was hoping for some help on a dilemma I faced the other day. (My higher math skills are not very good.)

I was playing a 10 hand Deuces Wild machine and was playing 6 coins which kicked in a multiplier. It was a 95.96% return machine otherwise (800, 200, 20, 10, 8, 4, 4, 3, 2, 1). I understand the return is lower playing the sixth coin.

So, on an 8 multiplier hand, I was dealt a wild royal (four to a royal with one deuce). The return was 8000 coins total (these were nickel plays), and the natural royal would have paid 32000 coins per hit. Given the 10 hands in play, I was tempted to go for the natural royal and figured a roughly 1 in 5 chance of hitting one (odds decreasing dramatically of hitting more than one). I took the bird in hand and stuck with the 10 wild royals.

I'm curious though as to what my expected return would have been on discarding the deuce and going for the natural royal. I can see on the strategy page what the expected value is when playing one hand, but does that change other than just by multiplying by ten when playing ten hands?

Thanks to anyone who can shed some light on this question!

--Greg

Quote: bkdealer

Hello. I have been a fan of Wizard of Odds for a long time, and I was hoping for some help on a dilemma I faced the other day. (My higher math skills are not very good.)

I was playing a 10 hand Deuces Wild machine and was playing 6 coins which kicked in a multiplier. It was a 95.96% return machine otherwise (800, 200, 20, 10, 8, 4, 4, 3, 2, 1). I understand the return is lower playing the sixth coin.

So, on an 8 multiplier hand, I was dealt a wild royal (four to a royal with one deuce). The return was 8000 coins total (these were nickel plays), and the natural royal would have paid 32000 coins per hit. Given the 10 hands in play, I was tempted to go for the natural royal and figured a roughly 1 in 5 chance of hitting one (odds decreasing dramatically of hitting more than one). I took the bird in hand and stuck with the 10 wild royals.

I'm curious though as to what my expected return would have been on discarding the deuce and going for the natural royal. I can see on the strategy page what the expected value is when playing one hand, but does that change other thanwhen playing ten hands?

Thanks to anyone who can shed some light on this question!

--Greg

No dilemma, just multiplying by ten or whatever, do what the strategy page tells you. If you do that you won drive yourself bonkers with what if's and dam I shouldn't of done that, I wish, I would have done that. Knowing you did the right thing will make you happy in the end. If you do the wrong thing and are unhappy you can't justify it.

In a one hand situation, I know the answer is clearly to hold the wild royal (as the expected value page tells you). The difference is like 25.0 coins vs. 3.4. Obvious. (On a full pay payout schedule, which this machine was not.)

However, in 10 hands, I am estimating a 1 in 48 chance for each hand for the natural. So, 10 hands should give me a 10 in 48 shot to hit one (and much higher odds to hit multiple). So, if that math holds, I have a better than one in five shot to hit 32000 coins plus whatever the other 9 hands hit. Or 4.x times what I hit holding my dealt wild royal.

Either I have it wrong, or the difference in expected value is much smaller given the 10 play vs. the one. I just want to know if it would make sense to go for the natural. Or how much I would be sacrificing to do so.

--Greg

Quote: bkdealer

In a one hand situation, I know the answer is clearly to hold the wild royal (as the expected value page tells you). The difference is like 25.0 coins vs. 3.4. Obvious. (On a full pay payout schedule, which this machine was not.)

However, in 10 hands, I am estimating a 1 in 48 chance for each hand for the natural. So, 10 hands should give me a 10 in 48 shot to hit one (and much higher odds to hit multiple). So, if that math holds, I have a better than one in five shot to hit 32000 coins plus whatever the other 9 hands hit. Or 4.x times what I hit holding my dealt wild royal.

Either I have it wrong, or the difference in expected value is much smaller given the 10 play vs. the one. I just want to know if it would make sense to go for the natural. Or how much I would be sacrificing to do so.

--Greg

It makes no difference how many hands you are playing; the correct play is the same. There is no difference between making the play once on 10-hand or 10 times in a row on single-hand. If it's the right play the first time you make it, it continues to be the right play for the next 9.

Quote: AxiomOfChoice

It makes no difference how many hands you are playing; the correct play is the same. There is no difference between making the play once on 10-hand or 10 times in a row on single-hand. If it's the right play the first time you make it, it continues to be the right play for the next 9.

Yeah, multiple hands makes no difference in your average result. Also it sounds like you are playing super times pay, and a multiplier doesn't change strategy at all either. And fyi, the 6th coin doesn't lower your overall return, it actually increases it about 0.28%.

As for the specific hand in question, it's closer when the wild royal is only 20 for 1, but it's still usually costing you about 5.65 nickels per hand normally (a little less with KQJT suited) to "go for it", but with the 8X multiplier that ups it to 45.2 nickels per hand, with 10 hands, 452 nickels. That's decent value here. And a $400 hit is nothing to sneeze at on a machine like this. Also if you would be fortunate enough to hit the Royal, then you have to also deal with a W2-G.

you played correctly in keeping the hand.
a bird in the hand is worth two in the bush.
what if you got 10 hands of nothing? hard on deuces wild but possible.
$400 aint bad on a nickel machine.
there will always be another hand.